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source/docinfo.ptx
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\DeclareMathOperator{\Geom}{Geom}
\DeclareMathOperator{\Poiss}{Poiss}
\DeclareMathOperator{\Exp}{Exp}
\DeclareMathOperator{\Norm}{N}
<!-- other notation -->
\DeclareMathOperator{\E}{E}
source/main.ptx
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<xi:include href="./notes/2-3.ptx" />
<xi:include href="./notes/2-5.ptx" />
<xi:include href="./notes/2-10.ptx" />
<xi:include href="./notes/2-19.ptx" />
<xi:include href="./notes/2-24.ptx" />
</chapter>
<chapter xml:id="quizzes">
source/notes/2-19.ptx
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<?xml version="1.0" encoding="UTF-8"?>
<section xml:id="notes-02-19">
<title>Thursday, Feb 19</title>
<introduction>
<p>
This is an outline of the topics we covered in class.
These notes are <em>not</em> a substitute for your own note-taking.
I highly recommend that you take your own notes during class.
If you ever miss a class for any reason, reach out to another student in class to get a copy of their notes.
</p>
</introduction>
<subsection>
<title>Central Limit Theorem</title>
<definition>
<statement>
<p>
A continuous random variable <m>X</m> has the <term>normal distribution</term> with mean <m>\mu</m> and variance <m>\sigma^2</m> if it has pdf:
<md>
<mrow> f(x; \mu, \sigma^2) = \frac{1}{\sqrt{2\pi \sigma^2}} e^{\frac{-(x - \mu)^2}{2\sigma^2}} </mrow>
</md>
We'll write <m>X \sim \Norm(\mu, \sigma^2)</m>.
</p>
<p>
The specific distribution <m>\Norm(0, 1)</m> is called the <term>standard normal distribution</term>, and its pdf has the notation:
<md>
<mrow> f(x; 0, 1) = \phi(x). </mrow>
</md>
</p>
</statement>
</definition>
<p>
If <m>X \sim \Norm(\mu, \sigma^2)</m>, then:
<md>
<mrow> \Pr(a \leq X \leq b) = \int_a^b f(x; \mu, \sigma^2)\ dx. </mrow>
</md>
Unfortunately, <m>e^{-x^2}</m> has no elementary antiderivative.
But we know <m>\int_a^b f(x)\ dx</m> represents the area under the graph <m>y = f(x; \mu, \sigma^2)</m>, and this area can be approximated to arbitrary precision.
</p>
<theorem xml:id="thm-CLT">
<statement>
<p>
Suppose <m>X_1, \dotsc, X_n</m> are <term>independent and identically distributed</term> (or <term>i.i.d.</term>) with finite expected value <m>\mu</m> and finite variance <m>\sigma^2</m>.
Define:
<md>
<mrow> S_n \amp = X_1 + X_2 + \dotsb + X_n </mrow>
<mrow> A_n \amp = \frac{X_1 + X_2 + \dotsb + X_n}{n} = \frac{S_n}{n} </mrow>
</md>
Then, for large enough <m>n</m>:
<md>
<mrow> S_n \amp \approx \Norm(n\mu, n\sigma^2) </mrow>
<mrow> A_n \amp \approx \Norm\left(\mu, \frac{\sigma^2}{n}\right) </mrow>
</md>
</p>
</statement>
</theorem>
<p>
We'll use two general rules of thumb for determining whether the number of measurements <m>n</m> is large enough for the approximation to be a good one:
<ol>
<li>
<p>
<m>n \geq 30</m>
</p>
</li>
<li>
<p>
If <m>S\sim \Bin(n, p)</m>, then <m>n</m> should be large enough so that there are at least 5 heads and 5 tails.
</p>
</li>
</ol>
</p>
<p>
We'll omit a proof of <xref ref="thm-CLT"/>, but we can at least check that the expected values and variances of <m>S_n, A_n</m> are correct.
