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Finished-ish variance.

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andyeisenberg
2026-01-01 21:31:09 +00:00
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<?xml version="1.0" encoding="UTF-8"?>
<!-- TODO make sure to include EV/Var of all standard distributions, perhaps include a table? -->
<chapter xml:id="ch-Expected-Value" xmlns:xi="http://www.w3.org/2001/XInclude">
<title>Expected Value and Variance</title>
@@ -676,15 +678,47 @@
<p>
We'll omit the proofs and just casually observe that these are reasonable properties.
<m>X + \alpha</m> is a shift of <m>X</m>, and shifting shouldn't change how spread out the distribution is.
<m>\alpha X</m> scaled <m>X</m> by a constant, so it should scale the spread of the distribution by the same constant.
<m>\alpha X</m> scaled <m>X</m> by a constant, so it should scale the spread of the distribution by the same constant.
But recall that the variance calculation involves squaring; it's an indirect measure of spread.
So it's reasonable that scaling the distribution by a constant should scale the variance by its square.
</p>
<p>
Let's put these properties to use in a pair of related examples:
</p>
<example xml:id="example-indicator-variance">
<statement>
<p>
Let <m>X</m> indicate an event <m>A</m> which has probability <m>p</m>.
In <xref ref="example-indicator-EV"/>, we saw <m>\E(X) = p</m>.
Notice that <m>X</m> takes the values 0 and 1 (with probabilities <m>1 - p</m> and <m>p</m>, respectively).
But <m>0^2 = 0</m>, and <m>1^1 = 1</m>, so the expected value calculation for <m>X^2</m> is precisely the same as for <m>X</m>!
<md>
<mrow> \E\left(X^2\right) \amp = 0^2 \cdot (1-p) + 1^2 \cdot p = p. </mrow>
</md>
So:
<md>
<mrow> \Var(X) \amp \E\left(X^2\right) - \left(\E(X)\right)^2 = p - p^2 = p(1 - p). </mrow>
</md>
</p>
</statement>
</example>
<example xml:id="example-binomial-variance">
<statement>
<p>
TODO
Recall the setup from <xref ref="example-binomial-EV"/>: a binomial random variable <m>S</m> is the sum <m>H_1 + H_2 + \dotsb + H_n</m> of <m>n</m> indicator random variables, each with parameter <m>p</m>.
The results of different coin flips are independent from each other, so the random variables <m>H_1, \dotsc, H_n</m> are independent from each other.
From <xref ref="example-indicator-variance"/>, we know that <m>\Var(H_i) = p(1-p)</m> for each <m>i</m>.
Therefore:
<md>
<mrow> \Var(S) \amp = \Var(H_1 + H_2 + \dotsb + H_n) </mrow>
<mrow> \amp = \Var(H_1) + \Var(H_2) + \dotsb + \Var(H_n) </mrow>
<mrow> \amp = \underbrace{p(1-p) + p(1-p) + \dotsb + p(1-p)}_{n \text{ times}} </mrow>
<mrow> \amp = np(1-p). </mrow>
</md>
Remember this formula! We'll make frequent use of it.
</p>
</statement>
</example>