From 1be515cd2bb8baa038a1faa1bc38579ad1734a72 Mon Sep 17 00:00:00 2001 From: andyeisenberg Date: Tue, 10 Feb 2026 07:59:07 -0500 Subject: [PATCH] Latest build deployed. --- .mapping.json | 2 +- _static/._39.47c68f7dc05f7c3e.js | Bin 0 -> 163 bytes _static/._39.47c68f7dc05f7c3e.js.LICENSE.txt | Bin 0 -> 163 bytes _static/._39.47c68f7dc05f7c3e.js.gz | Bin 0 -> 163 bytes _static/._39.47c68f7dc05f7c3e.js.map | Bin 0 -> 163 bytes _static/._39.47c68f7dc05f7c3e.js.map.gz | Bin 0 -> 163 bytes _static/._420.930dc253cc191e9e.js | Bin 0 -> 163 bytes _static/._420.930dc253cc191e9e.js.LICENSE.txt | Bin 0 -> 163 bytes _static/._420.930dc253cc191e9e.js.gz | Bin 0 -> 163 bytes _static/._420.930dc253cc191e9e.js.map | Bin 0 -> 163 bytes _static/._420.930dc253cc191e9e.js.map.gz | Bin 0 -> 163 bytes ..._prefix-runtime.0d5f811896ddeac1.bundle.js | Bin 0 -> 163 bytes ...efix-runtime.0d5f811896ddeac1.bundle.js.gz | Bin 0 -> 163 bytes ...fix-runtime.0d5f811896ddeac1.bundle.js.map | Bin 0 -> 163 bytes 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n{constructor(e){this.problem=e}inverseLISIndices(e){for(var t=[],i=0;in&&rh&&(h=c.length,l=c)}var d=[],u=0;for(let t=0;tl.length?d.push(t):e[t]==l[u]?u+=1:d.push(t);return d}grade(){var e=this.problem;e.clearFeedback(),this.correctLines=0,this.percentLines=0,this.incorrectIndents=0;var t,i=e.solution,s=e.answerLines();this.percentLines=Math.min(s.length,i.length)/Math.max(s.length,i.length),s.lengthe[s].indent&&this.indentLeft.push(t[s]);return this.incorrectIndents=this.indentLeft.length+this.indentRight.length,0==this.incorrectIndents}checkCorrectOrdering(e,t){let i=!0;this.correctLines=0,this.solutionLength=e.length;let s=Math.min(e.length,t.length);for(let n=0;n0&&u.length>0;)if(c.length<=u.length){a=c,c=[];for(let e of a)for(let t of o(e))if(l.has(t)||(c.push(t),l.set(t,e)),h.has(t))return[l,h,t]}else{a=u,u=[];for(let e of a)for(let t of r(e))if(h.has(t)||(u.push(t),h.set(t,e)),l.has(t))return[l,h,t]}throw new d("No path between "+t.toString()+" and 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a=0;a0&&(i.disabled=s),i}answerHash(){for(var e=[],t=this.answerBlocks(),i=0;ib&&(f-=1),t.lines[t.lines.length-1].indent==g.lines[0].indent&&(f+=1),f>=p&&(p=f,v=u)}for(t=d[v],g=d[v+1],u=0;u-1?r.push(t):a.push(s):r.push(t);n=r,r=[]}}if(this.recentAttempts<2?this.pairDistractors=!1:this.pairDistractors=!0,void 0===this.options.order){var c=[],d=[];for(s=0;s1)for(d=this.shuffled(d),p=0;p=0){for(var p=s;$.inArray(e[p-1],r)<0;)p--;var v=$.inArray(e[p-1],r),g=$.inArray(e[s],r);r.splice(g,1),r.splice(v,0,e[s])}return r}blockFromHash(e){for(var t=e.split("_"),i=[],s=0;s0&&l.setBlocksBefore(a),o=l,r++,n=0,s=[]}else s[0].distractor||n++,s=[];return i.length>0&&i[i.length-1].isSettled&&(this.lastBlockSettled=!0),console.log("settledblocksfromsource",i),i}getBlockById(e){for(var t=0;t1?$.i18n("msg_parson_correct",this.checkCount):$.i18n("msg_parson_correct_first_try");this.options.runnable&&(i+=" 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e=localStorage.getItem(this.adaptiveId+this.divid+"Count");null!=e&&"NaN"!=e||(e=0),this.checkCount=e,this.recentAttempts=localStorage.getItem(this.adaptiveId+"recentAttempts"),null!=this.recentAttempts&&"NaN"!=this.recentAttempts||(this.recentAttempts=3),localStorage.setItem(this.adaptiveId+"recentAttempts",this.recentAttempts),localStorage.setItem(this.adaptiveId+"Problem",this.divid),localStorage.setItem(this.adaptiveId+this.divid+"Count",this.checkCount),localStorage.setItem(this.adaptiveId+"Solved",!1)}usesIndentation(){return!this.noindent&&0!=this.solutionIndent()}distractorToRemove(){var e,t=this.enabledAnswerBlocks();for(let i=0;i{t.html($.i18n("msg_parson_not_solution"))}),10),e.lines[0].distractHelptext&&(e.view.setAttribute("data-toggle","tooltip"),e.view.setAttribute("title",e.lines[0].distractHelptext)),e.disable(),e.inSourceArea())$(e.view).css({"border-color":"#000","background-color":"#fff"}),$(e.view).animate({opacity:.3,"border-color":"#d3d3d3","background-color":"#efefef"},{duration:2e3,complete:function(){$(e.view).css({"border-color":"","background-color":""})}});else{var i=this.sourceArea.getBoundingClientRect(),s=e.pageXCenter()-1,n=e.pageYCenter(),r=i.left+window.pageXOffset+i.width/2,o=i.top+window.pageYOffset+e.view.getBoundingClientRect().height/2,a=e.slideUnderBlock();void 0!==a&&(o+=a.view.getBoundingClientRect().height+20,o+=parseInt($(a.view).css("top")));var l=this;$(e.view).css({"border-color":"#000","background-color":"#fff"}),$(e.view).animate({opacity:1},{duration:4*Math.sqrt(Math.pow(o-n,2)+Math.pow(r-s,2))+500,start:function(){l.moving=e,l.movingX=s,l.movingY=n,l.updateView()},progress:function(e,t,i){l.movingX=s*(1-t)+r*t,l.movingY=n*(1-t)+o*t,l.updateView()},complete:function(){delete l.moving,delete l.movingX,delete l.movingY,l.updateView(),$(e.view).animate({opacity:.3,"border-color":"#d3d3d3","background-color":"#efefef"},{duration:1e3,complete:function(){$(e.view).css({opacity:"","border-color":"","background-color":""}),$(e.view).addClass("disabled")}})}})}}removeIndentation(){var e=$(this.messageDiv);this.messageDiv.style.visibility="visible",e.attr("class","alert alert-info"),setTimeout((()=>{e.html($.i18n("msg_parson_provided_indent"))}),10);for(var t,i,s=200,n=0;n0&&t.indexOf(s)<0)return s;return null}getFurthestBlock(){for(var e=this.solutionBlocks(),t=this.answerBlocks(),i=0,s=0,n=null,r=null,o=0,a=null,l=0,h=0;h0&&(a=e[o-1],l=t.indexOf(a),(s=Math.abs(h-l))>i&&(i=s,n=r));return n}combineBlocks(){var e=this.solutionBlocks(),t=this.answerBlocks(),i=this.sourceBlocks(),s=$(this.messageDiv);this.messageDiv.style.visibility="visible",s.attr("class","alert alert-info"),setTimeout((()=>{s.html($.i18n("msg_parson_combined_blocks"))}),10);var n=null,r=null;null==(r=this.getSolutionBlockInSource())&&(r=this.getFurthestBlock());var o=e.indexOf(r);n=e[o-1];var a=t.indexOf(n),l=t.indexOf(r),h=!1;a>=0&&l>=0?h=a+1!==l:a<0&&l<0?h=(a=i.indexOf(n))+1!==(l=i.indexOf(r)):(h=!0,a<0&&(a=i.indexOf(n)),l<0&&(l=i.indexOf(r)));var c=l3?(alert($.i18n("msg_parson_will_combine")),this.combineBlocks(),this.logMove("combinedBlocks")):(alert($.i18n("msg_parson_three_blocks_left")),this.canHelp=!1):(alert($.i18n("msg_parson_remove_incorrect")),this.removeDistractor(e),this.logMove("removedDistractor-"+e.hash()))}helpMe(){this.clearFeedback(),this.numDistinct<3?alert($.i18n("msg_parson_atleast_three_attempts")):(this.gotHelp=!0,this.makeEasier())}dateFromTimestamp(e){var t=new Date(e);if(isNaN(t.getTime())){var i=e.split(/[- :]/);t=new Date(i[0],i[1]-1,i[2],i[3],i[4],i[5])}return t}shuffled(e){for(var t,i,s=e.length,n=e.slice();0!==s;)i=Math.floor(Math.random()*s),t=n[s-=1],n[s]=n[i],n[i]=t;return n}enterKeyboardMode(){$(this.keyboardTip).show(),$(this.sourceLabel).hide(),$(this.answerLabel).hide(),this.clearFeedback()}exitKeyboardMode(){$(this.keyboardTip).hide(),$(this.sourceLabel).show(),$(this.answerLabel).show()}clearFeedback(){$(this.answerArea).removeClass("incorrect correct");for(var e=this.answerArea.childNodes,t=0;t=i.left&&e<=i.right&&t>=i.top&&t<=i.bottom?"answer":e>=(i=this.sourceArea.getBoundingClientRect()).left&&e<=i.right&&t>=i.top&&t<=i.bottom?"source":"moving"}updateView(){var e,t,i,s=this.state,n=this.movingState(),r=!0,o=!0,a="moving"==n;s==n&&("rest"==n?(r=!1,o=!1):"source"==n?o=!1:"answer"==n?r=!1:"moving"==n&&(o=!1,r=!1)),void 0!==this.moving&&(e=$(this.moving.view).outerHeight(!0),$(this.moving.view).detach());var l,h,c=this.areaWidth-22;if(r){t=0;var d=this.sourceBlocks();if("source"==n){var u=!1,p=this.moving.pairedBin(),v=[];for(y=0;yor{")}1==_.length&&$(C).html("")}}if(o)if(t=0,i=this.areaWidth+this.indent*this.options.pixelsPerIndent-22,d=this.answerBlocks(),"answer"==n){u=!1,f=this.movingX-this.answerArea.getBoundingClientRect().left-window.pageXOffset-c/2-11;var A=Math.round(f/this.options.pixelsPerIndent);for(A<0?A=0:A>this.indent?A=this.indent:f=A*this.options.pixelsPerIndent,m=this.movingY-this.answerArea.getBoundingClientRect().top-window.pageYOffset,this.moving.indent=A,y=0;y=10&&(o+=" "),e[a].addLabel(o,0),s++}n+this.pairedBins.length>=10&&(this.areaWidth+=5,$(this.sourceArea).css({width:$(this.sourceArea).width()+5}),$(this.answerArea).css({width:$(this.answerArea).width()+5}))}resetView(){if(this.clearFeedback(),!0===this.options.scaffolding){const e=document.querySelector(`#copy-answer-button-${this.divid}`);e&&e.classList.add("copy-button-hide")}var e,t=document.body.scrollTop;for(let t=0;t{$.i18n().load({"pt-br":{msg_parson_check_me:"Verificar",msg_parson_reset:"Resetar",msg_parson_help:"Ajuda",msg_parson_too_short:"Seu programa Γ© muito curto. Adicione mais blocos.",msg_parson_drag_from_here:"Arraste daqui",msg_parson_drag_to_here:"Largue os blocos aqui",msg_parson_correct_first_try:"Perfeito! VocΓͺ levou apenas uma tentativa para resolver. Bom trabalho!",msg_parson_correct:"Perfeito! VocΓͺ levou $1 tentativas para resolver. Clique em Resetar para tentar resolver em uma tentativa.",msg_parson_wrong_indent:"Este bloco nΓ£o estΓ‘ indentado corretamente. Indente mais arrastando-o para a direita ou reduza a indentação arrastando para a esquerda.",msg_parson_wrong_indents:"Estes blocos nΓ£o estΓ£o indentados corretamente. Para indentar mais, arraste o bloco para a direita. Para reduzir a indentação, arraste para a esquerda.",msg_parson_wrong_order:"Blocos destacados no seu programa estΓ£o errados ou estΓ£o na ordem errada. Isso pode ser resolvido movendo, excluindo ou substituindo os blocos destacados.",msg_parson_arrow_navigate:"Use as teclas de setas para navegar. 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4 Confidence Intervals
diff --git a/backmatter.html b/backmatter.html index a4532c7..eb222f5 100755 --- a/backmatter.html +++ b/backmatter.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
4 Confidence Intervals
diff --git a/ch-Confidence-Intervals.html b/ch-Confidence-Intervals.html index 489fdc1..de482db 100755 --- a/ch-Confidence-Intervals.html +++ b/ch-Confidence-Intervals.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -135,7 +135,7 @@ eBookConfig.enable_chatcodes = false; -PrevUpNext +PrevUpNext
-PrevTopNext +PrevTopNext
diff --git a/ch-Expected-Value.html b/ch-Expected-Value.html index b6e3646..21e5a8b 100755 --- a/ch-Expected-Value.html +++ b/ch-Expected-Value.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
4 Confidence Intervals
diff --git a/ch-Hypothesis-Testing.html b/ch-Hypothesis-Testing.html index 1a233a6..b7d1492 100755 --- a/ch-Hypothesis-Testing.html +++ b/ch-Hypothesis-Testing.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
4 Confidence Intervals
diff --git a/ch-Introduction.html b/ch-Introduction.html index 855de54..d2d682f 100755 --- a/ch-Introduction.html +++ b/ch-Introduction.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
4 Confidence Intervals
diff --git a/ch-Likelihood.html b/ch-Likelihood.html new file mode 100755 index 0000000..fdfa767 --- /dev/null +++ b/ch-Likelihood.html @@ -0,0 +1,308 @@ + + + + + + + + + + + + + + + + +Likelihood + + + + + + + + + + + + + + + + + + + + +Skip to main content
+

