1
Independence is not needed
E[X + Y] = E[X] + E[Y] holds for any two random variables on the same space, however entangled.
Lesson · Discrete Probability · Random variables
Linearity of Expectation
The most useful fact in elementary probability. It needs no independence, which lets a complicated count be decomposed into indicators and summed term by term.
E[X + Y] = E[X] + E[Y] ALWAYS
E[c₁X₁ + … + cₙXₙ] = Σ cᵢE[Xᵢ]
E[XY] = E[X]E[Y] only when independent
Decompose into INDICATORS and sum.
No independence required
Core ideas
What this lesson establishes
Linearity is what makes hard expectations easy: split the variable into simple pieces and add.
2
Indicators are the workhorse
Write a count as a sum of indicators, one per thing being counted. Each indicator's expectation is a probability.
3
Products are different
E[XY] = E[X]E[Y] requires independence. Confusing the two rules is the one real trap here.
The material
Linearity
The theorem, why independence is irrelevant, and four applications.
Theorem · Linearity of expectation
For random variables X, Y on the same sample space and constants a, b: E[aX + bY] = aE[X] + bE[Y]. No independence assumption is needed.
Summing over outcomes, E[aX + bY] = Σ_s (aX(s) + bY(s))P(s) = a Σ_s X(s)P(s) + b Σ_s Y(s)P(s) = aE[X] + bE[Y].
Nothing about the joint behaviour of X and Y entered, which is exactly why independence is not required.
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The proof is three lines and the consequence is enormous. E[XY] by contrast genuinely depends on the joint distribution, and the two facts are worth keeping apart deliberately.
Indicator decomposition
Theorem · Expectation of an indicator
E[I_A] = 1 · P(A) + 0 · P(Ā) = P(A).
So if X counts how many of the events A₁, …, A_n occur, then X = I_{A₁} + ⋯ + I_{A_n} and E[X] = Σ P(Aᵢ). The events may overlap, depend on each other, or be governed by anything at all.
Example · Heads in n flips
X = Σ Iᵢ where Iᵢ indicates heads on flip i. Each has expectation p, so E[X] = np — without computing the binomial distribution (“Bernoulli Trials and the Binomial Distribution”) at all.
Example · Fixed points of a random permutation
X counts the elements left in place. X = Σ Iᵢ where Iᵢ indicates that i is fixed, and P(Iᵢ = 1) = 1/n. So E[X] = n · (1/n) = 1, for every n. The indicators are dependent — knowing n − 1 elements are fixed forces the last — and linearity does not care.
That result is worth pausing on. On average exactly one person gets their own hat back, whether there are 5 people or 5 million. Computing it from the derangement distribution of Combinatorics and Counting would be considerably harder.
Example · The coupon collector
There are n distinct coupons, one per box at random. How many boxes to collect them all? Let Xᵢ be the number of boxes bought while holding exactly i − 1 distinct coupons. Then E[Xᵢ] = n/(n − i + 1) (“The Geometric Distribution and Waiting Times” supplies that), so E[total] = n(1 + 1/2 + ⋯ + 1/n) ≈ n ln n. For n = 50 that is about 225 boxes.
Example · Matching pairs
Two shuffled decks are dealt face up side by side. The expected number of positions where the cards match is 52 · (1/52) = 1. Same computation as the hat problem, and again the indicators are dependent.
Theorem · Products need independence
If X and Y are independent then E[XY] = E[X]E[Y]. Without independence this fails: take Y = X with X a fair coin scored 0 or 1, so E[X] = 1/2 and E[XY] = E[X²] = 1/2 ≠ 1/4.
Interactive
Distribution, expectation, variance
Definition first, formula second
Method
Using linearity
Decompose, then add.
1
Write the variable as a sum
Usually one indicator per item, position or event being counted.
2
Compute the expectation of one piece
For an indicator this is a single probability, and symmetry often makes all the pieces identical.
3
Add, without checking independence
Linearity holds regardless. This is the step that saves the work.
4
Do not use it for products
E[XY] needs independence. If the quantity is a product, this technique does not apply.
Linearity
Score 0 / 0Choose the answer you can justify.
Expected number of matching pairs
Two shuffled decks of n cards, dealt side by side. How many positions match?
Let Iⱼ = 1 if position j matches, 0 otherwise. X = I₁ + … + Iₙ
P(Iⱼ = 1) = 1/n — whatever is in the first deck at position j,
the second deck has 1 chance in n of matching
By linearity:
E[X] = Σⱼ E[Iⱼ] = n · (1/n) = 1
The Iⱼ are NOT independent: if n−1 positions match, the last must.
Linearity does not care.
Computing this from the distribution of X would require the
derangement numbers of “Derangements” in Combinatorics and Counting — for one number.
The answer is 1 for every n, which connects directly to derangements: the distribution of the number of matches converges to a Poisson distribution with mean 1, and P(no match) → 1/e is exactly the derangement limit.
Common mistakes
1
Checking independence before addingUnnecessary, and the habit obscures how general linearity is. Add first.
2
Applying linearity to a productE[XY] depends on the joint distribution. Linearity says nothing about it.
3
Not looking for an indicator decompositionMost counting expectations become one line this way. If a distribution looks hard, ask what is being counted.
Check your understanding
✓
Finish when decomposition into indicators is your first move.
Compute the expected number of runs of consecutive heads in n fair flips, using one indicator per position for "a run starts here". The answer is (n + 1)/4. Adjacent indicators are dependent — a run cannot start at two consecutive positions — and linearity does not care.
Note: Linearity is the reason expectation is more tractable than probability. P(X = k) for a complicated X may be very hard while E[X] is a sum of n easy probabilities, which is why so many arguments compute the mean and never touch the distribution.