Given <m>\E(X_i) = \mu, \Var(X_i) = \sigma^2</m>:
<md>
<mrow> \E(S_n) \amp = \E(X_1 + \dotsb + X_n) </mrow>
<mrow> \amp = \E(X_1) + \dotsb + \E(X_n) </mrow>
<mrow> \amp = \mu + \dotsb + \mu </mrow>
<mrow> \amp = n \mu. \checkmark </mrow>
<mrow> \Var(S_n) \amp = \Var(X_1 + \dotsb + X_n) </mrow>
<mrow> \amp = \Var(X_1) + \dotsb + \Var(X_n) </mrow>
<mrow> \amp = \sigma^2 + \dotsb + \sigma^2 </mrow>
<mrow> \amp = n \sigma^2. \checkmark </mrow>
<mrow> \E(A_n) \amp = \E\left(\frac{S_n}{n}\right) </mrow>
<mrow> \amp = \frac{1}{n} \E(S_n) </mrow>
<mrow> \amp = \frac{1}{n} n \mu </mrow>
<mrow> \amp = \frac{1}{n} n \mu </mrow>
<mrow> \amp = \mu. \checkmark </mrow>
<mrow> \Var(A_n) \amp = \Var\left(\frac{S_n}{n}\right) </mrow>
<mrow> \amp = \frac{1}{n^2} \Var(S_n) </mrow>
<mrow> \amp = \frac{1}{n^2} n \sigma^2 </mrow>
<mrow> \amp = \frac{\sigma^2}{n}. \checkmark </mrow>
</md>
</p>
<example>
<statement>
<p>
Suppose we flip a fair coin 100 times, and let <m>S</m> be the number of heads.
Estimate <m>\Pr(40 \leq S \leq 60)</m>.
</p>
<p>
Since <m>S \sim \Bin(n, p)</m>, we know:
<md>
<mrow> \E(S) \amp = np = (100)(0.5) = 50 </mrow>
<mrow> \Var(S) \amp = np(1-p) = (100)(0.5)(1 - 0.5) = 25 </mrow>
</md>
So <m>S \approx \Norm(50, 25)</m>.
Then:
<md>
<mrow> \Pr(40 \leq S \leq 60) \approx \int_{40}^{60} f(x; 50, 25)\ dx, </mrow>
</md>
but we can't compute this integral.
</p>
<p>
Instead, we can <term>standardize</term> by shifting and scaling:
<md>
<mrow> Z = \frac{S - 50}{\sqrt{25}} = \frac{S - 50}{5}. </mrow>
</md>
Then <m>Z \sim \Norm(0, 1)</m>.
Values of <m>Z</m> are called <term><m>z</m>-scores</term>, sometimes indicated by an asterisk.
(I.e., if <m>a</m> is a value of <m>S</m>, then <m>a^*</m> is the corresponding <m>z</m>-score.) Now:
<md>
<mrow> \Pr(40 \leq S \leq 60) \amp \approx \Pr\left(\frac{40 - 50}{5} \leq Z \leq \frac{60 - 50}{5}\right) </mrow>
<mrow> \amp = \Pr(-2 \leq Z \leq 2) </mrow>
<mrow> \amp = \int_{-2}^2 \phi(x)\ dx. </mrow>
</md>
Now, we still can't compute the value of the integral.
However, there's a huge benefit to translating to the standard normal distribution, regardless of <em>which</em> normal distribution we used to approximate <m>S</m>.
We can approximate integrals like <m>\int_a^b \phi(x)\ dx</m> to abritrary precision and record the results in a table.
Then, we can look up the values when needed.
We don't need to redo our approximations for different normal distributions, we just standardize whatever normal distribution we come across.
</p>
<p>
So:
<md>
<mrow> \Pr(40 \leq S \leq 60) \amp \approx \Pr(-2 \leq Z \leq 2) </mrow>
<mrow> \amp = \Phi(2) - \Phi(-2) </mrow>
<mrow> \amp \approx 0.9772 - 0.0228 </mrow>
<mrow> \amp \approx 0.9544. </mrow>
</md>
</p>
</statement>
</example>
<p>
TODO: complex image introducing the idea of the <term>continuity correction</term>.
