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Notes up through 2-10

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<?xml version="1.0" encoding="UTF-8"?>
<section xml:id="notes-02-05">
<title>Thursday, Feb 5</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>Summary</title>
<p>
Here are the expected value and variance formulas for common distributions.
Some of these, we've shown justification for.
Others requires techniques beyond the scope of the class to justify.
</p>
<table>
<title>Expected Value and Variance Formulas</title>
<tabular halign="center">
<row bottom="minor">
<cell>Distribution</cell>
<cell>Parameters</cell>
<cell>Expected Value</cell>
<cell>Variance</cell>
</row>
<row>
<cell>Indicator</cell>
<cell><m>p</m></cell>
<cell><m>p</m></cell>
<cell><m>p(1-p)</m></cell>
</row>
<row>
<cell>Binomial</cell>
<cell><m>n, p</m></cell>
<cell><m>np</m></cell>
<cell><m>np(1-p)</m></cell>
</row>
<row>
<cell>Geometric</cell>
<cell><m>p</m></cell>
<cell><m>\frac{1}{p}</m></cell>
<cell><m>\frac{1-p}{p^2}</m></cell>
</row>
<row>
<cell>Poisson</cell>
<cell><m>\lambda</m></cell>
<cell><m>\lambda</m></cell>
<cell><m>\lambda</m></cell>
</row>
<row>
<cell>Exponential</cell>
<cell><m>\lambda</m></cell>
<cell><m>\frac{1}{\lambda}</m></cell>
<cell><m>\frac{1}{\lambda^2}</m></cell>
</row>
</tabular>
</table>
</subsection>
<subsection>
<title>Covariance</title>
<definition>
<statement>
<p>
If <m>X, Y</m> are random variables with expected values of <m>\mu_X, \mu_Y</m>, then the <term>covariance</term> of <m>X</m> and <m>Y</m> is:
<md>
<mrow> \Cov(X, Y) \amp = \E\left[ (X - \mu_X)(Y - \mu_Y)\right]. </mrow>
</md>
</p>
</statement>
</definition>
<p>
Observe that:
<md>
<mrow> \Cov(X, X) \amp = \E\left[(X - \mu_X)(X - \mu_X)\right] </mrow>
<mrow> \amp = \E\left[(X - \mu_X)^2\right] </mrow>
<mrow> \amp = \Var(X), </mrow>
</md>
so covariance generalizes the variance formula to two variables.
As with variance, there's an alternative formula more suited to doing computations:
<md>
<mrow> \Var(X) \amp = \E\left[(X - \mu_X)(X - \mu_X)\right] = \E(X^2) - \mu_X^2 </mrow>
<mrow> \Cov(X, Y) \amp = \E\left[(X - \mu_X)(Y - \mu_Y)\right] = \E(XY) - \mu_X\mu_Y </mrow>
</md>
</p>
<example>
<statement>
<p>
Consider the joint distribution table:
</p>
<table>
<title>Joint Distribution</title>
<tabular halign="center">
<row bottom="minor">
<cell right="minor"></cell>
<cell><m>X = 0</m></cell>
<cell><m>X = 1</m></cell>
</row>
<row>
<cell right="minor"><m>Y = 0</m></cell>
<cell>0.2</cell>
<cell>0.1</cell>
</row>
<row>
<cell right="minor"><m>Y = 1</m></cell>
<cell>0.05</cell>
<cell>0.65</cell>
</row>
</tabular>
</table>
<p>
From the table, we can calculate the marginal distributions:
<md>
<mrow> \Pr(X = 0) \amp = 0.25 \amp \Pr(Y = 0) \amp = 0.3 </mrow>
<mrow> \Pr(X = 1) \amp = 0.75 \amp \Pr(Y = 1) \amp = 0.7 </mrow>
</md>
So <m>\E(X) = \mu_X = 0.75</m> and <m>\E(Y) = \mu_Y = 0.7</m>.
Then:
<md>
<mrow> \E(XY) \amp = (0)(0)(0.2) + (1)(0)(0.1) + (0)(1)(0.05) + (1)(1)(0.65) </mrow>
<mrow> \amp = 0.65 </mrow>
<mrow> \Cov(X, Y) \amp = \E(XY) - \mu_X \mu_Y </mrow>
<mrow> \amp = 0.65 - (0.75)(0.7) </mrow>
<mrow> \amp = 0.125. </mrow>
</md>
</p>
</statement>
</example>
<p>
Question: the formula <m>\Var(X) = \E\left[(X - \mu_X)^2\right]</m> makes it clear that variance cannot be negative, since squares are nonnegative.
What about <m>\Cov(X, Y)</m>?
</p>
<example>
<statement>
<p>
In the previous example, since <m>X, Y</m> were both indicator random variables, the variances for each were simply equal to the sum of the second row/column.
Similarly, <m>\E(XY)</m> was equal to the <m>X = 1, Y = 1</m> entry in the table.
Using two indicator random variables, significantly simplifies the covariance calculation, so we can vary the table and recalculate covariance quickly.
Consider the following joint distribution:
</p>
<table>
<title>Joint Distribution</title>
<tabular halign="center">
<row bottom="minor">
<cell right="minor"></cell>
<cell><m>X = 0</m></cell>
<cell><m>X = 1</m></cell>
</row>
<row>
<cell right="minor"><m>Y = 0</m></cell>
<cell>0.1</cell>
<cell>0.4</cell>
</row>
<row>
<cell right="minor"><m>Y = 1</m></cell>
<cell>0.3</cell>
<cell>0.2</cell>
</row>
</tabular>
</table>
<p>
Then:
<md>
<mrow> \Cov(X, Y) = 0.2 - (0.6)(0.5) = -0.1 \lt 0. </mrow>
</md>
</p>
</statement>
</example>
<p>
Question: how do we interpret <m>\Cov(X, Y)</m>? <m>X - \mu_X</m> is positive when <m>X \gt \mu_X</m> and negative when <m>X \lt \mu_X</m>.
<m>Y - \mu_Y</m> is positive when <m>Y \gt \mu_Y</m> and negative when <m>Y \lt \mu_Y</m>.
So the product <m>(X - \mu_X)(Y - \mu_Y)</m> is positive when <m>X, Y</m> are both larger or both smaller than their expected values, and negative when one is larger and one is smaller.
That is, covariance tries to quantify the tendency of <m>X, Y</m> to get big/small at the same time.
</p>
</subsection>
</section>