Independence

Level 1 - Math II (Physics) topic page in Probability.

Principle

Independent events are events where knowing one occurred does not change the probability of the other. Independence is a probability statement, not just a statement that events sound unrelated.

Notation

\(A,B\)

events in the same probability model

\(P(A|B)\)

probability of A given B, defined when P(B)\gt 0

\(P(A\cap B)\)

probability that both A and B occur

\(P(A)P(B)\)

product of the separate probabilities

\(\varnothing\)

empty event

Method

Step 1: Compute the separate probabilities

Find \(P(A)\) and \(P(B)\).

Step 2: Compute the joint probability

Find \(P(A\cap B)\), the probability that both events occur.

Step 3: Compare with the product

If \(P(A\cap B)=P(A)P(B)\), the events are independent. If not, they are dependent.

Conditional definition

\[P(A|B)=P(A)\]

Use conditional probability

\[\frac{P(A\cap B)}{P(B)}=P(A)\]

Multiply by P(B)

\[P(A\cap B)=P(A)P(B)\]

Disjoint and independent are different ideas. If nonzero-probability events are disjoint, then one occurring makes the other impossible, so they are not independent.

Rules

Conditional independence

\[P(A|B)=P(A)\quad(P(B)\gt 0)\]

Product rule for independent events

\[P(A\cap B)=P(A)P(B)\]

Symmetric conditional form

\[P(B|A)=P(B)\quad(P(A)\gt 0)\]

Disjoint nonzero events are dependent

\[A\cap B=\varnothing,\ P(A)\gt 0,\ P(B)\gt 0\quad\Rightarrow\quad P(A\cap B)\ne P(A)P(B)\]

Examples

Question

Two fair dice are rolled. Let \(A\) be first die is 6 and \(B\) be second die is 6. Are \(A\) and \(B\) independent?

Answer

Here

\[P(A)=1/6\]

\[P(B)=1/6\]

and

\[P(A\cap B)=1/36\]

Since

\[P(A)P(B)=1/36\]

the events are independent.

Checks

Test independence with probabilities; do not infer it from separate-looking event names.
Disjoint events with positive probabilities are not independent.
Independent events can occur together; their intersection usually has positive probability.
If \(P(B)>0\), checking \(P(A|B)=P(A)\) is equivalent to checking \(P(A\cap B)=P(A)P(B)\).

Bayes Theorem

Partitions