</p>
<p>
We can make the approximation more accurate by extending the range of <m>S</m>-values by half a unit in each direction:
<md>
<mrow> \Pr(40 \leq S \leq 60) \amp \approx \Pr\left(\frac{39.5 - 50}{5} \leq Z \leq \frac{60.5 - 50}{5}\right) </mrow>
<mrow> \amp = \Pr(-2.1 \leq Z \leq 2.1) </mrow>
<mrow> \amp = \Phi(2.1) - \Phi(-2.1) </mrow>
<mrow> \amp \approx 0.9821 - 0.0179 </mrow>
<mrow> \amp \approx 0.9642. </mrow>
</md>
</p>
</subsection>
</section>
source/notes/2-24.ptx
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<?xml version="1.0" encoding="UTF-8"?>
<section xml:id="notes-02-24">
<title>Tuesday, Feb 24</title>
<introduction>
<p>
This is an outline of the topics we covered in class.
These notes are <em>not</em> a substitute for your own note-taking.
I highly recommend that you take your own notes during class.
If you ever miss a class for any reason, reach out to another student in class to get a copy of their notes.
</p>
</introduction>
<subsection>
<title>More Central Limit Theorem</title>
<example>
<statement>
<p>
Suppose we roll a fair D6 100 times, and let <m>m</m> be the average of the rolls.
Estimate <m>\Pr(3.45 \leq m \leq 3.55)</m>.
</p>
<p>
Let <m>R_1, \dotsc, R_{100}</m> be each roll's result.
Then we have previously calculated <m>\E(R_i) = 3.5, \Var(R_i) = \frac{35}{12}</m>.
We have:
<md>
<mrow> m \amp = \frac{R_1 + \dotsb + R_{100}}{100}. </mrow>
<mrow> \E(m) \amp = \E\left(\frac{R_1 + \dotsb + R_{100}}{100}\right) </mrow>
<mrow> \amp = \frac{1}{100}\left[\E(R_1) + \dotsb + \E(R_{100})\right] </mrow>
<mrow> \amp = \frac{1}{100}(100)(3.5) </mrow>
<mrow> \amp = 3.5 </mrow>
<mrow> \Var(m) \amp = \Var\left(\frac{R_1 + \dotsb + R_{100}}{100}\right) </mrow>
<mrow> \amp = \frac{1}{100^2} \left[\Var(R_1) + \dotsb + \Var(R_{100}) \right] </mrow>
<mrow> \amp = \frac{1}{100^2} (100)\left(\frac{35}{12}\right) </mrow>
<mrow> \amp = \frac{35}{1200} </mrow>
</md>
So, by the CLT, <m>m \approx \Norm\left(3.5, \frac{35}{1200}\right)</m>.
Even though <m>m</m> takes on decimal values, it's still a discrete random variable.
The sum of the rolls can only take on integer values, so <m>m</m> will take on the values <m>1.00, 1.01, \dotsc, 5.99, 6.00</m>.
Last time, we described the continuity correction as "extending the range by half a unit's width in each direction".
Here, the width of one unit is <m>0.01</m>.
So:
<md>
<mrow> \Pr(3.45 \leq m \leq 3.55) \amp \approx \Pr\left( \frac{3.4445 - 3.5}{\sqrt{35/1200}} \leq Z \leq \frac{3.555 - 3.5}{\sqrt{35/1200}}\right) </mrow>
<mrow> \amp \approx \Pr(-0.32 \leq Z \leq 0.32) </mrow>
<mrow> \amp = \Phi(0.32) - \Phi(-0.32) </mrow>
<mrow> \amp \approx 0.6255 - 0.3745 </mrow>
<mrow> \amp = 0.2510. </mrow>
</md>
</p>
</statement>
</example>
</subsection>
<subsection>
<title>Confidence Intervals</title>
<p>
Idea: suppose we collected (i.i.d.) measurements <m>X_1, \dotsc, X_n</m> which have some population mean <m>\mu</m> and variance <m>\sigma^2</m>.
We calculate the average:
<md>
<mrow> A_n = \frac{A_1 + \dotsb + A_n}{n} </mrow>
</md>
and we want to estimate the value of <m>\mu</m> from the collected data.
Previously, we've discussed the MLE, the single most likely estimate of the parameter.
Now, we'd like to give a range <m>[\mu_{\ell}, \mu_h]</m> where we can say something like: "we are 95% confident that <m>\mu \in [\mu_{\ell}, \mu_h]</m>."