Math 1044 Notes

+

Andy Eisenberg

+
+
+ +

+Chapter 4 Likelihood +

+
πŸ”—
+
+PrevTopNext +
+
+
+ +
+ + diff --git a/ch-Linear-Regression.html b/ch-Linear-Regression.html index 1c529f0..2abaa16 100755 --- a/ch-Linear-Regression.html +++ b/ch-Linear-Regression.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
4 Confidence Intervals
diff --git a/ch-Probability.html b/ch-Probability.html index 077e173..12a8c08 100755 --- a/ch-Probability.html +++ b/ch-Probability.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
4 Confidence Intervals
diff --git a/ch-Random-Variables.html b/ch-Random-Variables.html index 1dafa17..402cf42 100755 --- a/ch-Random-Variables.html +++ b/ch-Random-Variables.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
4 Confidence Intervals
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4 Confidence Intervals
diff --git a/frontmatter.html b/frontmatter.html index bf222a4..b179301 100755 --- a/frontmatter.html +++ b/frontmatter.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
4 Confidence Intervals
@@ -294,7 +301,7 @@ eBookConfig.enable_chatcodes = false;
Andy Eisenberg
Mathematics Department
Temple University
-
January 29, 2026
+
February 10, 2026
πŸ”—
PrevTopNext diff --git a/index.html b/index.html index 4835968..a9559b1 100755 --- a/index.html +++ b/index.html @@ -2,7 +2,7 @@ - + diff --git a/knowl/xref/example-exponential-MLE.html b/knowl/xref/example-exponential-MLE.html new file mode 100755 index 0000000..18ed7b1 --- /dev/null +++ b/knowl/xref/example-exponential-MLE.html @@ -0,0 +1,98 @@ + + + + + + + + + + + + + + + + + + + + +

