Lecture 13: conditional probability

Reading: Cameron 1.10, chapter 2
Last semester's notes
Definitions: independent events, conditional probability
Modeling with conditional probability
Statement of Bayes's rule

Independence

Definition: Two events $A$ and $B$ are independent if $Pr(A \cap B) = Pr(A)Pr(B)$ .

Important: Do not assume events are independent unless given a good reason to do so.

Example: Suppose I roll two 6-sided dice. For either die, the probability of getting any number from 1...6 is 1/6. What is the probability of getting a pair of twos?

Answer: it could be anything from 0 to 1/6. The dice could be taped together in a way that it is impossible to get (2,2) while still making the probability of any given roll for either die to be 1/6.

If we are also told that the two rolls are independent, then we can conclude that $\begin{aligned} Pr(\{(2,2)\}) &= Pr(\{(2,n)\mid n \in \{1,\dots,6\}\} \cap \{(n,2) \mid n \in \{1,\dots,6\}\}) \\ &= Pr(\{(2,n)\mid n \in \{1,\dots,6\}\})\cdot Pr(\{(n,2) \mid n \in \{1,\dots,6\}\}) \\ &= (1/6)(1/6) \\ \end{aligned}$

Conditional probability

Definition: If $A$ and $B$ are events, then the probability of A given B, written $Pr(A|B)$ is given by $Pr(A|B) ::= \frac{Pr(A \cap B)}{Pr(B)}$ Note that $Pr(A|B)$ is only defined if $Pr(B) \neq 0$ .

Intuitively, $Pr(A|B)$ is the probability of $A$ in a new sample space created by restricting our attention to the subset of the sample space where $B$ occurs. We divide by $Pr(B)$ so that $Pr(B|B) = 1$ .

Note: $A|B$ is not defined, only $Pr(A|B)$ ; this is an abuse of notation, but is standard.

Alternative definition of independence

Conditional probability can be used to give an equivalent definition of independence:

Claim: $A$ and $B$ are independent if and only if $Pr(A|B) = Pr(A)$ .

This is perhaps a more intuitive notion of independence: if I tell you that $B$ happened, it doesn't change your estimate of the probability that $A$ happens.

Proof: (⇒) Suppose $A$ and $B$ are independent. We wish to show $Pr(A|B) = Pr(A)$ . Well, by definition, $Pr(A|B) = Pr(A \cap B)/Pr(B)$ . Since $A$ and $B$ are independent, $Pr(A \cap B) = Pr(A)Pr(B)$ . Plugging this in, we see that $Pr(A|B) = Pr(A)Pr(B)/Pr(B) = Pr(A)$ .

(⇐) Suppose $Pr(A|B) = Pr(A)$ . Then $Pr(A) = Pr(A \cap B)/Pr(B)$ . Multiplying both sides by $Pr(B)$ gives the desired result.

Bayes's Rule

Bayes's rule is a simple way to compute $P(A|B)$ from $P(B|A)$ .

Claim: (Bayes's rule): $P(A|B) = P(B|A)P(A)/P(B)$ .

Proof: left as exercise.

Example: See next lecture

Probability trees

Using conditional probability, we can draw a tree to help discover the probabilities of various events. Each branch of the tree partitions part of the sample space into smaller parts.

For example: suppose that it rains with probability 30%. Suppose that when it rains, I bring my umbrella 3/4 of the time, while if it is not raining, I bring my umbrella with probability 1/10. Given that I bring my umbrella, what is the probability that it is raining?

One way to model this problem is with the sample space

$S = \{raining (r), not raining (nr)\} \times \{umbrella (u), no umbrella (nu)\} = \{(r,u), (nr,u), (r,nu), (nr,nu)\}$

Let $R$ be the event "it is raining". Then $R = \{(r,u), (r,nu)\}$ . Let $U$ be the event "I bring my umbrella". Then $U = \{(r,u), (nr,u)\}$ .

The problem tells us that $Pr(R) = 3/10$ . It also states that $Pr(U|R) = 3/4$ while $Pr(U|\bar{R}) = 1/10$ . We can use the following fact:

Fact: $Pr(\bar{A}|B) = 1 - Pr(A|B)$ . Proof left as exercise.

to conclude that $Pr(\bar{U}|R) = 1/4$ and $Pr(\bar{U}|\bar{R}) = 9/10$ .

We can draw a tree:

Probability tree (LaTeX source)

We can compute the probabilities of the events at the leaves by multiplying along the paths. For example, $Pr(\{(r,u)\}) = Pr(U \cap R) = Pr(R)Pr(U | R) = (3/10)(3/4) = (9/40)$

To answer our question, we are interested in $Pr(R|U) = Pr(U \cap R)/Pr(U) = (9/40)/(3/10) = 3/4$ .

Note we could also answer this using Bayes's rule and the law of total probability; it would amount to exactly the same calculation. The tree just helps organize all of the variables.