Summary

Builds on inclusion-exclusion to define disjoint events and partitions. Introduces conditional probability, derives the law of total probability, Bayes’ rule, and the base rate fallacy. Defines independence and pairwise independence.


Example — The Secretary Problem

𝑛 letters, 𝑛 envelopes. The secretary randomly puts letters into envelopes. What is (all letters in wrong envelope)?

Disjoint Events

Definition

Two events 𝐴 and 𝐵 are disjoint if 𝐴𝐵.

A collection of 𝑛 events 𝐴1,,𝐴𝑛 is disjoint if

𝐴𝑖𝐴𝑗𝑖,𝑗

This gives a simplification of inclusion-exclusion:

𝑃(𝐴1𝐴𝑛)=𝑛𝑖=1(𝐴𝑖)

There is nothing to exclude, since all intersections are empty.

Partitions

Definition

A collection of events 𝐴1,,𝐴𝑛 is called a partition of Ω if:

  1. 𝐴1,,𝐴𝑛 disjoint.
  2. 𝑛𝑖=1𝐴𝑖=Ω

Theorem — Law of Total Probability (Intersection Form)

If 𝐴1,,𝐴𝑛 is a partition of Ω, then event 𝐵Ω:

(𝐵)=𝑛𝑖=1(𝐵𝐴𝑖)

Corollary

For two events 𝐴 and 𝐵:

(𝐵)=(𝐴𝐵)+(𝐴𝐶𝐵)

This is true because 𝐴 and 𝐴𝐶 are a partition.

Conditional Probability

Consider two events 𝐴 and 𝐵 of a sample space Ω.

Definition

(𝐵|𝐴) (probability of B given A) is defined as

(𝐵|𝐴)=(𝐴𝐵)(𝐴)

under the assumption that (𝐴)>0.

Example

Rolling a die. 𝐴={The result at least 4},𝐵={Result is 5},𝐶={The result is 3}. (𝐵|𝐴)=1612=13.

Rearranging gives the multiplication rule:

(𝐴𝐵)=(𝐵|𝐴)(𝐴)

This might seem obvious, but sometimes one side is much easier to calculate than the other.

Law of Total Probability (Conditional Form)

Theorem — Law of Total Probability

Let 𝐴1,,𝐴𝑛 be a partition of Ω. Then 𝐵Ω:

(𝐵)=𝑛𝑖=1(𝐵|𝐴𝑖)(𝐴𝑖)

Corollary

If 𝐴,𝐵 are two events: (𝐵)=(𝐵|𝐴)(𝐴)+(𝐵|𝐴𝐶)(𝐴𝐶)

Base Rate Fallacy

Theorem - Bayes Rule

Let 𝐴 and 𝐵 be two events in Ω.

(𝐴|𝐵)=(𝐵|𝐴)(𝐴)(𝐵)

Example — Disease Testing

A rare disease affects 0.5% of the population. A test is 99% precise (i.e. (𝑃|𝐷)=0.99 and (𝑃𝐶|𝐷𝐶)=0.99). If a random person tests positive, what is the probability they have the disease?

Despite the 99% precise test, a positive result only means a 33.2% chance of having the disease. This is the base rate fallacy, as the low prevalence dominates.

Independence

Definition

An event 𝐵 is independent of 𝐴 if the info that 𝐴 happened is irrelevant for 𝐵 to happen.

(𝐵|𝐴)=(𝐵)

This implies (𝐵𝐴)(𝐴)=(𝐵)(𝐵𝐴)=(𝐴)(𝐵). The latter is less intuitive; it is more natural to say that the probability of something given something else equals its unconditional probability.

We write 𝐴𝐵 to denote that 𝐴 and 𝐵 are independent.

Example

Tossing two fair coins. 𝐴={First coin H}. 𝐵={Second coin T}. These events are independent.

(𝐴𝐵)=({HT})=14=(𝐴)(𝐵)

Example

Tossing three coins. 𝐴={First two coins HT}. 𝐵={Last two tosses HT}. These are not independent.

(𝐴𝐵)=()=0(𝐴)(𝐵)=1414=116

Definition

The events 𝐴1,,𝐴𝑛 are independent if

(𝐴𝑖1𝐴𝑖𝑘)=(𝐴𝑖1)(𝐴𝑖𝑘)𝑘{2,,𝑛},1𝑖1<<𝑖𝑘𝑛

Question

Is the above equivalent to (𝐴𝑖𝐴𝑗)=(𝐴𝑖)(𝐴𝑗)𝑖𝑗?

No — pairwise independence does not imply full independence. The pairwise condition is a special case (𝑘=2) of the definition, but higher-order intersections can still fail.

Example

Toss two coins. 𝐴={1st toss is H}, 𝐵={2nd toss is H}, 𝐶={both tosses match (HH or TT)}. 𝐴,𝐵,𝐶 are pairwise independent but NOT independent:

(𝐴𝐵𝐶)=({HH})=14(𝐴)(𝐵)(𝐶)=18

Definition

𝐴1,,𝐴𝑛 are pairwise independent if all pairs of events are independent.

All independent events are pairwise independent, but the converse is not always true, as shown above.

Question

Can an event 𝐴 be independent of itself?

𝐴𝐴(𝐴𝐴)=(𝐴)2(𝐴)=(𝐴)2. This holds only if (𝐴)=0 or (𝐴)=1.

Example

7 people sit at a round table at random. What’s the chance they sit in age order?

By circular permutations, there are (71)!=6! distinct seatings. There are 2 favorable arrangements (clockwise or counterclockwise age order).

26!=147!=1360