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<?xml version="1.0" encoding="UTF-8"?>
<section xml:id="notes-01-29">
<title>Thursday, Jan 29</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>Expected Value</title>
<example>
<statement>
<p>
Let <m>R</m> be the roll of a fair D6.
What is the average value of <m>R</m>? The sample space is <m>\Omega = \{1, 2, 3, 4, 5, 6\}</m>.
To find the average:
<md>
<mrow> \text{Avg} \amp = 1\left(\frac{1}{6}\right) + 2\left(\frac{1}{6}\right) + 3\left(\frac{1}{6}\right) + 4\left(\frac{1}{6}\right) + 5\left(\frac{1}{6}\right) + 6\left(\frac{1}{6}\right) </mrow>
<mrow> \amp = \frac{21}{6} = \frac{7}{2} = 3.5 </mrow>
</md>
That is: take each value of <m>R</m>, multiply by the probability, and add all the results together.
</p>
</statement>
</example>
<example>
<statement>
<p>
Given a distribution table:
</p>
<table>
<title>Distribution</title>
<tabular>
<row bottom="minor">
<cell><m>s</m></cell>
<cell><m>\Pr(S = s)</m></cell>
</row>
<row>
<cell>1</cell>
<cell>0.1</cell>
</row>
<row>
<cell>2</cell>
<cell>0.1</cell>
</row>
<row>
<cell>3</cell>
<cell>0.1</cell>
</row>
<row>
<cell>4</cell>
<cell>0.1</cell>
</row>
<row>
<cell>5</cell>
<cell>0.1</cell>
</row>
<row>
<cell>6</cell>
<cell>0.5</cell>
</row>
</tabular>
</table>
<p>
Then the average is:
<md>
<mrow> \text{Avg} = 1(0.1) + 2(0.1) + 3(0.1) + 4(0.1) + 5(0.1) + 6(0.5) = 4.5. </mrow>
</md>
</p>
</statement>
</example>
<definition xml:id="def-discrete-EV">
<statement>
<p>
Let <m>X</m> be a discrete random variable taking values <m>x_1, x_2, \dotsc, x_n</m> with probabilities <m>p_1, p_2, \dotsc, p_n</m>.
The <term>expected value</term> of <m>X</m> is:
<md>
<mrow> \E(X) = \sum_{i=1}^n x_ip_i = x_1p_1 + x_2p_2 + \dotsb + x_np_n. </mrow>
</md>
</p>
</statement>
</definition>
<example>
<statement>
<p>
Let <m>X</m> indicate <m>A</m>, with <m>\Pr(A) = p</m>.
Then:
<md>
<mrow> E(X) = 0 \cdot \underbrace{\Pr(X = 0)}_{1 - p} + 1 \cdot \underbrace{\Pr(X = 1)}_{p} = 0(1-p) + 1p = p. </mrow>
</md>
That is: "E(indicator random variable) = Pr(event that it indicates)".
</p>
</statement>
</example>
<example>
<statement>
<p>
Suppose we flip a fair coin 100 times.
Let <m>H_1, H_2, \dotsc, H_{100}</m> indicate heads on each flip.
Then <m>\Pr(H_i = 1) = \frac{1}{2}</m>, so <m>E(H_i) = \frac{1}{2}</m> for every <m>i</m>.
</p>
<p>
What if <m>X</m> indicates a run of 4 heads starting at flip 3? That is, flips 3, 4, 5, and 6 must come up heads, and all other flips can come up either heads or tails.
Since these flip results are independent, the probabilities multiply:
<md>
<mrow> \Pr(X = 1) = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{1}{16} = \E(X). </mrow>
</md>
</p>
</statement>
</example>
<definition xml:id="def-continuous-EV">
<statement>
<p>
Let <m>X \in [a, b]</m> be a continuous random variable with pdf <m>f(x)</m>.
The <term>expected value</term> of <m>X</m> is:
<md>
<mrow> \E(X) = \int_a^b x f(x)\ dx. </mrow>
</md>
</p>
</statement>
</definition>
<p>
It's worth putting this side-by-side with <xref ref="def-discrete-EV"/> to compare the structure of each formula.
These both say: "multiply each value of the random variable by the probability, then accumulate all of those products".
Expected value is a weighted average of random variable values, with the probabilities as the weights.
</p>
<example>
<statement>
<p>
Let <m>X \in [0, 1]</m> with pdf <m>f(x) = \frac{3}{2}\sqrt{x}</m>.
