276 lines
8.0 KiB
XML
276 lines
8.0 KiB
XML
<?xml version="1.0" encoding="UTF-8"?>
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<section xml:id="notes-02-03">
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<title>Tuesday, Feb 3</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>
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<title>Variance</title>
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<p>
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Question: How spread out are <m>X</m> values? One answer we might try is to measure the average distance from the average value:
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<md>
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<mrow> \E\left[|X - \E(X)|\right] </mrow>
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</md>
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To simplify notation, we'll write <m>\mu = \E(X)</m>.
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Also, it's often usefull to square a term rather than take absolute value when we want to ensure a positive output, so we'll define...
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</p>
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<definition xml:id="def-variance">
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<statement>
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<p>
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Let <m>X</m> be a random variable with <m>\E(X) = \mu</m>.
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The <term>variance</term> of <m>X</m> is:
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<md>
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<mrow> \sigma^2 = \Var(x) = \E\left[ (X - \mu)^2\right] </mrow>
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</md>
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<m>\sigma = \sqrt{\Var(X)}</m> is called the <term>standard deviation</term>.
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</p>
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</statement>
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</definition>
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<p>
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We can write an alternative formula here:
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<md>
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<mrow> \Var(X) \amp = \E\left[ (X - \mu)^2 \right] </mrow>
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<mrow> \amp = \E\left[ X^2 - 2\mu X + \mu^2 \right] </mrow>
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<mrow> \amp = \E(X^2) - 2\mu \E(X) + \E(\mu^2) </mrow>
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<mrow> \amp = \E(X^2) - 2\mu^2 + \mu^2 </mrow>
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<mrow> \amp = \E(X^2) - \mu^2. </mrow>
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</md>
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This formula is generally more useful for performing computations.
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</p>
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<example>
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<statement>
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<p>
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Let <m>X</m> indicate event <m>A</m> with <m>\Pr(A) = p</m>.
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</p>
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<sidebyside widths="30% 30%">
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<table>
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<title>Distribution for <m>X</m></title>
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<tabular halign="center">
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<row bottom="minor">
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<cell><m>k</m></cell>
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<cell><m>\Pr(X = k)</m></cell>
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</row>
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<row>
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<cell><m>0</m></cell>
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<cell><m>1 - p</m></cell>
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</row>
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<row>
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<cell><m>1</m></cell>
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<cell><m>p</m></cell>
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</row>
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</tabular>
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</table>
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<table>
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<title>Distribution for <m>X^2</m></title>
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<tabular halign="center">
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<row bottom="minor">
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<cell><m>k</m></cell>
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<cell><m>\Pr(X^2 = k)</m></cell>
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</row>
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<row>
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<cell><m>0^2</m></cell>
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<cell><m>1 - p</m></cell>
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</row>
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<row>
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<cell><m>1^2</m></cell>
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<cell><m>p</m></cell>
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</row>
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</tabular>
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</table>
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</sidebyside>
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<p>
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Then <m>\E(X) = p</m> and <m>\E(X^2) = p</m>, so:
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<md>
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<mrow> \Var(X) = \E(X^2) - \left(\E(X)\right)^2 = p - p^2 = p(1 - p). </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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Let <m>R</m> be the roll of a fair D6.
