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Notes up through 2-10

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2026-02-12 07:09:15 -05:00
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</p>
</statement>
</definition>
<p>
We can write an alternative formula here:
<md>
<mrow> \Var(X) \amp = \E\left[ (X - \mu)^2 \right] </mrow>
<mrow> \amp = \E\left[ X^2 - 2\mu X + \mu^2 \right] </mrow>
<mrow> \amp = \E(X^2) - 2\mu \E(X) + \E(\mu^2) </mrow>
<mrow> \amp = \E(X^2) - 2\mu^2 + \mu^2 </mrow>
<mrow> \amp = \E(X^2) - \mu^2. </mrow>
</md>
This formula is generally more useful for performing computations.
</p>
<example>
<statement>
<p>
Let <m>X</m> indicate event <m>A</m> with <m>\Pr(A) = p</m>.
</p>
<sidebyside widths="30% 30%">
<table>
<title>Distribution for <m>X</m></title>
<tabular halign="center">
<row bottom="minor">
<cell><m>k</m></cell>
<cell><m>\Pr(X = k)</m></cell>
</row>
<row>
<cell><m>0</m></cell>
<cell><m>1 - p</m></cell>
</row>
<row>
<cell><m>1</m></cell>
<cell><m>p</m></cell>
</row>
</tabular>
</table>
<table>
<title>Distribution for <m>X^2</m></title>
<tabular halign="center">
<row bottom="minor">
<cell><m>k</m></cell>
<cell><m>\Pr(X^2 = k)</m></cell>
</row>
<row>
<cell><m>0^2</m></cell>
<cell><m>1 - p</m></cell>
</row>
<row>
<cell><m>1^2</m></cell>
<cell><m>p</m></cell>
</row>
</tabular>
</table>
</sidebyside>
<p>
Then <m>\E(X) = p</m> and <m>\E(X^2) = p</m>, so:
<md>
<mrow> \Var(X) = \E(X^2) - \left(\E(X)\right)^2 = p - p^2 = p(1 - p). </mrow>
</md>
</p>
</statement>
</example>
<example>
<statement>
<p>
Let <m>R</m> be the roll of a fair D6.
</p>
<sidebyside widths="30% 30%">
<table>
<title>Distribution for <m>R</m></title>
<tabular halign="center">
<row bottom="minor">
<cell><m>k</m></cell>
<cell><m>\Pr(R = k)</m></cell>
</row>
<row>
<cell><m>1</m></cell>
<cell><m>1/6</m></cell>
</row>
<row>
<cell><m>2</m></cell>
<cell><m>1/6</m></cell>
</row>
<row>
<cell><m>3</m></cell>
<cell><m>1/6</m></cell>
</row>
<row>
<cell><m>4</m></cell>
<cell><m>1/6</m></cell>
</row>
<row>
<cell><m>5</m></cell>
<cell><m>1/6</m></cell>
</row>
<row>
<cell><m>6</m></cell>
<cell><m>1/6</m></cell>
</row>
</tabular>
</table>
<table>
<title>Distribution for <m>R^2</m></title>
<tabular halign="center">
<row bottom="minor">
<cell><m>k</m></cell>
<cell><m>\Pr(R^2 = k)</m></cell>
</row>
<row>
<cell><m>1^2</m></cell>
<cell><m>1/6</m></cell>
</row>
<row>
<cell><m>2^2</m></cell>
<cell><m>1/6</m></cell>
</row>
<row>
<cell><m>3^2</m></cell>
<cell><m>1/6</m></cell>
</row>
<row>
<cell><m>4^2</m></cell>
<cell><m>1/6</m></cell>
</row>
<row>
<cell><m>5^2</m></cell>
<cell><m>1/6</m></cell>
</row>
<row>
<cell><m>6^2</m></cell>
<cell><m>1/6</m></cell>
</row>
</tabular>
</table>
</sidebyside>
<p>
Then <m>\E(R) = \frac{7}{2}</m> and <m>\E(R^2) = \frac{91}{6}</m>, so:
<md>
<mrow> \Var(R) = \E(R^2) - \left(\E(R)\right)^2 = \frac{91}{6} - \frac{49}{4} = \frac{35}{12}. </mrow>
</md>
</p>
</statement>
</example>
<example>
<statement>
<p>
Let <m>X \in [0, 1]</m> with pdf <m>f(x) = 2x</m>.
<md>
<mrow> \E(X) \amp = \int_0^1 x \cdot 2x\ dx </mrow>
<mrow> \amp = \int_0^1 2x^2\ dx </mrow>
<mrow> \amp = \frac{2x^3}{3}\bigg|_0^1 </mrow>
<mrow> \amp = \frac{2}{3} - 0 </mrow>
<mrow> \amp = \frac{2}{3} </mrow>
<mrow> \E(X^2) \amp = \int_0^1 x^2\cdot 2x\ dx </mrow>
<mrow> \amp = \int_0^1 2x^3\ dx </mrow>
<mrow> \amp = \frac{2x^4}{4}\bigg|_0^1 </mrow>
<mrow> \amp = \frac{1}{2} - 0 </mrow>
<mrow> \amp = \frac{1}{2} </mrow>
<mrow> \Rightarrow \quad \Var(X) \amp = \E(X^2) - \left(\E(X)\right)^2 </mrow>
<mrow> \amp = \frac{1}{2} - \frac{4}{9} </mrow>
<mrow> \amp = \frac{1}{18}. </mrow>
</md>
</p>
</statement>
</example>
<p>
WARNING!!! In general, variance is <em>not</em> linear! That is,
<md>
<mrow> \Var(X + Y) \amp \neq \Var(X) + \Var(Y) </mrow>
<mrow>\Var(kX) \amp \neq k \Var(X)</mrow>
</md>
But there are still relevant properties we can state here:
</p>
<theorem>
<statement>
<p>
Let <m>X</m> be a random variable and <m>k\in \R</m>.
Then:
<md>
<mrow> \Var(X + k) \amp = \Var(X) \amp \Var(kX) \amp = k^2\Var(X). </mrow>
</md>
Moreover, if <m>Y</m> is another random variable and <m>X, Y</m> are independent, then:
<md>
<mrow> \Var(X + Y) = \Var(X) + \Var(Y). </mrow>
</md>
</p>
</statement>
</theorem>
<example>
<statement>
<p>
Let <m>H_1, H_2, \dotsc, H_n</m> be indicators with parameter <m>p</m>.
Let <m>S = H_1 + \dotsb + H_n</m>, so <m>S \sim \Bin(n, p)</m>.
We know <m>\Var(H_i) = p(1-p)</m>, and <m>H_1, \dotsc, H_n</m> are independent.
So:
<md>
<mrow> \Var(S) \amp = \Var(H_1 + \dotsb + H_n) </mrow>
<mrow> \amp = \Var(H_1) + \dotsb + \Var(H_n) </mrow>
<mrow> \amp = p(1-p) + \dotsb + p(1-p) </mrow>
<mrow> \amp = np(1-p) </mrow>
</md>
</p>
</statement>
</example>
</subsection>
</section>