Definition 34.
A random variable is a function \(X \colon \Omega \to \R\text{.}\)
Section Tuesday, Jan 20
This is an outline of the topics we covered in class. These notes are not 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.
Subsection HW 1 Q5
Write \(F\) for the event that there’s a fire and \(S\) for the event that there’s visible smoke. Then the information we’re given can be interpreted as:
\begin{align*}
\Pr(F) \amp = 0.01 \\
\Pr(S) \amp = 0.1 \\
\Pr(S\mid F) \amp = 0.9
\end{align*}
In this case, we can use the simpler version of Bayes’ Theorem:
\begin{gather*}
\Pr(F \mid S) = \frac{\Pr(S \mid F)\Pr(F)}{\Pr(S)} = \dotsb
\end{gather*}
Unlike our usual diagnostic testing examples, we do have access to the denominator probability here.
Subsection Sec 2.1: Random Variables
The idea is that \(X\) is a variable representing a real number value which depends on the outcome of an experiment.
Example 35.
An experiment consists of planting 50 seeds in a garden, then growing them for 3 months. Let \(H_i\) be the height of plant \(i\text{.}\) Let \(D\) be the number of seeds that didn’t sprout. Let
\begin{align*}
A \amp = \text{avg height of all 50 plants} \\
\amp = \frac{H_1 + H_2 + \dotsb + H_{50}}{50}
\end{align*}
A random variable has its own probability distribution.
Example 36.
Roll a fair D6 twice. Let \(S\) be the sum of the rolls. Then \(\Omega = \{(1, 1), (1, 2), \dotsc, (6, 6)\}\) has 36 elements. Since the die is fair, the distribution on \(\Omega\) is uniform, i.e., \(\Pr(\omega) = \frac{1}{36}\) for any \(\omega \in \Omega\text{.}\)
| \(x\) | \(\Pr(S = x)\) |
| 2 | 1/36 |
| 3 | 2/36 |
| 4 | 3/36 |
| \(\vdots\) | \(\vdots\) |
| 7 | 6/36 |
| 8 | 5/36 |
| \(\vdots\) | |
| 12 | 1/36 |
Example 38.
Let \(\Omega\) be a sample space and \(A \subset \Omega\) an event. Let
\begin{align*}
X = \begin{cases} 1 \amp x \in A \\ 0 \amp x \notin A \end{cases}
\end{align*}
\(X\) is called an indicator random variable, and we say "\(X\) indicates \(A\)".
The distribution on \(X\) is:
\begin{align*}
\Pr(X = 1) \amp = \Pr(\{ x \mid x \in A\}) = \Pr(A) \\
\Pr(X = 0) \amp = 1 - \Pr(A)
\end{align*}
Example 39.
Suppose we flip a coin \(n\) times. Let \(H_i\) indicate heads on flip \(i\text{.}\) Let \(S\) be the total number of heads in all flips. Then:
\begin{gather*}
S = H_1 + H_2 + \dotsb + H_{n}
\end{gather*}
If \(p\) is the probability of the coin coming up heads on a flip, then \(S\) has the binomial distribution with parameters \(n, p\text{.}\) We’ll use the notation \(S \sim \Bin(n, p)\) and:
\begin{gather*}
b(k) = b(k; n, p) = \Pr(S = k).
\end{gather*}
Suppose the coin has \(p = 0.3\) and we flip it \(n = 4\) times. Find \(b(2) = b(2; 4, 0.3)\text{.}\)
The relevant flip sequences are:
\begin{gather*}
THHT, HHTT, TTHH, THTH, HTHT, HTTH
\end{gather*}
Each individual sequence has a probability of \((0.3)(0.3)(0.7)(0.7) = 0.0441\text{.}\) So the total probability is:
\begin{gather*}
b(2; 4, 0.3) = (6)(0.0441) = 0.2646.
\end{gather*}
That is, (number of flip sequences)(probability of each sequence).