+Example 4.1.11. +

+
+
Suppose we observe a cell and measure the time \(T\) until a toxin molecule leaves the cell. Then \(T \sim \Exp(\lambda)\) for some unknown rate parameter \(\lambda\text{:}\) +
+
+\begin{gather*} +f(t) = \lambda e^{-\lambda t}. +\end{gather*} +
+
Suppose we observe a molecule leave the cell at \(t = 0.3\) minutes. Then:
+
+\begin{align*} +L(\lambda) \amp = \lambda e^{-0.3\lambda} \\ +L'(\lambda) \amp = e^{-0.3\lambda} + \lambda e^{-0.3t} (-0.3) \\ +\amp = e^{-0.3\lambda}\left[ 1 - 0.3\lambda\right] +\end{align*} +
+
The critical point is \(\lambda = \frac{1}{0.3} \approx 3.33\text{.}\) Since \(\L' \gt 0\) on \((0, 3.33)\) and \(\L' \lt 0\) on \((3.33, \infty)\text{,}\) there is a global maximum at \(\lambda \approx 3.33\text{,}\) which is therefore the MLE.
+
πŸ”—
+
The specific time \(0.3\) minutes doesn’t particularly matter in this calculation. Whatever the time \(t\text{,}\) essentially the same calculation will result in a MLE of \(\lambda = 1/t\text{.}\) But what if we collect multiple pieces of data?
πŸ”—
+
Table 4.1.12. Time Data
πŸ”—
+ + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
MoleculeWaiting TimeRate Estimation
1\(0.3\)\(3.33\)
2\(0.8\)\(1.25\)
3\(0.5\)\(2.00\)
4\(0.6\)\(1.67\)
5\(0.9\)\(1.11\)
How should take all of this data into account in our maximum likelihood estimation? We might consider taking the average of all of the separate rate estimations, which would give \(1.87\text{.}\) Is this the most likely? We need some mathematical justification.
πŸ”—
+
+
To account for multiple, independent data points, we should multiply the probability densities for each in the creation of our likelihood function:
+
+\begin{align*} +\L(\lambda) \amp = \left(\lambda e^{-0.3\lambda}\right)\left(\lambda e^{-0.8\lambda}\right)\left(\lambda e^{-0.5\lambda}\right)\left(\lambda e^{-0.6\lambda}\right)\left(\lambda e^{-0.9\lambda}\right) \\ +\amp = \lambda^5 e^{-0.3\lambda - 0.8\lambda - 0.5\lambda - 0.6\lambda - 0.9\lambda } \\ +\amp = \lambda^5 e^{-3.1\lambda} +\end{align*} +
+
Now we can find the maximum:
+
+\begin{align*} +\L'(\lambda) \amp = 5 \lambda^4 e^{-3.1\lambda} + \lambda^5 e^{-3.1\lambda} (-3.1) \\ +\amp = 5 \lambda^4 e^{-3.1\lambda} + \lambda^5 e^{-3.1\lambda} (-3.1) \\ +\amp = \lambda^4 e^{-3.1\lambda} \left[5 - 3.1\lambda \right] +\end{align*} +
+
The only critical point is \(5/3.1 \approx 1.61\text{.}\) Since \(\L' \gt 0\) on \((0, 1.61)\) and \(\L' \lt 0\) on \((1.61, \infty)\text{,}\) there is a global maximum at \(\lambda = 1.61\text{,}\) which is therefore the MLE.
+
πŸ”—
+
It may seem less clear how to generalize this calculation for other tables of data. Observe that the value \(3.1\) is the sum of the five times in the table, so \(3.1/5\) is the average time. The MLE turned out to be the reciprocal of the average time (just as the MLE with only one data point was the reciprocal of that one time). Notice that this does not match the guess we made previously of averaging the individual rate estimations for each data point.
πŸ”—
+
+
πŸ”—
in-context + + diff --git a/knowl/xref/fact-MLE-binomial.html b/knowl/xref/fact-MLE-binomial.html new file mode 100755 index 0000000..b066df6 --- /dev/null +++ b/knowl/xref/fact-MLE-binomial.html @@ -0,0 +1,28 @@ + + + + + + + + + + + + + + + + + + + + +in-context + + diff --git a/lunr-pretext-search-index.js b/lunr-pretext-search-index.js index d9f9c13..766b7b2 100755 --- a/lunr-pretext-search-index.js +++ b/lunr-pretext-search-index.js @@ -403,7 +403,7 @@ var ptx_lunr_docs = [ "type": "Section", "number": "2.1", "title": "Discrete Random Variables", - "body": " Discrete Random Variables Discrete Random Variables Let be a sample space. A random variable is a function which assigns a number to each outcome. Suppose an experiment consists of flipping a fair coin 4 times. Let be the number of heads. The actual outcomes in the experiment are heads\/tails sequences of length 4, such as and . The random variable assigns a numerical measurement to the outcomes, such as and . For any real number value , we can consider the event consisting of outcomes such that . For example, the flip sequences , and all have 3 heads, so . Indicator Random Variable Let be an event in the sample space . An indicator random variable for is the random variable such that Indicator random variables are also called Bernoulli random variables , although we'll prefer the former term since it more clearly states the purpose of these random variables: to indicate whether or not a particular event has occurred. We'll use the language \" indicates \" to mean that is an indicator random variable for the event . In this case, , and . Consider the set of all real numbers which are actually output by a random variable . If does not contain any interval of values, then the random variable is called discrete . Otherwise, it's called continuous . Any random variable defined on a finite sample space is discrete the set of outputs of cannot contain an interval if it only has finitely many values. Suppose an experiment consists of growing a plant in a new fertilizer, and a random variable measures the height of the plant after a set growing period. Now could conceivably take on an interval's worth of values (such as, e.g., any real number between 10 inches and 20 inches), so this random variable would be continuous. The distinction between discrete and continuous random variables is not simply the distinction of whether the sample space is finite or infinite. Consider the experiment in which we flip a coin repeatedly until we first see a coin come up heads. Let be the number of times the coin is flipped. Then there are infinitely many possible values of (1, 2, 3, and so on), but there's no interval of real numbers which are all possible outputs of . The possible outputs of are discrete (in the non-technical, English sense of the word: separated). There are several particular types of distributions that appear repeatedly, since they model common behaviors. Indicator random variables, for example, will appear any time a random process generates a yes\/no or success\/failure type of answer to a question. Below are some other common distributions. The Binomial Distribution Suppose an event occurs with probability . If we perform independent trials, let be the random variable which counts the number of trials in which the event occurred. Then has the binomial distribution with parameters and . We'll write to denote this. For each value , we'll write for . (If we want to keep track of the parameter values, we may write .) Suppose we have a coin which comes up heads with probability . If we flip the coin three times and let be the number of heads, then is binomially distributed with parameters and . Then has the distribution below. 0 0.216 1 0.432 2 0.288 3 0.064 To understand how these probabilities are calculated, consider a particular heads\/tails sequence, such as . If the probability of heads is 0.4, then the probability of tails is 0.6. Since the results of different coin flips are independent from each other, the probability of seeing precisely the sequence would be the product . The probability of seeing precisely would be , the same product of terms in a different order. So, if we knew exactly the number of flip sequences of length 3 with exactly 2 heads, we could calculate the probability of seeing 2 heads as: The probability for each sequence is straightforward to generalize: if there are flips and of them are heads, then of them are tails. So the product we should calculate will have copies of the parameter and copies of the complementary probability : We'll write the symbol to mean the number of ways to choose things out of things. The number is read \" choose \" and called a binomial coefficient . The term \"binomial coefficient\" is used here because these numbers form coefficients of a polynomial in an important theorem about binomials called the Binomial Theorem . The binomial coefficients will precisely tell us the number of flip sequences of length with precisely heads (to build such a sequence, we must choose out of flips in the sequence to be heads; the rest must be tails). So, now we need a way to find the numbers . Let be a positive integer. Then the factorial of , written , is the product of the positive integers up to , i.e.: Binomial Coefficient Formula For any , We won't provide a justification of this formula here, but we'll put it to use to calculate probabilities in the binomial distribution. For example, the number of flip sequences of length 3 with exactly 2 heads is: Since we previously said the probability of such a sequence is 0.096, we can now find the probability of seeing exactly 2 heads in 3 flips of the coin: which matches the value in . More generally: Binomial Probability Formula Geometric Distribution Suppose an event occurs with probability . If we perform independent trials repeatedly, let be the random variable which counts the number of trials performed until we see the event occur for the first time. ( counts the final trial in which the event occurs.) Then has the geometric distribution with parameter . We'll write to denote this. For each value , we'll write for . (If we want to keep track of the parameter value, we may write .) Suppose we flip a coin repeatedly until we first see heads. The number of flips is geometrically distributed with parameter , which gives the probability of a flip coming up heads. Then the probability of tails is . For any particular , in order to have , we must see exactly the flip sequence , where the number of tails is exactly . The probability of seeing this particular sequence is then: Geometric Probability Formula Poisson Distribution A Poisson process is one in which some event occurs randomly at a constant probabilistic rate over time. Suppose we observe a Poisson process for a fixed amount of time , and let count the number of occurrences of the event. Then has the Poisson distribution with parameters and . We'll write to denote this. For each value , we'll write for . (If we want to keep track of the parameter values, we may write . See for an important detail about units.) Poisson Probability Formula The term is potentially misleading here depending on the units given for each value. The idea is that the rate information might be given for a different period of time than the observation period, so the rate should be scaled before calculating a probability. But the rate and observation time must be using the same time unit before multiplying. See the following example. Suppose we observe traffic along a particular stretch of highway which typically has 200 cars pass per hour. Let be the number of cars seen in a 2-hour observation period. Since the parameter cars per hour is given for a 1-hour period, we should scale it to find the rate for a 2-hour period: cars per 2-hours. Then, for example: Now consider the random variable counting the number of cars seen in a 30-minute observation period. We should not calculate , since measures time in hours and measures time in minutes. We should first convert hours, then we can find our appropriately scaled rate information: cars per half-hour. Then, for example: Note that when we write the notation , we assume that and are expressed with the same time units already. An experiment consists of flipping a biased coin 20 times. If the coin comes up heads with probability , find the probability of seeing 5 heads. Find the probability of seeing up to (and including) 3 heads. Let be the number of heads. Then , so: An experiment consists of flipping a coin repeatedly until we first see heads. If the coin comes up heads with probability 0.4, what is the probability we'll see our first heads within three flips? What about precisely on the third flip? Let be the number of flips until we see heads. Then is geometric with parameter , so: Which flip has the highest chance of being the first flip to come up heads? , so every additional flip multiplies the probability by 0.6. Therefore, the highest value for occurs when , in which case A particular store has an average of 20 customers each hour. During a 4-hour afternoon shift, what is the probability of serving 80 customers. Let be the number of customers seen during the afternoon shift. Since the store averages 20 customers per hour, it will average 80 per 4-hours. So will have distribution . Therefore: " + "body": " Discrete Random Variables Discrete Random Variables Let be a sample space. A random variable is a function which assigns a number to each outcome. Suppose an experiment consists of flipping a fair coin 4 times. Let be the number of heads. The actual outcomes in the experiment are heads\/tails sequences of length 4, such as and . The random variable assigns a numerical measurement to the outcomes, such as and . For any real number value , we can consider the event consisting of outcomes such that . For example, the flip sequences , and all have 3 heads, so . Indicator Random Variable Let be an event in the sample space . An indicator random variable for is the random variable such that Indicator random variables are also called Bernoulli random variables , although we'll prefer the former term since it more clearly states the purpose of these random variables: to indicate whether or not a particular event has occurred. We'll use the language \" indicates \" to mean that is an indicator random variable for the event . In this case, , and . Consider the set of all real numbers which are actually output by a random variable . If does not contain any interval of values, then the random variable is called discrete . Otherwise, it's called continuous . Any random variable defined on a finite sample space is discrete the set of outputs of cannot contain an interval if it only has finitely many values. Suppose an experiment consists of growing a plant in a new fertilizer, and a random variable measures the height of the plant after a set growing period. Now could conceivably take on an interval's worth of values (such as, e.g., any real number between 10 inches and 20 inches), so this random variable would be continuous. The distinction between discrete and continuous random variables is not simply the distinction of whether the sample space is finite or infinite. Consider the experiment in which we flip a coin repeatedly until we first see a coin come up heads. Let be the number of times the coin is flipped. Then there are infinitely many possible values of (1, 2, 3, and so on), but there's no interval of real numbers which are all possible outputs of . The possible outputs of are discrete (in the non-technical, English sense of the word: separated). There are several particular types of distributions that appear repeatedly, since they model common behaviors. Indicator random variables, for example, will appear any time a random process generates a yes\/no or success\/failure type of answer to a question. Below are some other common distributions. The Binomial Distribution Suppose an event occurs with probability . If we perform independent trials, let be the random variable which counts the number of trials in which the event occurred. Then has the binomial distribution with parameters and . We'll write to denote this. For each value , we'll write for . (If we want to keep track of the parameter values, we may write .) Suppose we have a coin which comes up heads with probability . If we flip the coin three times and let be the number of heads, then is binomially distributed with parameters and . Then has the distribution below. 0 0.216 1 0.432 2 0.288 3 0.064 To understand how these probabilities are calculated, consider a particular heads\/tails sequence, such as . If the probability of heads is 0.4, then the probability of tails is 0.6. Since the results of different coin flips are independent from each other, the probability of seeing precisely the sequence would be the product . The probability of seeing precisely would be , the same product of terms in a different order. So, if we knew exactly the number of flip sequences of length 3 with exactly 2 heads, we could calculate the probability of seeing 2 heads as: The probability for each sequence is straightforward to generalize: if there are flips and of them are heads, then of them are tails. So the product we should calculate will have copies of the parameter and copies of the complementary probability : We'll write the symbol to mean the number of ways to choose things out of things. The number is read \" choose \" and called a binomial coefficient . The term \"binomial coefficient\" is used here because these numbers form coefficients of a polynomial in an important theorem about binomials called the Binomial Theorem . The binomial coefficients will precisely tell us the number of flip sequences of length with precisely heads (to build such a sequence, we must choose out of flips in the sequence to be heads; the rest must be tails). So, now we need a way to find the numbers . Let be a positive integer. Then the factorial of , written , is the product of the positive integers up to , i.e.: Binomial Coefficient Formula For any , We won't provide a justification of this formula here, but we'll put it to use to calculate probabilities in the binomial distribution. For example, the number of flip sequences of length 3 with exactly 2 heads is: Since we previously said the probability of such a sequence is 0.096, we can now find the probability of seeing exactly 2 heads in 3 flips of the coin: which matches the value in . More generally: Binomial Probability Formula Geometric Distribution Suppose an event occurs with probability . If we perform independent trials repeatedly, let be the random variable which counts the number of trials performed until we see the event occur for the first time. ( counts the final trial in which the event occurs.) Then has the geometric distribution with parameter . We'll write to denote this. For each value , we'll write for . (If we want to keep track of the parameter value, we may write .) Suppose we flip a coin repeatedly until we first see heads. The number of flips is geometrically distributed with parameter , which gives the probability of a flip coming up heads. Then the probability of tails is . For any particular , in order to have , we must see exactly the flip sequence , where the number of tails is exactly . The probability of seeing this particular sequence is then: Geometric Probability Formula Poisson Distribution A Poisson process is one in which some event occurs randomly at a constant probabilistic rate over time. Suppose we observe a Poisson process for a fixed amount of time , and let count the number of occurrences of the event. Then has the Poisson distribution with parameters and . We'll write to denote this. For each value , we'll write for . (If we want to keep track of the parameter values, we may write . See for an important detail about units.) Poisson Probability Formula The term is potentially misleading here depending on the units given for each value. The idea is that the rate information might be given for a different period of time than the observation period, so the rate should be scaled before calculating a probability. But the rate and observation time must be using the same time unit before multiplying. See