</p>
<p>
From the CLT, if <m>n</m> is large enough, we approximate <m>A_n \approx \Norm\left(\mu, \frac{\sigma^2}{n}\right)</m>.
We can think of an interval centered at <m>\mu</m> as a collection of numbers within a certain distance <m>d</m> of <m>\mu</m>.
We'd like to choose a distance so that our measured <m>A_n</m> is within <m>d</m> of <m>\mu</m>.
We can shift our point of view here: if <m>A_n</m> is within <m>d</m> of <m>\mu</m>, then <m>\mu</m> is within <m>d</m> of <m>A_n</m>.
</p>
<p>
TODO: image
</p>
<p>
To pick the distance <m>d</m>, we want to make our interval large enough so that there's only a small amount of area under the normal curve outside of the interval.
So:
<ol>
<li>
<p>
Pick some value <m>\alpha \in (0, 1)</m> (representing the area under the curve outside the interval)
</p>
</li>
<li>
<p>
Let <m>c_{\alpha} = \Phi^{-1}\left(1 - \frac{\alpha}{2}\right)</m> (this is the standardized <m>z</m>-score of the right-hand side of the interval)
</p>
</li>
<li>
<p>
Then, the interval will be:
<md>
<mrow> \left[A_n - c_{\alpha} \sqrt{\frac{\sigma^2}{n}}, A_n + c_{\alpha} \sqrt{\frac{\sigma^2}{n}}\right]. </mrow>
</md>
</p>
</li>
</ol>
</p>
<p>
For <m>\alpha = 0.05</m>, we'll have <m>c_{\alpha} = \Phi^{-1}\left(1 - \frac{0.05}{2}\right) = \Phi^{-1}(0.975) = 1.96</m>.
So:
</p>
<theorem>
<statement>
<p>
The 95% confidence limits are given by:
<md>
<mrow> \mu_{\ell} \amp = A_n - 1.96 \sqrt{\frac{\sigma^2}{n}} </mrow>
<mrow> \mu_{h} \amp = A_n + 1.96 \sqrt{\frac{\sigma^2}{n}} </mrow>
</md>
</p>
</statement>
</theorem>
<definition>
<statement>
<p>
The term <m>\sqrt{\frac{\sigma^2}{n}}</m> is called the <term>standard error of the mean</term>.
(It's the standard deviation of <m>A_n</m>.)
</p>
</statement>
</definition>
<p>
We don't know the true value of <m>\sigma^2</m>, just like we don't know the true value of <m>\mu</m>.
So we'll have to use our sample of measurements to estimate <m>\sigma^2</m>, and then use that estimate to give us our range of values for <m>\mu</m>.
</p>
<definition>
<statement>
<p>
Let <m>X_1, \dotsc, X_n</m> be (i.i.d.) measurements.
Then the <term>sample mean</term> is:
<md>
<mrow> A_n = \frac{X_1 + \dotsb + X_n}{n} </mrow>
</md>
and the <term>sample variance</term> is:
<md>
<mrow> s^2 = \frac{\sum (X_i - A_n)^2}{n - 1} = \frac{\left(\sum X_i^2\right) - n(A_n^2)}{n - 1} </mrow>
</md>
</p>
</statement>
</definition>
<p>
Why <m>n - 1</m> here? It turns out that, as defined above, <m>s^2</m> is an unbiased estimator for <m>\sigma^2</m>, whereas it wouldn't be if we divided by <m>n</m> instead.
(We'll ommit both the calculation justifying this and a more conceptual explanation for now.)
</p>
<p>
So, we can adjust the confidence limits from the theorem:
<md>
<mrow> \mu_{\ell} \amp = A_n - 1.96 \sqrt{\frac{s^2}{n}} </mrow>
<mrow> \mu_{h} \amp = A_n + 1.96 \sqrt{\frac{s^2}{n}} </mrow>
</md>
</p>
<example>
<statement>
<p>
Suppose we're given measurements <m>X_1, X_2, X_3, X_4, X_5</m> below.