Then:
<md>
<mrow> \E(X) \amp = \int_0^1 x f(x)\ dx </mrow>
<mrow> \amp = \int_0^1 x \frac{3}{2} x^{1/2}\ dx </mrow>
<mrow> \amp = \int_0^1 \frac{3}{2} x^{3/2}\ dx </mrow>
<mrow> \amp = \frac{3}{2} \frac{2x^{5/2}}{5}\bigg|_0^1 </mrow>
<mrow> \amp = \frac{3}{5} \cdot 1^{5/2} - \frac{3}{5} \cdot 0^{5/2} </mrow>
<mrow> \amp = \frac{3}{5}. </mrow>
</md>
What would happen if you forgot the <m>x</m>? Then the calculation would become:
<md>
<mrow> \int_0^1 f(x)\ dx = 1 = \text{total probability} \amp </mrow>
</md>
This mistake will often be easy to catch.
For example, this random variable takes values between 0 and 1, so it doesn't seem very likely that the average value is 1!
</p>
</statement>
</example>
<example>
<statement>
<p>
Find <m>\E(X^2)</m> given a distribution for <m>X</m>.
</p>
<table>
<title>Distribution for <m>X</m></title>
<tabular>
<row bottom="minor">
<cell><m>x</m></cell>
<cell><m>\Pr(X = x)</m></cell>
</row>
<row>
<cell>-1</cell>
<cell>0.4</cell>
</row>
<row>
<cell>1</cell>
<cell>0.1</cell>
</row>
<row>
<cell>2</cell>
<cell>0.2</cell>
</row>
<row>
<cell>3</cell>
<cell>0.3</cell>
</row>
</tabular>
</table>
<p>
We can start by writing a distribution table for <m>X^2</m>:
</p>
<table>
<title>Distribution for <m>X^2</m></title>
<tabular>
<row bottom="minor">
<cell><m>x</m></cell>
<cell><m>\Pr(X^2 = x)</m></cell>
</row>
<row>
<cell>1</cell>
<cell>0.5</cell>
</row>
<row>
<cell>4</cell>
<cell>0.2</cell>
</row>
<row>
<cell>9</cell>
<cell>0.3</cell>
</row>
</tabular>
</table>
<p>
Then:
<md>
<mrow> \E(X^2) 1(0.5) + 4(0.2) + 9(0.3) = 4. </mrow>
</md>
</p>
</statement>
</example>
<theorem>
<statement>
<p>
If <m>X</m> takes on values <m>x_1, \dotsc, x_n</m> with probabilities <m>p_1, \dotsc, p_n</m>, and <m>h</m> is any function, then:
<md>
<mrow> \E(h(X)) = \sum_{i=1}^n h(x_i)p_i. </mrow>
</md>
</p>
</statement>
</theorem>
<example>
<statement>
<p>
Revisiting the previous example, the theorem says we don't need to first create the distribution table for <m>X^2</m>.
We can use the distribution table for <m>X</m>, and just apply the square to each value:
<md>
<mrow> \E(X^2) = (-1)^2(0.4) + (1)^2(0.1) + (2)^2(0.2) + (3)^2(0.3) = 4. </mrow>
</md>
</p>
</statement>
</example>
<example>
<statement>
<p>
The situation with continuous random variables is similar.
Let <m>X\in [0, 1]</m> with pdf <m>f(x) = \frac{3}{2}\sqrt{x}</m>.
Then:
<md>
<mrow> \E(X^2) \amp = \int_0^1 x^2 f(x)\ dx </mrow>
<mrow> \amp = \int_0^1 x^2 \frac{3}{2} x^{1/2}\ dx </mrow>
<mrow> \amp = \int_0^1 \frac{3}{2} x^{5/2}\ dx </mrow>
<mrow> \amp = \frac{3}{2}\cdot \frac{2x^{7/2}}{7}\bigg|_0^1 </mrow>
<mrow> \amp = \frac{3}{7}\cdot 1^{7/2} - 0\cdot 0^{7/2} </mrow>
<mrow> \amp = \frac{3}{7}. </mrow>
</md>
Notice, in particular, that <m>\E(X^2) \neq \E(X).</m>
</p>
</statement>
</example>
<theorem xml:id="thm-linearity">
<statement>
<p>
If <m>X, Y</m> are random variables with finite expected value and <m>k \in \R</m>, then:
<md>
<mrow> \E(X+Y) \amp = E(X) + E(Y) </mrow>
<mrow> \E(kX) \amp = kE(X) </mrow>
</md>
</p>
</statement>
</theorem>
<example>
<statement>
<p>
Let <m>N</m> be the number of heads in <m>n</m> coin flips with bias <m>p</m>.
