Linearity of expectation
E[aX + bY] = aE[X] + bE[Y] holds always — no independence required. This is remarkable and enormously useful, because it lets you decompose a complicated count into simple pieces whose dependence you can completely ignore. Problems that look intractable by distribution become one-line calculations.
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Method: how to approach it
The order below is what examiners expect to see, and each step carries its own marks.
- Decompose the quantity into a sumWrite X = X₁ + X₂ + … where each piece is simple, typically an indicator.
- Compute each expectation separatelyFor an indicator, E[Xᵢ] is just the probability of the corresponding event.
- Add them upE[X] = Σ E[Xᵢ]. Ignore all dependence between the pieces — that is the whole point.
- Do NOT do this for varianceVar only adds for uncorrelated variables. Linearity is a property of expectation alone.
Worked example
n letters are placed at random into n addressed envelopes. Find the expected number of correct matches.
- Let Xᵢ = 1 if letter i lands in its own envelope, and 0 otherwise. Then X = ΣXᵢ.
- Each letter is equally likely to land in any envelope, so P(Xᵢ = 1) = 1/n.
- E[Xᵢ] = 1/n for each i.
- By linearity, E[X] = n × (1/n).
Answer. E[X] = 1, regardless of n. The Xᵢ are strongly dependent, but linearity does not care.
Where marks get dropped
These are the specific errors that cost credit on linearity of expectation questions — QED's rubric penalises each of them separately.
- Believing independence is required. It is not — that is what distinguishes linearity from the product rule.
- Applying the same additivity to variance. Var(X + Y) needs a covariance term unless the variables are uncorrelated.
- Trying to compute the full distribution first. Linearity exists precisely so you can avoid that.
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Linearity of expectation — frequently asked questions
Why does it work without independence?
Because expectation is a sum over outcomes, and summation is linear. The joint distribution never enters the calculation.
What is the classic application?
Coupon collecting, the matching problem above, and expected running times of randomised algorithms — all decompose into indicators.
Does it extend to infinitely many terms?
Yes under mild conditions (absolute convergence), which is what makes E[X] = Σ P(X ≥ k) valid for non-negative integer variables.
The rest of Probability
Events, conditional probability, random variables. Each subtopic below has its own method, worked example and mark-losing traps.
- 1Sample spaces & events
- 2Conditional probability & independence
- 3Bayes’ theorem
- 4Random variables & expected value
- 5Variance & standard deviation
- 6Binomial & uniform distributions
- 7Law of total probability & tree diagrams
- 8Geometric & Poisson distributions
- 9Joint, marginal & conditional distributions
- 10Markov & Chebyshev bounds
- 11Equally likely outcomes & counting
- 12Linearity of expectation
- 13Indicator random variables
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