</p>
<tabular halign="center">
<row header="yes" bottom="minor">
<cell><m>i</m></cell>
<cell><m>X_i</m></cell>
</row>
<row>
<cell><m>1</m></cell>
<cell><m>19.2</m></cell>
</row>
<row>
<cell><m>2</m></cell>
<cell><m>20.1</m></cell>
</row>
<row>
<cell><m>3</m></cell>
<cell><m>21.3</m></cell>
</row>
<row>
<cell><m>4</m></cell>
<cell><m>20.7</m></cell>
</row>
<row>
<cell><m>5</m></cell>
<cell><m>19.8</m></cell>
</row>
</tabular>
<p>
We're just practicing the computation here, so we'll pretend that 5 measurements is large enough for the CLT to apply.
<md>
<mrow> A_n \amp = \frac{19.2 + \dotsb + 19.8}{5} = 20.22 </mrow>
<mrow>\sum X_i^2 \amp = 19.2^2 + \dotsb + 19.8^2 = 2046.87 </mrow>
<mrow> s^2 \amp = \frac{2046.87 - 5(20.22^2)}{4} = 0.657 </mrow>
<mrow> \sqrt{\frac{s^2}{n}} \amp = \sqrt{\frac{0.657}{5}} \approx 0.362 </mrow>
<mrow> \mu_{\ell} \amp = 20.22 - 1.96(0.362) \approx 19.51 </mrow>
<mrow> \mu_{h} \amp = 20.22 + 1.96(0.362) \approx 20.93 </mrow>
</md>
</p>
</statement>
</example>
<p>
This way of computing confidence limits only applies if we can think of the parameter we're estimating as a mean.
</p>
<example>
<statement>
<p>
Suppose we flip a coin 100 times and see 40 heads.
Find a 98% confidence interval for the bias <m>p</m>.
</p>
<p>
Let <m>H_1, \dotsc, H_{100}</m> indicate heads on each flip.
Let <m>S_{100} = H_1 + \dotsb + H_{100}</m> and <m>A_n = \frac{S_n}{n}</m>.
Then:
<md>
<mrow> \E(A_n) \amp = \E\left(\frac{S_n}{n}\right) = \frac{1}{n}\E(S_n) = \frac{1}{n} np = p </mrow>
</md>
So a 98% confidence interval for <m>p</m> is also a 98% confidence interval for <m>p</m>.
</p>
<p>
In this setting, we can simplify the computation of <m>s^2</m> and <m>\sqrt{\frac{s^2}{n}}</m>, using our knowledge of the binomial distribution <m>S_n</m>.
<md>
<mrow> \Var(S_n) \amp = np(1 - p) </mrow>
<mrow> \Var(A_n) \amp = \Var\left(\frac{S_n}{n}\right) = \frac{1}{n^2} \Var(S_n) = \frac{1}{n^2} np(1-p) = \frac{p(1-p)}{n} </mrow>
</md>
So <m>s^2 = p(1 - p)</m> and <m>\sqrt{\frac{s^2}{n}} = \sqrt{\frac{p(1-p)}{n}}</m>.
We don't know the value of <m>p</m>, so we'll use the MLE <m>\widehat{p} = \frac{40}{100} = 0.4</m>.
So:
<md>
<mrow> \sqrt{\frac{s^2}{n}} = \sqrt{\frac{(0.4)(0.6)}{100}} \approx 0.049. </mrow>
</md>
</p>
<p>
For a 98% confidence interval, we have <m>\alpha = 1 - 0.98 = 0.02</m>, so:
<md>
<mrow> c_{\alpha} = \Phi^{-1}\left(1 - \frac{\alpha}{2}\right) = \Phi^{-1}(0.99) \approx 2.33. </mrow>
</md>
Then, the confidence limits are:
<md>
<mrow> p_{\ell} \amp = 0.4 - 2.33(0.049) \approx 0.286 </mrow>
<mrow> p_{h} \amp = 0.4 + 2.33(0.049) \approx 0.514 </mrow>
</md>
For comparison, the 95% confidence limits are:
<md>
<mrow> p_{\ell} \amp = 0.4 - 1.96(0.049) \approx 0.304 </mrow>
<mrow> p_{h} \amp = 0.4 + 1.96(0.049) \approx 0.496 </mrow>
</md>
</p>
</statement>
</example>
</subsection>
</section>