the following example. Suppose we observe traffic along a particular stretch of highway which typically has 200 cars pass per hour. Let be the number of cars seen in a 2-hour observation period. Since the parameter cars per hour is given for a 1-hour period, we should scale it to find the rate for a 2-hour period: cars per 2-hours. Then, for example: Now consider the random variable counting the number of cars seen in a 30-minute observation period. We should not calculate , since measures time in hours and measures time in minutes. We should first convert hours, then we can find our appropriately scaled rate information: cars per half-hour. Then, for example: Note that when we write the notation , we assume that and are expressed with the same time units already. An experiment consists of flipping a biased coin 20 times. If the coin comes up heads with probability , find the probability of seeing 5 heads. Find the probability of seeing up to (and including) 3 heads. Let be the number of heads. Then , so: An experiment consists of flipping a coin repeatedly until we first see heads. If the coin comes up heads with probability 0.4, what is the probability we'll see our first heads within three flips? What about precisely on the third flip? Let be the number of flips until we see heads. Then is geometric with parameter , so: Which flip has the highest chance of being the first flip to come up heads? , so every additional flip multiplies the probability by 0.6. Therefore, the highest value for occurs when , in which case A particular store has an average of 20 customers each hour. During a 4-hour afternoon shift, what is the probability of serving 80 customers. Let be the number of customers seen during the afternoon shift. Since the store averages 20 customers per hour, it will average 80 per 4-hours. So will have distribution . Therefore: " }, { "id": "def-RV", @@ -592,7 +592,7 @@ var ptx_lunr_docs = [ "type": "Section", "number": "2.2", "title": "Continuous Random Variables", - "body": " Continuous Random Variables Probability Density Functions We have to think about probability differently for continuous random variables. If is a continuous random variable taking values in an interval , it doesn't quite make sense to ask, as we have for discrete random variables, to find the probabilities of each individual outcome. When we roll a fair 6-sided die, there are 6 equally likely outcomes, so they each get assigned the probability . If we try to select a random number in the interval in a similarly \"fair\" way (i.e., with each outcome equally likely), then what probability should we assign to each outcome? If , then we have a problem: if you add up the probabilities of all the (infinitely many) outcomes, you get an infinite total, whereas the total probability in any sample space should be 1. If , then we also have a problem: if you add up the probabilities of all the outcomes, you get total of 0, whereas the total probability in any sample space should be 1. We need a new way to model and calculate probabilities for continuous random variables. Let be a continuous random variable taking values in the interval . The probability density function (or pdf ) for is a function defined on satisfying the properties: for all , and . It's important to keep in mind that the output values of a pdf are probability densities , not probabilities. You can (somewhat casually) think of the units of values as \"probability per \". You may not have discussed the units of the differential very much in your Calculus 1 course, but has the same units as and is being multiplied by the integrand when you take an integral. So, although is not a probability, the result of computing is a probability. A pdf's purpose in life is to be integrated, which tells us probabilities for a continuous random variable: Let take values in the interval . We say that has the uniform distribution if, for any subinterval , the probability that lies within is equal to the proportion of the total sample space taken by . That is: In this case, has the pdf . We'll ignore for the moment how to find this pdf and instead we'll quickly check that it behaves the way we want it to: as we wanted. Let be a continuous random variable taking values in the interval with pdf for some constant . Because of the restriction that the total probability must be 1, we can determine the value of : Therefore . Now we can calculate probabilities, e.g.: Cumulative Distribution Functions When a random variable takes values in an interval , it's often useful to consider how much probability has been accumulated up to some specified value . For each value of , we can calculate this accumulated probability. Since there's an association between inputs (the values) and output (the accumulated probability), we can naturally build a function. Let be a continuous random variable taking values in an interval . The cumulative distribution function (or cdf ) is the function defined on such that How do we actually calculate these cumulative probabilities? Luckily, these are precisely the sort of probabilities we can use the pdf to find: (Note that on the far left side of the equation, we used the letter to refer to some point within up to which we're trying to accumulate probability. So that value must also appear at the top of the integral sign. We can't then use the same letter for the integrand and differential inside the integral; we can't use the same letter to refer to two different variables at the same time. So, we have changed the letter inside the integral to .) This equation may look familiar from Calculus 1. If you review the First Fundamental Theorem of Calculus , you'll see precisely the relationship between the cdf and the pdf : Let's see how we can put this relationship to use. Suppose is uniformly distributed in the interval . We previously said in that, for any subinterval , we should have In particular, for any , we would have: Therefore, taking the derivative: So, we can see that the idea we wanted to capture in probability led to a formula that gave accumulated probability, which then led to the pdf. Both the cdf and the pdf contain essentially all of the information we need to answer probability questions about . However, keep in mind that these functions output different mathematical objects. outputs probabilities already; must be integrated over a subinterval to find probabilities. The Exponential Distribution Suppose we observe a Poisson process in which an event occurs with rate parameter . Let be the time until the event is first observed. Then has the exponential distribution with parameter . We'll write to denote this. Exponential Density Function The binomial, geometric, Poisson, and exponential distributions are conceptually grouped together. Consider the following two dichotomies: Is time being measured discretely or continuously? Does the random variable count occurrences of an event, or does it measure time until an event? There are four combinations of answers to these two questions. Binomial \"Time\" is discrete; the random variables counts occurrences. Geometric \"Time\" is discrete; the random variable measures \"time\" until an event. Poisson Time is continuous; the random variable counts occurrences. Exponential Time is continuous; the random variable measures time until an event. When we talk about discrete \"time\", we don't always mean something measured in actual units of time. We might; for example, an experiment may involve a bacterial colony growing in a petri dish over time, but we only check in and measure something about the bacterial colony once each day. In this case, since our measurements of time must occur in whole numbers of days, time is being measured in a discrete way (that is, not involving an interval of real numbers). Consider, instead, an experiment in which we flip a coin repeatedly until we first see heads appear. The number of flips doesn't have units of time, but counting flips is analogous to measuring time until an event occurs. The flip count is a discrete stand-in for something analogous to time. We can consider this experiment to be a discrete analog of a Poisson process. It's also worth including a warning here about the Poisson distribution, specifically. The dichotomy mentioned above about whether time is being measured discretely or continuously is not the same as the consideration of whether the random variable itself is discrete or continuous. In a Poisson process, time is being measured continuously. But the Poisson random variable counts the number of occurrences of the event, and this number must be a whole number. So a Poisson random variable is a discrete random variable. Normal Distribution Consider the parametrized family of functions: The function is called a normal density function with parameters and . A continuous random variable with pdf is said to have a normal distribution . In the specific case that and , we call the resulting distribution the standard normal distribution . It's an unfortunate reality that the normal density functions can't easily be integrated. (In fact, it is provably impossible to write nice formulas for their antiderivatives). So we won't worry much about probability computations like for normal distributions. The normal distributions will play a central role in later chapters (after which you can return here and appreciate this pun). A continuous random variable taking values in has p.d.f. for some constant . What is the value of ? Find A continuous random variable taking values in has p.d.f. . Find the c.d.f. . Use your c.d.f. to find . . A continuous random variable taking values in has c.d.f. . Find the p.d.f. . Let . Find and . . . " + "body": " Continuous Random Variables Probability Density Functions We have to think about probability differently for continuous random variables. If is a continuous random variable taking values in an interval , it doesn't quite make sense to ask, as we have for discrete random variables, to find the probabilities of each individual outcome. When we roll a fair 6-sided die, there are 6 equally likely outcomes, so they each get assigned the probability . If we try to select a random number in the interval in a similarly \"fair\" way (i.e., with each outcome equally likely), then what probability should we assign to each outcome? If , then we have a problem: if you add up the probabilities of all the (infinitely many) outcomes, you get an infinite total, whereas the total probability in any sample space should be 1. If , then we also have a problem: if you add up the probabilities of all the outcomes, you get total of 0, whereas the total probability in any sample space should be 1. We need a new way to model and calculate probabilities for continuous random variables. Let be a continuous random variable taking values in the interval . The probability density function (or pdf ) for is a function defined on satisfying the properties: for all , and . It's important to keep in mind that the output values of a pdf are probability densities , not probabilities. You can (somewhat casually) think of the units of values as \"probability per \". You may not have discussed the units of the differential very much in your Calculus 1 course, but has the same units as and is being multiplied by the integrand when you take an integral. So, although is not a probability, the result of computing is a probability. A pdf's purpose in life is to be integrated, which tells us probabilities for a continuous random variable: Let take values in the interval . We say that has the uniform distribution if, for any subinterval , the probability that lies within is equal to the proportion of the total sample space taken by . That is: In this case, has the pdf . We'll ignore for the moment how to find this pdf and instead we'll quickly check that it behaves the way we want it to: as we wanted. Let be a continuous random variable taking values in the interval with pdf for some constant . Because of the restriction that the total probability must be 1, we can determine the value of : Therefore . Now we can calculate probabilities, e.g.: Cumulative Distribution Functions When a random variable takes values in an interval , it's often useful to consider how much probability has been accumulated up to some specified value . For each value of , we can calculate this accumulated probability. Since there's an association between inputs (the values) and output (the accumulated probability), we can naturally build a function. Let be a continuous random variable taking values in an interval . The cumulative distribution function (or cdf ) is the function defined on such that How do we actually calculate these cumulative probabilities? Luckily, these are precisely the sort of probabilities we can use the pdf to find: (Note that on the far left side of the equation, we used the letter to refer to some point within up to which we're trying to accumulate probability. So that value must also appear at the top of the integral sign. We can't then use the same letter for the integrand and differential inside the integral; we can't use the same letter to refer to two different variables at the same time. So, we have changed the letter inside the integral to .) This equation may look familiar from Calculus 1. If you review the First Fundamental Theorem of Calculus , you'll see precisely the relationship between the cdf and the pdf : Let's see how we can put this relationship to use. Suppose is uniformly distributed in the interval . We previously said in that, for any subinterval , we should have In particular, for any , we would have: Therefore, taking the derivative: So, we can see that the idea we wanted to capture in probability led to a formula that gave accumulated probability, which then led to the pdf. Both the cdf and the pdf contain essentially all of the information we need to answer probability questions about . However, keep in mind that these functions output different mathematical objects. outputs probabilities already; must be integrated over a subinterval to find probabilities. The Exponential Distribution Suppose we observe a Poisson process in which an event occurs with rate parameter . Let be the time until the event is first observed. Then has the exponential distribution with parameter . We'll write to denote this. Exponential Density Function The binomial, geometric, Poisson, and exponential distributions are conceptually grouped together. Consider the following two dichotomies: Is time being measured discretely or continuously? Does the random variable count occurrences of an event, or does it measure time until an event? There are four combinations of answers to these two questions. Binomial \"Time\" is discrete; the random variables counts occurrences. Geometric \"Time\" is discrete; the random variable measures \"time\" until an event. Poisson Time is continuous; the random variable counts occurrences. Exponential Time is continuous; the random variable measures time until an event. When we talk about discrete \"time\", we don't always mean something measured in actual units of time. We might; for example, an experiment may involve a bacterial colony growing in a petri dish over time, but we only check in and measure something about the bacterial colony once each day. In this case, since our measurements of time must occur in whole numbers of days, time is being measured in a discrete way (that is, not involving an interval of real numbers). Consider, instead, an experiment in which we flip a coin repeatedly until we first see heads appear. The number of flips doesn't have units of time, but counting flips is analogous to measuring time until an event occurs. The flip count is a discrete stand-in for something analogous to time. We can consider this experiment to be a discrete analog of a Poisson process. It's also worth including a warning here about the Poisson distribution, specifically. The dichotomy mentioned above about whether time is being measured discretely or continuously is not the same as the consideration of whether the random variable itself is discrete or continuous. In a Poisson process, time is being measured continuously. But the Poisson random variable counts the number of occurrences of the event, and this number must be a whole number. So a Poisson random variable is a discrete random variable. Normal Distribution Consider the parametrized family of functions: The function is called a normal density function with parameters and . A continuous random variable with pdf is said to have a normal distribution . In the specific case that and , we call the resulting distribution the standard normal distribution . It's an unfortunate reality that the normal density functions can't easily be integrated. (In fact, it is provably impossible to write nice formulas for their antiderivatives). So we won't worry much about probability computations like for normal distributions. The normal distributions will play a central role in later chapters (after which you can return here and appreciate this pun). A continuous random variable taking values in has p.d.f. for some constant . What is the value of ? Find A continuous random variable taking values in has p.d.f. . Find the c.d.f. . Use your c.d.f. to find . . A continuous random variable taking values in has c.d.f. . Find the p.d.f. . Let . Find and . . . " }, { "id": "def-pdf", @@ -1188,12 +1188,165 @@ var ptx_lunr_docs = [ "title": "", "body": " Suppose we roll a fair, 4-sided die two times. Let be the sum of the rolls, and let be the product of the rolls. Find the covariance of and . [Note: Somewhat tedious.] " }, +{ + "id": "sec-Likelihood", + "level": "1", + "url": "sec-Likelihood.html", + "type": "Section", + "number": "4.1", + "title": "Likelihood", + "body": " Likelihood So far, we've been concerned with probability theory. Starting with a probability distribution and some parameter values, we've tried to answer questions like: What's the probability of seeing certain experimental results? Statistics is concerned with going in the other direction: Upon seeing the experimental results, can we determine the type of underlying probability distribution? Can we determine its parameters? An estimator is a value of a parameter computed from a sample of data. Suppose we find a coin on the street and don't know whether or not it's fair. We want to know the probability of the coin coming up heads. We might, for example, flip the coin times and count the number of heads. Then, we'll estimate . We'll refer to this as a common sense estimator. (Other distributions and parameter types will have different notions of \"common sense\".) An estimator is, itself, a random variable: it produces a numerical value based on the results of an experiment. We'll use notation like for a random variable which is an estimator for a parameter . (Similarly, would denote an estimator for a parameter called .) An estimator is called unbiased if . Suppose we have a coin with parameter , which we'll flip times and count the number of heads. We use the unbiased estimator . In this case, notice that , so we know , although we don't know the value of . (We probably do know the value of ; after all, we're flipping the coin!) Now: It's worth pausing