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</p>
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<sidebyside widths="30% 30%">
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<table>
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<title>Distribution for <m>R</m></title>
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<tabular halign="center">
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<row bottom="minor">
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<cell><m>k</m></cell>
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<cell><m>\Pr(R = k)</m></cell>
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</row>
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<row>
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<cell><m>1</m></cell>
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<cell><m>1/6</m></cell>
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</row>
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<row>
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<cell><m>2</m></cell>
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<cell><m>1/6</m></cell>
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</row>
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<row>
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<cell><m>3</m></cell>
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<cell><m>1/6</m></cell>
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</row>
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<row>
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<cell><m>4</m></cell>
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<cell><m>1/6</m></cell>
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</row>
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<row>
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<cell><m>5</m></cell>
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<cell><m>1/6</m></cell>
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</row>
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<row>
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<cell><m>6</m></cell>
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<cell><m>1/6</m></cell>
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</row>
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</tabular>
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</table>
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<table>
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<title>Distribution for <m>R^2</m></title>
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<tabular halign="center">
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<row bottom="minor">
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<cell><m>k</m></cell>
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<cell><m>\Pr(R^2 = k)</m></cell>
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</row>
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<row>
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<cell><m>1^2</m></cell>
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<cell><m>1/6</m></cell>
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</row>
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<row>
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<cell><m>2^2</m></cell>
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<cell><m>1/6</m></cell>
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</row>
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<row>
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<cell><m>3^2</m></cell>
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<cell><m>1/6</m></cell>
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</row>
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<row>
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<cell><m>4^2</m></cell>
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<cell><m>1/6</m></cell>
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</row>
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<row>
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<cell><m>5^2</m></cell>
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<cell><m>1/6</m></cell>
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</row>
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<row>
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<cell><m>6^2</m></cell>
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<cell><m>1/6</m></cell>
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</row>
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</tabular>
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</table>
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</sidebyside>
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<p>
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Then <m>\E(R) = \frac{7}{2}</m> and <m>\E(R^2) = \frac{91}{6}</m>, so:
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<md>
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<mrow> \Var(R) = \E(R^2) - \left(\E(R)\right)^2 = \frac{91}{6} - \frac{49}{4} = \frac{35}{12}. </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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Let <m>X \in [0, 1]</m> with pdf <m>f(x) = 2x</m>.
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<md>
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<mrow> \E(X) \amp = \int_0^1 x \cdot 2x\ dx </mrow>
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<mrow> \amp = \int_0^1 2x^2\ dx </mrow>
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<mrow> \amp = \frac{2x^3}{3}\bigg|_0^1 </mrow>
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<mrow> \amp = \frac{2}{3} - 0 </mrow>
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<mrow> \amp = \frac{2}{3} </mrow>
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<mrow> \E(X^2) \amp = \int_0^1 x^2\cdot 2x\ dx </mrow>
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<mrow> \amp = \int_0^1 2x^3\ dx </mrow>
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<mrow> \amp = \frac{2x^4}{4}\bigg|_0^1 </mrow>
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<mrow> \amp = \frac{1}{2} - 0 </mrow>
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<mrow> \amp = \frac{1}{2} </mrow>
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<mrow> \Rightarrow \quad \Var(X) \amp = \E(X^2) - \left(\E(X)\right)^2 </mrow>
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<mrow> \amp = \frac{1}{2} - \frac{4}{9} </mrow>
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<mrow> \amp = \frac{1}{18}. </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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WARNING!!! In general, variance is <em>not</em> linear! That is,
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<md>
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<mrow> \Var(X + Y) \amp \neq \Var(X) + \Var(Y) </mrow>
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<mrow>\Var(kX) \amp \neq k \Var(X)</mrow>
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</md>
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But there are still relevant properties we can state here:
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</p>
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<theorem>
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<statement>
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<p>
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Let <m>X</m> be a random variable and <m>k\in \R</m>.
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Then:
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<md>
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<mrow> \Var(X + k) \amp = \Var(X) \amp \Var(kX) \amp = k^2\Var(X). </mrow>
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</md>
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Moreover, if <m>Y</m> is another random variable and <m>X, Y</m> are independent, then:
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<md>
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<mrow> \Var(X + Y) = \Var(X) + \Var(Y). </mrow>
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</md>
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</p>
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</statement>
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</theorem>
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<example>
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<statement>
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<p>
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Let <m>H_1, H_2, \dotsc, H_n</m> be indicators with parameter <m>p</m>.
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Let <m>S = H_1 + \dotsb + H_n</m>, so <m>S \sim \Bin(n, p)</m>.
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We know <m>\Var(H_i) = p(1-p)</m>, and <m>H_1, \dotsc, H_n</m> are independent.
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So:
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<md>
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<mrow> \Var(S) \amp = \Var(H_1 + \dotsb + H_n) </mrow>
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<mrow> \amp = \Var(H_1) + \dotsb + \Var(H_n) </mrow>
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<mrow> \amp = p(1-p) + \dotsb + p(1-p) </mrow>
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<mrow> \amp = np(1-p) </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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</subsection>
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</section> |