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Mathematics · University

Conditional probability and expectation

P(A∣B)=P(A and B)/P(B) requires P(B)>0.

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Prerequisites

Fractions, sums and elementary probability.

Study notes

Conditioning

P(A∣B)=P(A and B)/P(B) requires P(B)>0.

Independence

A and B are independent when P(A and B)=P(A)P(B); independence is not the same as disjointness.

Expectation

Expectations add even without independence, provided the relevant expectations exist.

Variance

Variance adds for independent variables; in general include covariance terms.

Worked examples

Given P(A and B)=0.12 and P(B)=0.4, find P(A∣B).

Divide 0.12 by 0.4.

0.3

X takes values 0,2,5 with probabilities 0.2,0.5,0.3. Find E[X].

Compute 0(0.2)+2(0.5)+5(0.3)=2.5.

2.5

Check your understanding

  1. Two events have positive probability and are disjoint. Are they independent?

    1. Always
    2. Only if their names differ
    3. No
    4. Independence is undefined for events

  2. Independent X and Y have variances 4 and 9. Find Var(2X−Y).

  3. Show by an example that E[XY] need not equal E[X]E[Y].

  4. Does E[X+Y]=E[X]+E[Y] require independence?

    1. Yes in every case
    2. Only if the means are zero
    3. Only for normal distributions
    4. No, assuming expectations exist

Answers and marking guidance

Answers and marking notes
  1. No

    Their intersection probability is zero while the product of their positive probabilities is positive.

  2. 25

    Variance scales by the square of coefficients, giving 4×4+9=25.

  3. Let X=Y be a fair 0/1 variable. Then E[XY]=E[X]=1/2, while E[X]E[Y]=1/4.

    Credit a dependent example and both computed quantities.

  4. No, assuming expectations exist

    Linearity of expectation does not require independence.

Further reading and review references

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