for a moment to be appropriately impressed with ourselves. We still don't know the true value of . But we managed to show that our common sense method of estimating gives, on average, the correct value. We'd like to be able to collect some data and use that data to estimate the values of whatever parameters our distribution has. Perhaps as a starting point, it would be good to identify the single most likely value of a parameter: Suppose we collect some data to estimate a parameter . The likelihood function is: The value of which maximizes the function is called the maximum likelihood estimation , or MLE , of . Suppose we flip a coin 100 times. Let be the number of heads. Then for some unknown value of : In this case, is the data that we collect, and is the value of the parameter. Suppose we see heads. Then: If we want to know the most likely value of the parameter , then we should maximize over the interval . We can see that when , and the endpoints of the interval we're maximizing over are , so we can build a table: Looking for the Maximum Note that we don't need to know the exact value of in order to see that it's strictly positive, and therefore the maximum value of . So, the MLE of is . We can see in the previous example that the MLE of is also the common sense estimation of . This will be the case in general for the binomial distribution, so we won't need to redo this work over and over: Binomial MLE If we see heads in coin flips, then the MLE of the bias is . As we remember from Calculus 1, maximizing a continuous function works slightly differently over a closed interval (like the previous example) or an open interval (like the next example). Suppose a radioactive material emits particles as it decays. Let count the particles emitted. Then for some unknown rate parameter : Suppose we observe a sample of material for 1 hour and count 8 particles emitted. Then: To maximize over the interval , we start by finding critical points. The only critical point is , but we have not justified that this is the location of a global maximum. A critical point is only a potential location of a local minimum or maximum. But with a bit more justification: observe that is positive on the interval and negative on the interval . Therefore the function increases on and decreases on . So it must reach its maximum at , which is therefore the MLE. As with the binomial distribution, the calculation will be essentially the same regardless of the specific number of particles observed, so: Poisson MLE If we observe a Poisson process and see events occur, then the MLE of the rate parameter is . The situation for continuous random variables is similar, but slightly different. To build the likelihood function, we should use the pdf of the continuous random variable. So the likelihood function will give the probabiliy density given the data collected, rather than the probability itself. Suppose we observe a cell and measure the time until a toxin molecule leaves the cell. Then for some unknown rate parameter : Suppose we observe a molecule leave the cell at minutes. Then: The critical point is . Since on and on , there is a global maximum at , which is therefore the MLE. The specific time minutes doesn't particularly matter in this calculation. Whatever the time , essentially the same calculation will result in a MLE of . But what if we collect multiple pieces of data? Time Data Molecule Waiting Time Rate Estimation 1 2 3 4 5 How should take all of this data into account in our maximum likelihood estimation? We might consider taking the average of all of the separate rate estimations, which would give . Is this the most likely? We need some mathematical justification. To account for multiple, independent data points, we should multiply the probability densities for each in the creation of our likelihood function: Now we can find the maximum: The only critical point is . Since on and on , there is a global maximum at , which is therefore the MLE. It may seem less clear how to generalize this calculation for other tables of data. Observe that the value is the sum of the five times in the table, so is the average time. The MLE turned out to be the reciprocal of the average time (just as the MLE with only one data point was the reciprocal of that one time). Notice that this does not match the guess we made previously of averaging the individual rate estimations for each data point. Exponential MLE If we observe a Poisson process and see events occur after waiting times , then the MLE of the rate parameter is . Suppose a coin has an unknown probability of coming up heads. We perform the experiment in five independent trials, during which it takes 4, 5, 4, 3, and 6 flips to see our first heads in each trial. What is the maximum likelihood estimation for the probability of the coin coming up heads on a flip? Suppose a coin has an unknown probability of coming up heads. We perform the experiment in independent trials, during which it takes flips to see our first heads in each trial. Find a \"common sense\" MLE formula for the geometric distribution. You could set up a calculation analogous to . Or, you could consider . A particular store owner wants to approximate the average hourly rate at which customers come into the store. They observe 80 customers enter during a particular 4-hour shift. What is the maximum likelihood estimation for the hourly customer rate? A radioactive material emits particles at an unknown probabilistic rate particles per minute. We observe particles emitted at times 1.1, 1.7, 1.3, 2.2, 1.9, and 1.8 minutes. Write the likelihood function based on this data. What is the maximum likelihood estimation for ? Suppose a parameter takes values in with likelihood function . Find the maximum likelihood estimation of . " +}, +{ + "id": "def-estimator", + "level": "2", + "url": "sec-Likelihood.html#def-estimator", + "type": "Definition", + "number": "4.1.1", + "title": "", + "body": " An estimator is a value of a parameter computed from a sample of data. " +}, +{ + "id": "sec-Likelihood-4", + "level": "2", + "url": "sec-Likelihood.html#sec-Likelihood-4", + "type": "Example", + "number": "4.1.2", + "title": "", + "body": " Suppose we find a coin on the street and don't know whether or not it's fair. We want to know the probability of the coin coming up heads. We might, for example, flip the coin times and count the number of heads. Then, we'll estimate . We'll refer to this as a common sense estimator. (Other distributions and parameter types will have different notions of \"common sense\".) " +}, +{ + "id": "def-unbiased", + "level": "2", + "url": "sec-Likelihood.html#def-unbiased", + "type": "Definition", + "number": "4.1.3", + "title": "", + "body": " An estimator is called unbiased if . " +}, +{ + "id": "sec-Likelihood-7", + "level": "2", + "url": "sec-Likelihood.html#sec-Likelihood-7", + "type": "Example", + "number": "4.1.4", + "title": "", + "body": " Suppose we have a coin with parameter , which we'll flip times and count the number of heads. We use the unbiased estimator . In this case, notice that , so we know , although we don't know the value of . (We probably do know the value of ; after all, we're flipping the coin!) Now: It's worth pausing for a moment to be appropriately impressed with ourselves. We still don't know the true value of . But we managed to show that our common sense method of estimating gives, on average, the correct value. " +}, +{ + "id": "def-likelihood", + "level": "2", + "url": "sec-Likelihood.html#def-likelihood", + "type": "Definition", + "number": "4.1.5", + "title": "", + "body": " Suppose we collect some data to estimate a parameter . The likelihood function is: The value of which maximizes the function is called the maximum likelihood estimation , or MLE , of . " +}, +{ + "id": "sec-Likelihood-10", + "level": "2", + "url": "sec-Likelihood.html#sec-Likelihood-10", + "type": "Example", + "number": "4.1.6", + "title": "", + "body": " Suppose we flip a coin 100 times. Let be the number of heads. Then for some unknown value of : In this case, is the data that we collect, and is the value of the parameter. Suppose we see heads. Then: If we want to know the most likely value of the parameter , then we should maximize over the interval . We can see that when , and the endpoints of the interval we're maximizing over are , so we can build a table: Looking for the Maximum Note that we don't need to know the exact value of in order to see that it's strictly positive, and therefore the maximum value of . So, the MLE of is . " +}, +{ + "id": "fact-MLE-binomial", + "level": "2", + "url": "sec-Likelihood.html#fact-MLE-binomial", + "type": "Fact", + "number": "4.1.8", + "title": "Binomial MLE.", + "body": " Binomial MLE If we see heads in coin flips, then the MLE of the bias is . " +}, +{ + "id": "sec-Likelihood-14", + "level": "2", + "url": "sec-Likelihood.html#sec-Likelihood-14", + "type": "Example", + "number": "4.1.9", + "title": "", + "body": " Suppose a radioactive material emits particles as it decays. Let count the particles emitted. Then for some unknown rate parameter : Suppose we observe a sample of material for 1 hour and count 8 particles emitted. Then: To maximize over the interval , we start by finding critical points. The only critical point is , but we have not justified that this is the location of a global maximum. A critical point is only a potential location of a local minimum or maximum. But with a bit more justification: observe that is positive on the interval and negative on the interval . Therefore the function increases on and decreases on . So it must reach its maximum at , which is therefore the MLE. " +}, +{ + "id": "fact-MLE-Poisson", + "level": "2", + "url": "sec-Likelihood.html#fact-MLE-Poisson", + "type": "Fact", + "number": "4.1.10", + "title": "Poisson MLE.", + "body": " Poisson MLE If we observe a Poisson process and see events occur, then the MLE of the rate parameter is . " +}, +{ + "id": "example-exponential-MLE", + "level": "2", + "url": "sec-Likelihood.html#example-exponential-MLE", + "type": "Example", + "number": "4.1.11", + "title": "", + "body": " Suppose we observe a cell and measure the time until a toxin molecule leaves the cell. Then for some unknown rate parameter : Suppose we observe a molecule leave the cell at minutes. Then: The critical point is . Since on and on , there is a global maximum at , which is therefore the MLE. The specific time minutes doesn't particularly matter in this calculation. Whatever the time , essentially the same calculation will result in a MLE of . But what if we collect multiple pieces of data? Time Data Molecule Waiting Time Rate Estimation 1 2 3 4 5 How should take all of this data into account in our maximum likelihood estimation? We might consider taking the average of all of the separate rate estimations, which would give . Is this the most likely? We need some mathematical justification. To account for multiple, independent data points, we should multiply the probability densities for each in the creation of our likelihood function: Now we can find the maximum: The only critical point is . Since on and on , there is a global maximum at , which is therefore the MLE. It may seem less clear how to generalize this calculation for other tables of data. Observe that the value is the sum of the five times in the table, so is the average time. The MLE turned out to be the reciprocal of the average time (just as the MLE with only one data point was the reciprocal of that one time). Notice that this does not match the guess we made previously of averaging the individual rate estimations for each data point. " +}, +{ + "id": "fact-MLE-exponential", + "level": "2", + "url": "sec-Likelihood.html#fact-MLE-exponential", + "type": "Fact", + "number": "4.1.13", + "title": "Exponential MLE.", + "body": " Exponential MLE If we observe a Poisson process and see events occur after waiting times , then the MLE of the rate parameter is . " +}, +{ + "id": "sec-Likelihood-20-1", + "level": "2", + "url": "sec-Likelihood.html#sec-Likelihood-20-1", + "type": "Exercise", + "number": "4.1.1", + "title": "", + "body": " Suppose a coin has an unknown probability of coming up heads. We perform the experiment in five independent trials, during which it takes 4, 5, 4, 3, and 6 flips to see our first heads in each trial. What is the maximum likelihood estimation for the probability of the coin coming up heads on a flip? " +}, +{ + "id": "sec-Likelihood-20-2", + "level": "2", + "url": "sec-Likelihood.html#sec-Likelihood-20-2", + "type": "Exercise", + "number": "4.1.2", + "title": "", + "body": " Suppose a coin has an unknown probability of coming up heads. We perform the experiment in independent trials, during which it takes flips to see our first heads in each trial. Find a \"common sense\" MLE formula for the geometric distribution. You could set up a calculation analogous to . Or, you could consider . " +}, +{ + "id": "sec-Likelihood-20-3", + "level": "2", + "url": "sec-Likelihood.html#sec-Likelihood-20-3", + "type": "Exercise", + "number": "4.1.3", + "title": "", + "body": " A particular store owner wants to approximate the average hourly rate at which customers come into the store. They observe 80 customers enter during a particular 4-hour shift. What is the maximum likelihood estimation for the hourly customer rate? " +}, +{ + "id": "sec-Likelihood-20-4", + "level": "2", + "url": "sec-Likelihood.html#sec-Likelihood-20-4", + "type": "Exercise", + "number": "4.1.4", + "title": "", + "body": " A radioactive material emits particles at an unknown probabilistic rate particles per minute. We observe particles emitted at times 1.1, 1.7, 1.3, 2.2, 1.9, and 1.8 minutes. Write the likelihood function based on this data. What is the maximum likelihood estimation for ? " +}, +{ + "id": "sec-Likelihood-20-5", + "level": "2", + "url": "sec-Likelihood.html#sec-Likelihood-20-5", + "type": "Exercise", + "number": "4.1.5", + "title": "", + "body": " Suppose a parameter takes values in with likelihood function . Find the maximum likelihood estimation of . " +}, { "id": "sec-CLT", "level": "1", "url": "sec-CLT.html", "type": "Section", - "number": "4.1", + "number": "4.2", "title": "Central Limit Theorem", "body": " Central Limit Theorem Text of section. Suppose a coin has probability 0.4 of coming up heads, and we flip the coin 100 times. Let be the number of heads. Estimate the probability that . Suppose a fair die is rolled 100 times, and let be the average value of the rolls. Estimate the probability that . The heights of men in the US have a mean of 69 in and a variance of about 9 in , and the heights of women in the US have a mean of 63.5 in with a variance of 6.25 in . Suppose the heights of 30 men are sampled, and a sample mean is taken. Find the expected value and variance of . Estimate the probability that . What if the sampled group was women? " }, @@ -1202,7 +1355,7 @@ var ptx_lunr_docs = [ "level": "2", "url": "sec-CLT.html#exercises-CLT-1", "type": "Exercise", - "number": "4.1.1", + "number": "4.2.1", "title": "", "body": " Suppose a coin has probability 0.4 of coming up heads, and we flip the coin 100 times. Let be the number of heads. Estimate the probability that . " }, @@ -1211,7 +1364,7 @@ var ptx_lunr_docs = [ "level": "2", "url": "sec-CLT.html#exercises-CLT-2", "type": "Exercise", - "number": "4.1.2", + "number": "4.2.2", "title": "", "body": " Suppose a fair die is rolled 100 times, and let be the average value of the rolls. Estimate the probability that . " }, @@ -1220,7 +1373,7 @@ var ptx_lunr_docs = [ "level": "2", "url": "sec-CLT.html#exercises-CLT-3", "type": "Exercise", - "number": "4.1.3", + "number": "4.2.3", "title": "", "body": " The heights of men in the US have a mean of 69 in and a variance of about 9 in , and the heights of women in the US have a mean of 63.5 in with a variance of 6.25 in . Suppose the heights of 30 men are sampled, and a sample mean is taken. Find the expected value and variance of . Estimate the probability that . What if the sampled group was women? " }, @@ -1229,7 +1382,7 @@ var ptx_lunr_docs = [ "level": "1", "url": "sec-Confidence-Intervals.html", "type": "Section", - "number": "4.2", + "number": "4.3", "title": "Confidence Intervals", "body": " Confidence Intervals Text of section. Suppose we flip a coin 100 times and count 60 heads. Let be the (unknown) probability that the coin comes up heads on a flip. Give an approximate 95% confidence interval for the value of . Suppose in a sample of 100 people, 12 are left-handed. Give a 95% confidence interval for the proportion of left-handed people. Give a 90% confidence interval. The weights of five mice are measured and recorded below. Give a 95% confidence interval for the sample mean weight of mice. (Pretend 5 measurements is large enough for the CLT to apply.) mouse 1 2 3 4 5 weight (g) 26 32 33 20 29 The heights of five plants are measured and recorded below. Give a 95% confidence interval around the sample mean for the heights of the plants. (Pretend 5 measurements is large enough for the CLT to apply.) plant 1 2 3 4 5 height (in) 15 14 18 21 17 " }, @@ -1238,7 +1391,7 @@ var ptx_lunr_docs = [ "level": "2", "url": "sec-Confidence-Intervals.html#exercises-Confidence-Intervals-1", "type": "Exercise", - "number": "4.2.1", + "number": "4.3.1", "title": "", "body": " Suppose we flip a coin 100 times and count 60 heads. Let be the (unknown) probability that the coin comes up heads on a flip. Give an approximate 95% confidence interval for the value of . " }, @@ -1247,7 +1400,7 @@ var ptx_lunr_docs = [ "level": "2", "url": "sec-Confidence-Intervals.html#exercises-Confidence-Intervals-2", "type": "Exercise", - "number": "4.2.2", + "number": "4.3.2", "title": "", "body": " Suppose in a sample of 100 people, 12 are left-handed. Give a 95% confidence interval for the proportion of left-handed people. Give a 90% confidence interval. " }, @@ -1256,7 +1409,7 @@ var ptx_lunr_docs = [ "level": "2", "url": "sec-Confidence-Intervals.html#exercises-Confidence-Intervals-3", "type": "Exercise", - "number": "4.2.3", + "number": "4.3.3", "title": "", "body": " The weights of five mice are measured and recorded below. Give a 95% confidence interval for the sample mean weight of mice. (Pretend 5 measurements is large enough for the CLT to apply.) mouse 1 2 3 4 5 weight (g) 26 32 33 20 29 " }, @@ -1265,7 +1418,7 @@ var ptx_lunr_docs = [ "level": "2", "url": "sec-Confidence-Intervals.html#exercises-Confidence-Intervals-4", "type": "Exercise", - "number": "4.2.4", + "number": "4.3.4", "title": "", "body": " The heights of five plants are measured and recorded below. Give a 95% confidence interval around the sample mean for the heights of the plants. (Pretend 5 measurements is large enough for the CLT to apply.) plant 1 2 3 4 5 height (in) 15 14 18 21 17 " }, diff --git a/my-great-book.html b/my-great-book.html index 0b1cc93..8ad0b5b 100755 --- a/my-great-book.html +++ b/my-great-book.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
4 Confidence Intervals
diff --git a/sec-CLT.html b/sec-CLT.html index 8e4e5a7..d0cb187 100755 --- a/sec-CLT.html +++ b/sec-CLT.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -135,7 +135,7 @@ eBookConfig.enable_chatcodes = false;
-PrevUpNext +PrevUpNext

