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2026-01-22 10:15:41 -05:00
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source/notes/1-15.ptx
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<?xml version="1.0" encoding="UTF-8"?>
<section xml:id="notes-01-15">
<title>Thursday 1/15</title>
<title>Thursday, Jan 15</title>
<introduction>
<p>
source/notes/1-20.ptx
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<?xml version="1.0" encoding="UTF-8"?>
<section xml:id="notes-01-20">
<title>Tuesday 1/20</title>
<title>Tuesday, Jan 20</title>
<introduction>
<p>
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</definition>
<p>
The idea is that <m>X</m> is a variable representing a real number value which depends on the outcome of an experiment.
The idea is that <m>X</m> is a variable representing a real number value which depends on the outcome of an experiment.
</p>
<example>
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<p>
An experiment consists of planting 50 seeds in a garden, then growing them for 3 months.
Let <m>H_i</m> be the height of plant <m>i</m>.
Let <m>D</m> be the number of seeds that didn't sprout.
Let
<md>
<mrow> A \amp = \text{avg height of all 50 plants} </mrow>
<mrow> \amp = \frac{H_1 + H_2 + \dotsb + H_{50}}{50} </mrow>
</md>
</p>
</statement>
</example>
<p>
A random variable has its own probability distribution.
</p>
<example>
<statement>
<p>
Roll a fair D6 twice.
Let <m>S</m> be the sum of the rolls.
Then <m>\Omega = \{(1, 1), (1, 2), \dotsc, (6, 6)\}</m> has 36 elements.
Since the die is fair, the distribution on <m>\Omega</m> is uniform, i.e., <m>\Pr(\omega) = \frac{1}{36}</m> for any <m>\omega \in \Omega</m>.
</p>
<p>
<m>S</m> takes on the values <m>2, 3, 4, \dotsc, 12</m>, with probabilities:
</p>
<table>
<title>Distribution for <m>S</m></title>
<tabular halign="center">
<row>
<cell><m>x</m></cell>
<cell><m>\Pr(S = x)</m></cell>
</row>
<row>
<cell>2</cell>
<cell>1/36</cell>
</row>
<row>
<cell>3</cell>
<cell>2/36</cell>
</row>
<row>
<cell>4</cell>
<cell>3/36</cell>
</row>
<row>
<cell><m>\vdots</m></cell>
<cell><m>\vdots</m></cell>
</row>
<row>
<cell>7</cell>
<cell>6/36</cell>
</row>
<row>
<cell>8</cell>
<cell>5/36</cell>
</row>
<row>
<cell><m>\vdots</m></cell>
<cell></cell>
</row>
<row>
<cell>12</cell>
<cell>1/36</cell>
</row>
</tabular>
</table>
<p>
Note that the distribution on <m>\Omega</m> is uniform, but the distribution on <m>S</m> is not.
</p>
</statement>
</example>
<example>
<statement>
<p>
Let <m>\Omega</m> be a sample space and <m>A \subset \Omega</m> an event.
Let
<md>
<mrow> X = \begin{cases} 1 \amp x \in A \\ 0 \amp x \notin A \end{cases} </mrow>
</md>
<m>X</m> is called an <term>indicator random variable</term>, and we say "<m>X</m> <term>indicates</term> <m>A</m>".
</p>
<p>
The distribution on <m>X</m> is:
<md>
<mrow> \Pr(X = 1) \amp = \Pr(\{ x \mid x \in A\}) = \Pr(A) </mrow>
<mrow> \Pr(X = 0) \amp = 1 - \Pr(A) </mrow>
</md>
</p>
</statement>
</example>
<example>
<statement>
<p>
Suppose we flip a coin <m>n</m> times.
Let <m>H_i</m> indicate heads on flip <m>i</m>.
Let <m>S</m> be the total number of heads in all flips.
Then:
<md>
<mrow> S = H_1 + H_2 + \dotsb + H_{n} </mrow>
</md>
If <m>p</m> is the probability of the coin coming up heads on a flip, then <m>S</m> has the <term>binomial distribution</term> with parameters <m>n, p</m>.
We'll use the notation <m>S \sim \Bin(n, p)</m> and:
<md>
<mrow> b(k) = b(k; n, p) = \Pr(S = k). </mrow>
</md>
</p>
<p>
Suppose the coin has <m>p = 0.3</m> and we flip it <m>n = 4</m> times.
Find <m>b(2) = b(2; 4, 0.3)</m>.
</p>
<p>
The relevant flip sequences are:
<md>
<mrow> THHT, HHTT, TTHH, THTH, HTHT, HTTH </mrow>
</md>
Each individual sequence has a probability of <m>(0.3)(0.3)(0.7)(0.7) = 0.0441</m>.
So the total probability is:
<md>
<mrow> b(2; 4, 0.3) = (6)(0.0441) = 0.2646. </mrow>
</md>
That is, (number of flip sequences)(probability of each sequence).
</p>
</statement>
</example>