Then <m>N\sim \Bin(n, p)</m>.
<md>
<mrow> \E)N) = \sum_{i=0}^n i\cdot b(i; n, p) = \sum_{i = 0}^n i \cdot {n \choose i} p^i (1-p)^{n - i}. </mrow>
</md>
Gross.
</p>
<p>
Instead of calculating this directly, define <m>H_1, H_2, \dotsc, H_n</m> to indicate heads on each flip.
Then:
<md>
<mrow> N \amp = H_1 + H_2 + \dotsb + H_n </mrow>
<mrow> \E(N) \amp = \E(H_1 + H_2 + \dotsb + H_n) </mrow>
<mrow> \E(N) \amp = \E(H_1) + \E(H_2) + \dotsb + \E(H_n) </mrow>
<mrow> \amp = p + p + \dotsb + p </mrow>
<mrow> \amp = np. </mrow>
</md>
</p>
</statement>
</example>
<example>
<statement>
<p>
Flip a fair coin 100 times.
Then, the expected number of heads is:
<md>
<mrow> \underbrace{(100)}_{n}\underbrace{(0.5)}_{p} = 50. </mrow>
</md>
</p>
</statement>
</example>
<example>
<statement>
<p>
Flip a fair coin 100 times.
What is the expected number of runs of 4 heads? As in the binomial EV calculation, define indicator random variables <m>R_1, R_2, \dotsc, R_{97}</m> each indicating a run of 4 heads starting at the specified flip.
Then <m>\E(R_i) = \frac{1}{16}</m> for each <m>i</m>.
Let <m>T</m> be the number of runs of 4 heads.
Then:
<md>
<mrow> \E(T) \amp = \E(R_1 + \dotsb + R_{97}) </mrow>
<mrow> \amp = \E(R_1) + \dotsb + (R_{97}) </mrow>
<mrow> \amp = \frac{1}{16} + \dotsb + \frac{1}{16} </mrow>
<mrow> \amp = \frac{97}{16} \approx 6. </mrow>
</md>
</p>
</statement>
</example>
<example>
<statement>
<p>
Consider a geometric distribution <m>T \sim \Geom(p)</m>.
</p>
<table>
<title>Geometric Distribution</title>
<tabular>
<row bottom="minor">
<cell><m>k</m></cell>
<cell><m>\Pr(T = k)</m></cell>
<cell>with <m>p = \frac{1}{2}</m></cell>
</row>
<row>
<cell>1</cell>
<cell><m>p</m></cell>
<cell><m>\frac{1}{2}</m></cell>
</row>
<row>
<cell>2</cell>
<cell><m>(1-p)p</m></cell>
<cell><m>\frac{1}{4}</m></cell>
</row>
<row>
<cell>3</cell>
<cell><m>(1-p)^2p</m></cell>
<cell><m>\frac{1}{8}</m></cell>
</row>
<row>
<cell>4</cell>
<cell><m>(1-p)^3p</m></cell>
<cell><m>\frac{1}{16}</m></cell>
</row>
<row>
<cell><m>\vdots</m></cell>
<cell><m>\vdots</m></cell>
<cell><m>\vdots</m></cell>
</row>
</tabular>
</table>
<p>
Then <m>\E(T) = (1)\left(\frac{1}{2}\right) + (2)\left(\frac{1}{4}\right) + (3)\left(\frac{1}{8}\right) + \dotsb</m> is an infinite summation.
</p>
<p>
Instead, consider the following argument.
If we flip a coin until we see heads, we either see heads on flip 1 or not.
In the first case, <m>T = 1</m>.
In the second case, what is the average value of <m>T</m>? Starting at flip 2, it will take on average <m>\E(T)</m> flips to see heads.
Since we already flipped the coin once, the total number of flips will be <m>1 + \E(T)</m>.
So, we can write:
<md>
<mrow> \E(T) \amp = 1 \left(\frac{1}{2}\right) + (1 + \E(T)) \left(1 - \frac{1}{2}\right) </mrow>
<mrow> \E(T) \amp = \frac{1}{2} + \frac{1}{2} + \frac{\E(T)}{2} </mrow>
<mrow> \E(T) - \frac{\E(T)}{2} \amp = 1 </mrow>
<mrow> \frac{\E(T)}{2} \amp = 1 </mrow>
<mrow> \E(T) \amp = 2. </mrow>
</md>
</p>
</statement>
</example>
</subsection>
</section>