-Section 4.1 Central Limit Theorem +Section 4.2 Central Limit Theorem

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1.

Suppose a coin has probability 0.4 of coming up heads, and we flip the coin 100 times. Let \(S\) be the number of heads. Estimate the probability that \(34 \leq S \leq 44\text{.}\)
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Suppose a fair die is rolled 100 times, and let \(m\) be the average value of the rolls. Estimate the probability that \(3.45 \leq m \leq 3.55\text{.}\)
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The heights of men in the US have a mean of 69 in and a variance of about 9 in\(^2\text{,}\) and the heights of women in the US have a mean of 63.5 in with a variance of 6.25 in\(^2\text{.}\)
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(a)
Suppose the heights of 30 men are sampled, and a sample mean \(m\) is taken. Find the expected value and variance of \(m\text{.}\)
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diff --git a/sec-Chi-Squared.html b/sec-Chi-Squared.html index 2854fc5..b863c09 100755 --- a/sec-Chi-Squared.html +++ b/sec-Chi-Squared.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
4 Confidence Intervals
diff --git a/sec-Conditional-Probability.html b/sec-Conditional-Probability.html index 65f0d7c..b5f9a0a 100755 --- a/sec-Conditional-Probability.html +++ b/sec-Conditional-Probability.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
4 Confidence Intervals
diff --git a/sec-Confidence-Intervals.html b/sec-Confidence-Intervals.html index 9cca550..6400737 100755 --- a/sec-Confidence-Intervals.html +++ b/sec-Confidence-Intervals.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
4 Confidence Intervals
@@ -290,7 +297,7 @@ eBookConfig.enable_chatcodes = false;
  • Backmatter

    • -Section 4.2 Confidence Intervals +Section 4.3 Confidence Intervals

    Text of section.
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    @@ -300,18 +307,18 @@ eBookConfig.enable_chatcodes = false;

    1.

    Suppose we flip a coin 100 times and count 60 heads. Let \(p\) be the (unknown) probability that the coin comes up heads on a flip. Give an approximate 95% confidence interval for the value of \(p\text{.}\)
    πŸ”—
    -
    πŸ”—
    πŸ”—

    3.

