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.

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: \Pr(F) \amp = 0.01 \Pr(S) \amp = 0.1 \Pr(S\mid F) \amp = 0.9 In this case, we can use the simpler version of Bayes' Theorem: \Pr(F \mid S) = \frac{\Pr(S \mid F)\Pr(F)}{\Pr(S)} = \dotsb Unlike our usual diagnostic testing examples, we do have access to the denominator probability here.

Sec 2.1: Random Variables

A random variable is a function X \colon \Omega \to \R.

The idea is that X is a variable representing a real number value which depends on the outcome of an experiment.

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. Let D be the number of seeds that didn't sprout. Let A \amp = \text{avg height of all 50 plants} \amp = \frac{H_1 + H_2 + \dotsb + H_{50}}{50}

A random variable has its own probability distribution.

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.

S takes on the values 2, 3, 4, \dotsc, 12, with probabilities:

Distribution for <m>S</m> 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

Note that the distribution on \Omega is uniform, but the distribution on S is not.

Let \Omega be a sample space and A \subset \Omega an event. Let X = \begin{cases} 1 \amp x \in A \\ 0 \amp x \notin A \end{cases} X is called an indicator random variable, and we say "X indicates A".

The distribution on X is: \Pr(X = 1) \amp = \Pr(\{ x \mid x \in A\}) = \Pr(A) \Pr(X = 0) \amp = 1 - \Pr(A)

Suppose we flip a coin n times. Let H_i indicate heads on flip i. Let S be the total number of heads in all flips. Then: S = H_1 + H_2 + \dotsb + H_{n} If p is the probability of the coin coming up heads on a flip, then S has the binomial distribution with parameters n, p. We'll use the notation S \sim \Bin(n, p) and: b(k) = b(k; n, p) = \Pr(S = k).

Suppose the coin has p = 0.3 and we flip it n = 4 times. Find b(2) = b(2; 4, 0.3).

The relevant flip sequences are: THHT, HHTT, TTHH, THTH, HTHT, HTTH Each individual sequence has a probability of (0.3)(0.3)(0.7)(0.7) = 0.0441. So the total probability is: b(2; 4, 0.3) = (6)(0.0441) = 0.2646. That is, (number of flip sequences)(probability of each sequence).