Linearity of Expectation

E[X+Y] = E[X] + E[Y] always holds — no independence required — making it the single most useful tool in quant interviews.

1. The idea

Linearity says $E [a X + bY + c] = a E [X] + b E [Y] + c$ for any random variables $X, Y$ and constants $a, b, c$ . Crucially, this holds whether or not $X$ and $Y$ are independent, and whether or not they are even defined on a nice distribution.

This makes linearity the workhorse for messy sums: instead of finding the distribution of $X + Y$ , split it into simple pieces, take each expectation separately, and add. Complicated dependence between terms is irrelevant to the calculation.

The technique pairs naturally with indicator variables (next lesson): write a complicated count as a sum of 0/1 variables, and linearity turns the whole problem into single-event probabilities.

2. Why it matters

Nearly every 'expected number of X' interview question is solved by decomposing into a sum and applying linearity — it converts a hard joint-distribution problem into several easy single-variable ones.

3. Core mathematics

Linearity

E [i = 1 \sum n a_{i} X_{i} + c] = i = 1 \sum n a_{i} E [X_{i}] + c

Holds regardless of independence or dependence.

Special case

E [X + Y] = E [X] + E [Y]

Product needs independence

E [X Y] = E [X] E [Y] only if X ⊥ Y

Linearity applies to sums, not products.

4. How to recognize it

Think of this when…

You need the expected value of a sum, total, or count
Variables may be dependent, so you cannot use independence-based tricks
The problem asks 'expected number of' something that can be broken into pieces
You are tempted to find a full joint distribution — linearity likely avoids that

5. Worked example

Worked example

Roll a fair die 3 times. Let $S$ be the sum of the three rolls. Find $E [S]$ .

1
Let $X_{1}, X_{2}, X_{3}$ be the three rolls, so $S = X_{1} + X_{2} + X_{3}$ .
$S = X_{1} + X_{2} + X_{3}$
2
Each $X_{i}$ has the same expectation, $E [X_{i}] = 3.5$ , regardless of the others.
$E [X_{i}] = 3.5$
3
Apply linearity directly — the rolls are independent here, but that is not required.
$E [S] = E [X_{1}] + E [X_{2}] + E [X_{3}] = 3 (3.5) = 10.5$
Answer: $E [S] = 10.5$

6. Common mistakes

Believing linearity requires independence — it never does
Applying the same additive rule to variance, where independence does matter
Trying to compute $E [X Y]$ via linearity when it actually requires independence or covariance
Forgetting to split a compound sum into its simplest additive pieces before applying linearity

7. Quick check

Does linearity of expectation require independence?

If $E [X] = 4$ and $E [Y] = - 1$ , what is $E [3 X - 2 Y + 5]$ ?

8. Practice

5 QuantSolve questions use this concept.

Practice this concept

9. Related methods

Linearity of Expectation

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