    The weights of five mice are measured and recorded below. Give a 95% confidence interval for the sample mean weight of mice. (Pretend 5 measurements is large enough for the CLT to apply.)
    πŸ”—
    -
    Table 4.2.1.
    πŸ”—
    +
    Table 4.3.1.
    πŸ”—
    @@ -329,9 +336,9 @@ eBookConfig.enable_chatcodes = false; -
    mouse \(i\) 20 29
    πŸ”—

    4.

    +
    πŸ”—

    4.

    The heights of five plants are measured and recorded below. Give a 95% confidence interval around the sample mean for the heights of the plants. (Pretend 5 measurements is large enough for the CLT to apply.)
    πŸ”—
    -
    Table 4.2.2.
    πŸ”—
    +
    Table 4.3.2.
    πŸ”—
    @@ -349,7 +356,7 @@ eBookConfig.enable_chatcodes = false; -
    plant \(i\) 21 17
    πŸ”—
    πŸ”—
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    +
    πŸ”—
    πŸ”—
    πŸ”—
    PrevTopNext
    diff --git a/sec-Continuous-RVs.html b/sec-Continuous-RVs.html index 363743b..6a4fcec 100755 --- a/sec-Continuous-RVs.html +++ b/sec-Continuous-RVs.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
    4 Confidence Intervals
    diff --git a/sec-Correlation.html b/sec-Correlation.html index 5780b4e..90d1074 100755 --- a/sec-Correlation.html +++ b/sec-Correlation.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
    4 Confidence Intervals
    diff --git a/sec-Covariance.html b/sec-Covariance.html index 6d9d7f4..f672825 100755 --- a/sec-Covariance.html +++ b/sec-Covariance.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
    4 Confidence Intervals
    diff --git a/sec-Discrete-RVs.html b/sec-Discrete-RVs.html index 3c5a198..4b14c29 100755 --- a/sec-Discrete-RVs.html +++ b/sec-Discrete-RVs.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
    4 Confidence Intervals
    diff --git a/sec-Example.html b/sec-Example.html new file mode 100755 index 0000000..40ff1aa --- /dev/null +++ b/sec-Example.html @@ -0,0 +1,392 @@ + + + + + + + + + + + + + + + + +Example + + + + + + + + + + + + + + + + + + + + +Skip to main content
    +

    Math 1044 Notes

    +

    Andy Eisenberg

    +
    +
    +
    +

    +Section 4.4 Example +

    +
    Welcome to PreTeXt. Let’s type up some math. According to Pythagoras, \(a^2 + b^2 = c^2\text{.}\) But he also thought \(\sqrt{2}\) was rational, so maybe we shouldn’t take his word for it. Let’s see a proof by picture.
    πŸ”—
    +
    +

    Proof.

    +Right triangle with squares appended to each side.
    Right triangle with legs labeled \(a\) and \(b\) and hypotenuse labeled \(c\text{.}\) Squares are appended to each side of the triangle.
    πŸ”—
    +
    +
    +
    πŸ”—

    +Definition 4.4.2. +

    +
    A real number \(x \in \R\) is called rational if there are integers \(a, b\) so that \(x = \frac{a}{b}\text{.}\) Otherwise, \(x\) is called irrational, no matter what Pythagoras says.
    πŸ”—
    +
    +
    πŸ”—

    +Example 4.4.4. +

    +
    Is \(\sqrt{2}\) rational?
    πŸ”—
    +
    +
    +
    Hint 1.
    +
    Why does \(\pi = \frac{\sqrt{2}}{1}\) not show that \(\sqrt{2}\) is rational?
    πŸ”—
    +
    +
    πŸ”—
    +
    Hint 2.
    +
    What would happen if you assumed \(\sqrt{2}\) is rational?
    πŸ”—
    +
    +
    πŸ”—
    +
    Answer.
    + +
    πŸ”—
    +
    Solution.
    +
    +
    Suppose that \(\sqrt{2} = \frac{a}{b}\) for relatively prime integers \(a, b\text{.}\) Then:
    +
    +\begin{align*} +2 \amp = \frac{a^2}{b^2}\\ +2b^2 \amp = a^2 +\end{align*} +
    +
    But, after taking the prime factorization of both sides, the left side has an odd number of prime factors and the right side has an even number. This is a contradiction, so \(\sqrt{2}\) is irrational.
    +
    πŸ”—
    +
    +
    πŸ”—
    +
    +
    +
    πŸ”—

    +Exercise 4.4.5. +

    +Exercises Exercises +

    +
    πŸ”—
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    +
    +PrevTopNext +
    +
    +
    + +
    + + diff --git a/sec-Expected-Value.html b/sec-Expected-Value.html index cac1349..3ca2cae 100755 --- a/sec-Expected-Value.html +++ b/sec-Expected-Value.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
    4 Confidence Intervals
    diff --git a/sec-Independent-Events.html b/sec-Independent-Events.html index 0482d53..2f518dd 100755 --- a/sec-Independent-Events.html +++ b/sec-Independent-Events.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
    4 Confidence Intervals
    diff --git a/sec-Introduction.html b/sec-Introduction.html index 68fc61c..fe4d24c 100755 --- a/sec-Introduction.html +++ b/sec-Introduction.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
    4 Confidence Intervals
    diff --git a/sec-Joint-Distributions.html b/sec-Joint-Distributions.html index 2bcd100..52225c4 100755 --- a/sec-Joint-Distributions.html +++ b/sec-Joint-Distributions.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
    4 Confidence Intervals
    diff --git a/sec-Likelihood.html b/sec-Likelihood.html new file mode 100755 index 0000000..403b412 --- /dev/null +++ b/sec-Likelihood.html @@ -0,0 +1,554 @@ + + + + + + + + + + + + + + + + +Likelihood + + + + + + + + + + + + + + + + + + + + +Skip to main content
    +

    Math 1044 Notes

    +

    Andy Eisenberg

    +
    +
    + +

    +Section 4.1 Likelihood +

    +
    So far, we’ve been concerned with probability theory. Starting with a probability distribution and some parameter values, we’ve tried to answer questions like: What’s the probability of seeing certain experimental results? Statistics is concerned with going in the other direction: Upon seeing the experimental results, can we determine the type of underlying probability distribution? Can we determine its parameters?
    πŸ”—
    +
    +

    +Definition 4.1.1. +

    +
    An estimator is a value of a parameter computed from a sample of data.
    πŸ”—
    +
    +
    πŸ”—

    +Example 4.1.2. +

    +
    Suppose we find a coin on the street and don’t know whether or not it’s fair. We want to know the probability \(p\) of the coin coming up heads. We might, for example, flip the coin \(n\) times and count the number \(k\) of heads. Then, we’ll estimate \(p = \frac{k}{n}\text{.}\) We’ll refer to this as a common sense estimator. (Other distributions and parameter types will have different notions of "common sense".)
    πŸ”—
    +
    +
    πŸ”—
    An estimator is, itself, a random variable: it produces a numerical value based on the results of an experiment. We’ll use notation like \(\est{p}\) for a random variable which is an estimator for a parameter \(p\text{.}\) (Similarly, \(\est{\lambda}\) would denote an estimator for a parameter called \(\lambda\text{.}\))
    πŸ”—
    +
    +

    +Definition 4.1.3. +

    +
    An estimator \(\est{p}\) is called unbiased if \(\E(\est{p}) = p\text{.}\)
    πŸ”—
    +
    +
    πŸ”—

    +Example 4.1.4. +

    +
    +
    Suppose we have a coin with parameter \(p\text{,}\) which we’ll flip \(n\) times and count the number \(k\) of heads. We use the unbiased estimator \(\est{p} = \frac{k}{n}\text{.}\) In this case, notice that \(k \sim \Bin(n, p)\text{,}\) so we know \(\E(k) = np\text{,}\) although we don’t know the value of \(p\text{.}\) (We probably do know the value of \(n\text{;}\) after all, we’re flipping the coin!) Now:
    +
    +\begin{gather*} +\E(\est{p}) = \E\left(\frac{k}{n}\right) = \frac{1}{n} \cdot \E(k) = \frac{1}{n} \cdot np = p. +\end{gather*} +
    +
    It’s worth pausing for a moment to be appropriately impressed with ourselves. We still don’t know the true value of \(p\text{.}\) But we managed to show that our common sense method of estimating \(p\) gives, on average, the correct value.
    +
    πŸ”—
    +
    +
    πŸ”—
    We’d like to be able to collect some data and use that data to estimate the values of whatever parameters our distribution has. Perhaps as a starting point, it would be good to identify the single most likely value of a parameter:
    πŸ”—
    +
    +

    +Definition 4.1.5. +

    +
    +
    Suppose we collect some data to estimate a parameter \(p\text{.}\) The likelihood function is:
    +
    +\begin{gather*} +\L(p) = \Pr(\text{data} \mid \text{parameter value is } p). +\end{gather*} +
    +
    The value of \(p\) which maximizes the function \(\L(p)\) is called the maximum likelihood estimation, or MLE, of \(p\text{.}\) +
    +
    πŸ”—
    +
    +
    πŸ”—

    +Example 4.1.6. +

    +
    +
    Suppose we flip a coin 100 times. Let \(N\) be the number of heads. Then \(N\sim \Bin(100, p)\) for some unknown value of \(p\text{:}\) +
    +
    +\begin{align*} +\Pr(N = k) \amp = b(k; 100, p) = {100 \choose k} p^k (1 - p)^{100 - k} +\end{align*} +
    +
    In this case, \(k\) is the data that we collect, and \(p\) is the value of the parameter. Suppose we see \(52\) heads. Then:
    +
    +\begin{gather*} +\L(p) = {100 \choose 52} p^{52} (1 - p)^{48}. +\end{gather*} +
    +
    If we want to know the most likely value of the parameter \(p\text{,}\) then we should maximize \(\L(p)\) over the interval \(0 \leq p \leq 1\text{.}\) +
    +
    +\begin{align*} +\L'(p) \amp = {100 \choose 52} \left[ 52 p^{51}(1 - p)^{48} + p^{52} 48 (1 - p)^{47}(-1)\right] \\ +\amp = {100 \choose 52} p^{51} (1 - p)^{47} \left[ 52 (1 - p) - 48p \right] \\ +\amp = {100 \choose 52} p^{51} (1 - p)^{47} \left[ 52 - 100 p \right]. +\end{align*} +
    +
    We can see that \(\L'(p) = 0\) when \(p = 0, 1, \frac{52}{100}\text{,}\) and the endpoints of the interval we’re maximizing over are \(p = 0, 1\text{,}\) so we can build a table:
    +
    πŸ”—
    +
    Table 4.1.7. Looking for the Maximum
    πŸ”—
    + + + + + + + + + + + + + + + + +
    \(p\)\(\L(p)\)
    \(0\)\(0\)
    \(52/100\)\(\gt 0\)
    \(1\)\(0\)
    Note that we don’t need to know the exact value of \(\L(52/100)\) in order to see that it’s strictly positive, and therefore the maximum value of \(\L(p)\text{.}\) So, the MLE of \(p\) is \(52/100\text{.}\)
    πŸ”—
    +
    +
    πŸ”—
    We can see in the previous example that the MLE of \(p\) is also the common sense estimation of \(p\text{.}\) This will be the case in general for the binomial distribution, so we won’t need to redo this work over and over:
    πŸ”—
    +
    +
    As we remember from Calculus 1, maximizing a continuous function works slightly differently over a closed interval (like the previous example) or an open interval (like the next example).
    πŸ”—
    +
    +

