Conditional Expectation

E[X|Y] is a random variable in Y, and averaging it over Y's distribution recovers E[X] — the law of total expectation.

1. The idea

$E [X ∣ Y = y]$ is the average of $X$ restricted to the event $Y = y$ . Viewed as a function of $Y$ before $Y$ is observed, $E [X ∣ Y]$ is itself a random variable.

The law of total expectation states $E [X] = E [E [X ∣ Y]]$ : average the conditional means over the distribution of $Y$ . This lets you compute a hard unconditional expectation by conditioning on whichever variable makes the problem simple.

Conditioning on the first step of a process (a coin flip, a die roll, an arrival) is one of the most common ways to set up a solvable recursion for a global expectation.

2. Why it matters

Conditioning on the right variable turns an intractable direct calculation into a weighted average of simple cases — the core move behind recursive expectation and optimal stopping.

3. Core mathematics

Law of total expectation

E [X] = E [E [X ∣ Y]] = y \sum P (Y = y) E [X ∣ Y = y]

Conditional expectation as a variable

E [X ∣ Y] is a function of Y

4. How to recognize it

Think of this when…

A natural variable (a coin flip, a category, a first roll) splits the problem into cases
Each case is easy to compute an expectation within, but the overall expectation is not
The phrase 'condition on' or 'given the first outcome' fits naturally

5. Worked example

Worked example

A box has 2 fair coins and 1 two-headed coin. A coin is picked at random and flipped twice. Find the expected number of heads.

1
Condition on which coin type $C$ is picked. $P (fair) = 2/3$ , $P (two-headed) = 1/3$ .
$E [X] = P (fair) E [X ∣ fair] + P (two-headed) E [X ∣ two-headed]$
2
Given a fair coin, 2 flips give $E [X ∣ fair] = 2 (0.5) = 1$ . Given the two-headed coin, both flips are heads: $E [X ∣ two-headed] = 2$ .
$E [X ∣ fair] = 1, E [X ∣ two-headed] = 2$
3
Combine with total expectation.
$E [X] = \frac{2}{3} (1) + \frac{1}{3} (2) = \frac{4}{3}$
Answer: $E [X] = 4/3$

6. Common mistakes

Treating $E [X ∣ Y]$ as a number instead of a random variable before $Y$ is known
Forgetting to weight each conditional case by its own probability
Conditioning on a variable that does not actually simplify each case
Confusing $E [X ∣ Y = y]$ (a number) with $E [X ∣ Y]$ (a random variable)

7. Quick check

If $E [X ∣ Y = 1] = 5$ , $E [X ∣ Y = 2] = 9$ , and $P (Y = 1) = P (Y = 2) = 0.5$ , what is $E [X]$ ?

8. Practice

3 QuantSolve questions use this concept.

Practice this concept

9. Related methods

Conditional Expectation

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