198 lines
5.8 KiB
XML
198 lines
5.8 KiB
XML
<?xml version="1.0" encoding="UTF-8"?>
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<section xml:id="notes-01-20">
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<title>Tuesday, Jan 20</title>
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<introduction>
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<p>
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This is an outline of the topics we covered in class.
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These notes are <em>not</em> a substitute for your own note-taking.
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I highly recommend that you take your own notes during class.
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If you ever miss a class for any reason, reach out to another student in class to get a copy of their notes.
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</p>
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</introduction>
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<subsection xml:id="subsec-hw1-review">
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<title>HW 1 Q5</title>
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<p>
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Write <m>F</m> for the event that there's a fire and <m>S</m> for the event that there's visible smoke.
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Then the information we're given can be interpreted as:
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<md>
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<mrow> \Pr(F) \amp = 0.01 </mrow>
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<mrow> \Pr(S) \amp = 0.1 </mrow>
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<mrow> \Pr(S\mid F) \amp = 0.9 </mrow>
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</md>
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In this case, we can use the simpler version of Bayes' Theorem:
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<md>
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<mrow> \Pr(F \mid S) = \frac{\Pr(S \mid F)\Pr(F)}{\Pr(S)} = \dotsb </mrow>
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</md>
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Unlike our usual diagnostic testing examples, we do have access to the denominator probability here.
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</p>
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</subsection>
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<subsection xml:id="subsec-Discrete-Random-Variables">
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<title>Sec 2.1: Random Variables</title>
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<definition xml:id="def-RV">
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<statement>
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<p>
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A <term>random variable</term> is a function <m>X \colon \Omega \to \R</m>.
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</p>
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</statement>
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</definition>
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<p>
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The idea is that <m>X</m> is a variable representing a real number value which depends on the outcome of an experiment.
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</p>
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<example>
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<statement>
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<p>
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An experiment consists of planting 50 seeds in a garden, then growing them for 3 months.
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Let <m>H_i</m> be the height of plant <m>i</m>.
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Let <m>D</m> be the number of seeds that didn't sprout.
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Let
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<md>
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<mrow> A \amp = \text{avg height of all 50 plants} </mrow>
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<mrow> \amp = \frac{H_1 + H_2 + \dotsb + H_{50}}{50} </mrow>
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</md>
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</p>
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</statement>
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</example>
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<p>
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A random variable has its own probability distribution.
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</p>
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<example>
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<statement>
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<p>
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Roll a fair D6 twice.
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Let <m>S</m> be the sum of the rolls.
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Then <m>\Omega = \{(1, 1), (1, 2), \dotsc, (6, 6)\}</m> has 36 elements.
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Since the die is fair, the distribution on <m>\Omega</m> is uniform, i.e., <m>\Pr(\omega) = \frac{1}{36}</m> for any <m>\omega \in \Omega</m>.
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</p>
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<p>
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<m>S</m> takes on the values <m>2, 3, 4, \dotsc, 12</m>, with probabilities:
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</p>
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<table>
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<title>Distribution for <m>S</m></title>
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<tabular halign="center">
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<row>
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<cell><m>x</m></cell>
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<cell><m>\Pr(S = x)</m></cell>
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</row>
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<row>
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<cell>2</cell>
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<cell>1/36</cell>
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</row>
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<row>
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<cell>3</cell>
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<cell>2/36</cell>
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</row>
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<row>
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<cell>4</cell>
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<cell>3/36</cell>
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</row>
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<row>
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<cell><m>\vdots</m></cell>
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<cell><m>\vdots</m></cell>
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</row>
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<row>
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<cell>7</cell>
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<cell>6/36</cell>
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</row>
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<row>
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<cell>8</cell>
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<cell>5/36</cell>
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</row>
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<row>
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<cell><m>\vdots</m></cell>
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<cell></cell>
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</row>
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<row>
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<cell>12</cell>
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<cell>1/36</cell>
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</row>
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</tabular>
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</table>
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<p>
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Note that the distribution on <m>\Omega</m> is uniform, but the distribution on <m>S</m> is not.
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</p>
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</statement>
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</example>
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<example>
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<statement>
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<p>
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Let <m>\Omega</m> be a sample space and <m>A \subset \Omega</m> an event.
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Let
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<md>
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<mrow> X = \begin{cases} 1 \amp x \in A \\ 0 \amp x \notin A \end{cases} </mrow>
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</md>
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<m>X</m> is called an <term>indicator random variable</term>, and we say "<m>X</m> <term>indicates</term> <m>A</m>".
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</p>
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<p>
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The distribution on <m>X</m> is:
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<md>
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<mrow> \Pr(X = 1) \amp = \Pr(\{ x \mid x \in A\}) = \Pr(A) </mrow>
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<mrow> \Pr(X = 0) \amp = 1 - \Pr(A) </mrow>
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</md>
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</p>
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</statement>
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</example>
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<example>
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<statement>
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<p>
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Suppose we flip a coin <m>n</m> times.
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Let <m>H_i</m> indicate heads on flip <m>i</m>.
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Let <m>S</m> be the total number of heads in all flips.
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Then:
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<md>
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<mrow> S = H_1 + H_2 + \dotsb + H_{n} </mrow>
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</md>
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If <m>p</m> is the probability of the coin coming up heads on a flip, then <m>S</m> has the <term>binomial distribution</term> with parameters <m>n, p</m>.
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We'll use the notation <m>S \sim \Bin(n, p)</m> and:
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<md>
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<mrow> b(k) = b(k; n, p) = \Pr(S = k). </mrow>
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</md>
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</p>
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<p>
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Suppose the coin has <m>p = 0.3</m> and we flip it <m>n = 4</m> times.
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Find <m>b(2) = b(2; 4, 0.3)</m>.
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</p>
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<p>
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The relevant flip sequences are:
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<md>
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<mrow> THHT, HHTT, TTHH, THTH, HTHT, HTTH </mrow>
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</md>
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Each individual sequence has a probability of <m>(0.3)(0.3)(0.7)(0.7) = 0.0441</m>.
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So the total probability is:
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<md>
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<mrow> b(2; 4, 0.3) = (6)(0.0441) = 0.2646. </mrow>
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</md>
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That is, (number of flip sequences)(probability of each sequence).
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</p>
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</statement>
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</example>
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</subsection>
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</section> |