    +Example 4.1.9. +

    +
    +
    Suppose a radioactive material emits particles as it decays. Let \(N\) count the particles emitted. Then \(N \sim \Poiss(\lambda)\) for some unknown rate parameter \(\lambda\text{:}\) +
    +
    +\begin{gather*} +\Pr(N = k) = p(k; \lambda) = \frac{\lambda^k}{k!} e^{-\lambda} +\end{gather*} +
    +
    Suppose we observe a sample of material for 1 hour and count 8 particles emitted. Then:
    +
    +\begin{gather*} +\L(\lambda) = \frac{\lambda^8}{8!} e^{-\lambda} +\end{gather*} +
    +
    To maximize \(\L(\lambda)\) over the interval \(0 \lt \lambda \lt \infty\text{,}\) we start by finding critical points.
    +
    +\begin{align*} +\L'(\lambda) \amp = \frac{1}{8!} \left[ 8 \lambda^7 e^{-\lambda} + \lambda^8 e^{-\lambda} (-1)\right] \\ +\amp = \frac{1}{8!} \lambda^7 e^{-\lambda} \left[ 8 - e^{-\lambda} \right] +\end{align*} +
    +
    The only critical point is \(\lambda = 8\text{,}\) but we have not justified that this is the location of a global maximum. A critical point is only a potential location of a local minimum or maximum. But with a bit more justification: observe that \(\L'\) is positive on the interval \((0, 8)\) and negative on the interval \((8, \infty)\text{.}\) Therefore the function \(\L\) increases on \((0, 8)\) and decreases on \((8, \infty)\text{.}\) So it must reach its maximum at \(\lambda = 8\text{,}\) which is therefore the MLE.
    +
    πŸ”—
    +
    +
    πŸ”—
    As with the binomial distribution, the calculation will be essentially the same regardless of the specific number of particles observed, so:
    πŸ”—
    +
    +
    The situation for continuous random variables is similar, but slightly different. To build the likelihood function, we should use the pdf of the continuous random variable. So the likelihood function will give the probabiliy density given the data collected, rather than the probability itself.
    πŸ”—
    +
    +

    +Example 4.1.11. +

    +
    +
    Suppose we observe a cell and measure the time \(T\) until a toxin molecule leaves the cell. Then \(T \sim \Exp(\lambda)\) for some unknown rate parameter \(\lambda\text{:}\) +
    +
    +\begin{gather*} +f(t) = \lambda e^{-\lambda t}. +\end{gather*} +
    +
    Suppose we observe a molecule leave the cell at \(t = 0.3\) minutes. Then:
    +
    +\begin{align*} +L(\lambda) \amp = \lambda e^{-0.3\lambda} \\ +L'(\lambda) \amp = e^{-0.3\lambda} + \lambda e^{-0.3t} (-0.3) \\ +\amp = e^{-0.3\lambda}\left[ 1 - 0.3\lambda\right] +\end{align*} +
    +
    The critical point is \(\lambda = \frac{1}{0.3} \approx 3.33\text{.}\) Since \(\L' \gt 0\) on \((0, 3.33)\) and \(\L' \lt 0\) on \((3.33, \infty)\text{,}\) there is a global maximum at \(\lambda \approx 3.33\text{,}\) which is therefore the MLE.
    +
    πŸ”—
    +
    The specific time \(0.3\) minutes doesn’t particularly matter in this calculation. Whatever the time \(t\text{,}\) essentially the same calculation will result in a MLE of \(\lambda = 1/t\text{.}\) But what if we collect multiple pieces of data?
    πŸ”—
    +
    Table 4.1.12. Time Data
    πŸ”—
    + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + +
    MoleculeWaiting TimeRate Estimation
    1\(0.3\)\(3.33\)
    2\(0.8\)\(1.25\)
    3\(0.5\)\(2.00\)
    4\(0.6\)\(1.67\)
    5\(0.9\)\(1.11\)
    How should take all of this data into account in our maximum likelihood estimation? We might consider taking the average of all of the separate rate estimations, which would give \(1.87\text{.}\) Is this the most likely? We need some mathematical justification.
    πŸ”—
    +
    +
    To account for multiple, independent data points, we should multiply the probability densities for each in the creation of our likelihood function:
    +
    +\begin{align*} +\L(\lambda) \amp = \left(\lambda e^{-0.3\lambda}\right)\left(\lambda e^{-0.8\lambda}\right)\left(\lambda e^{-0.5\lambda}\right)\left(\lambda e^{-0.6\lambda}\right)\left(\lambda e^{-0.9\lambda}\right) \\ +\amp = \lambda^5 e^{-0.3\lambda - 0.8\lambda - 0.5\lambda - 0.6\lambda - 0.9\lambda } \\ +\amp = \lambda^5 e^{-3.1\lambda} +\end{align*} +
    +
    Now we can find the maximum:
    +
    +\begin{align*} +\L'(\lambda) \amp = 5 \lambda^4 e^{-3.1\lambda} + \lambda^5 e^{-3.1\lambda} (-3.1) \\ +\amp = 5 \lambda^4 e^{-3.1\lambda} + \lambda^5 e^{-3.1\lambda} (-3.1) \\ +\amp = \lambda^4 e^{-3.1\lambda} \left[5 - 3.1\lambda \right] +\end{align*} +
    +
    The only critical point is \(5/3.1 \approx 1.61\text{.}\) Since \(\L' \gt 0\) on \((0, 1.61)\) and \(\L' \lt 0\) on \((1.61, \infty)\text{,}\) there is a global maximum at \(\lambda = 1.61\text{,}\) which is therefore the MLE.
    +
    πŸ”—
    +
    It may seem less clear how to generalize this calculation for other tables of data. Observe that the value \(3.1\) is the sum of the five times in the table, so \(3.1/5\) is the average time. The MLE turned out to be the reciprocal of the average time (just as the MLE with only one data point was the reciprocal of that one time). Notice that this does not match the guess we made previously of averaging the individual rate estimations for each data point.
    πŸ”—
    +
    +
    πŸ”—

    +Exercises Exercises +

    +

    1.

    +
    Suppose a coin has an unknown probability of coming up heads. We perform the experiment in five independent trials, during which it takes 4, 5, 4, 3, and 6 flips to see our first heads in each trial. What is the maximum likelihood estimation for the probability of the coin coming up heads on a flip?
    πŸ”—
    +
    +
    πŸ”—

    2.

    +
    Suppose a coin has an unknown probability of coming up heads. We perform the experiment in \(n\) independent trials, during which it takes \(k_1, k_2, \dotsc, k_n\) flips to see our first heads in each trial. Find a "common sense" MLE formula for the geometric distribution.
    πŸ”—
    +
    +
    Hint.
    +
    You could set up a calculation analogous to ExampleΒ 4.1.11. Or, you could consider FactΒ 4.1.8.
    πŸ”—
    +
    +
    πŸ”—
    +
    +
    πŸ”—

    3.

    +
    A particular store owner wants to approximate the average hourly rate at which customers come into the store. They observe 80 customers enter during a particular 4-hour shift. What is the maximum likelihood estimation for the hourly customer rate?
    πŸ”—
    +
    +
    πŸ”—

    4.

    +
    A radioactive material emits particles at an unknown probabilistic rate \(\lambda\) particles per minute. We observe particles emitted at times 1.1, 1.7, 1.3, 2.2, 1.9, and 1.8 minutes. Write the likelihood function \(\mathcal{L}(\lambda)\) based on this data. What is the maximum likelihood estimation for \(\lambda\text{?}\)
    πŸ”—
    +
    +
    πŸ”—

    5.

    +
    Suppose a parameter \(\theta\) takes values in \([0, 1]\) with likelihood function \(\mathcal{L}(\theta) = \sqrt{\theta} - \theta^2\text{.}\) Find the maximum likelihood estimation of \(\theta\text{.}\)
    πŸ”—
    +
    +
    πŸ”—
    πŸ”—
    πŸ”—
    +
    +PrevTopNext +
    +
    +
    + +
    + + diff --git a/sec-Linear-Regression.html b/sec-Linear-Regression.html index 46dbc4a..552c47a 100755 --- a/sec-Linear-Regression.html +++ b/sec-Linear-Regression.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
    4 Confidence Intervals
    diff --git a/sec-One-Sample-Tests.html b/sec-One-Sample-Tests.html index 2dde7ff..6ab3d81 100755 --- a/sec-One-Sample-Tests.html +++ b/sec-One-Sample-Tests.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
    4 Confidence Intervals
    diff --git a/sec-Power.html b/sec-Power.html index 041ad00..a42a5d5 100755 --- a/sec-Power.html +++ b/sec-Power.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
    4 Confidence Intervals
    diff --git a/sec-Probability.html b/sec-Probability.html index 777c709..26f94b1 100755 --- a/sec-Probability.html +++ b/sec-Probability.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
    4 Confidence Intervals
    diff --git a/sec-Set-Theory.html b/sec-Set-Theory.html index f3b93d1..fce0a2d 100755 --- a/sec-Set-Theory.html +++ b/sec-Set-Theory.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
    4 Confidence Intervals
    diff --git a/sec-Two-Sample-Tests.html b/sec-Two-Sample-Tests.html index d59d4bb..fba228c 100755 --- a/sec-Two-Sample-Tests.html +++ b/sec-Two-Sample-Tests.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
    4 Confidence Intervals
    diff --git a/sec-Variance.html b/sec-Variance.html index fae702a..b52abae 100755 --- a/sec-Variance.html +++ b/sec-Variance.html @@ -62,7 +62,7 @@ window.MathJax = { "[pretext]/mathjaxknowl3.js" ], "paths": { - "pretext": "_static/pretext/js/lib" + "pretext": "_static/pretext/js" } }, "startup": { @@ -79,7 +79,7 @@ window.MathJax = { - + @@ -101,16 +101,16 @@ eBookConfig.allow_pairs = false; eBookConfig.enableScratchAC = false; eBookConfig.build_info = ""; eBookConfig.python3 = null; -eBookConfig.runestone_version = '7.11.11'; +eBookConfig.runestone_version = '7.11.13'; eBookConfig.jobehost = ''; eBookConfig.proxyuri_runs = ''; eBookConfig.proxyuri_files = ''; eBookConfig.enable_chatcodes = false; - + - @@ -151,6 +151,9 @@ eBookConfig.enable_chatcodes = false; \DeclareMathOperator{\E}{E} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Cov}{Cov} +\newcommand{\L}{\mathcal{L}} + +\newcommand{\est}{\widehat} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&} @@ -244,12 +247,16 @@ eBookConfig.enable_chatcodes = false;
    4 Confidence Intervals