Expected Value: Fundamentals

The weighted average of a random variable's outcomes, and how to compute it for discrete and continuous variables.

1. The idea

The expected value of a random variable $X$ is its probability-weighted average outcome. For a discrete variable it is $E [X] = \sum_{x} x P (X = x)$ ; for a continuous variable with density $f$ , it is $E [X] = \int x f (x) d x$ .

$E [X]$ is not a prediction of any single outcome — it is the long-run average if you could repeat the experiment many times. A fair die shows $E [X] = 3.5$ , a value the die can never actually land on.

Expectation extends to functions of $X$ : $E [g (X)] = \sum_{x} g (x) P (X = x)$ or $\int g (x) f (x) d x$ . This is the law of the unconscious statistician — you do not need the distribution of $g (X)$ itself, only that of $X$ .

2. Why it matters

Every pricing, sizing, and risk question in a quant interview reduces to computing an expectation. Fluency with the definition lets you set up the right sum or integral instead of guessing.

3. Core mathematics

Discrete expectation

E [X] = x \sum x P (X = x)

Continuous expectation

E [X] = \int_{- \infty}^{\infty} x f (x) d x

Function of a random variable

E [g (X)] = x \sum g (x) P (X = x)

Law of the unconscious statistician.

Nonnegativity

X \geq 0 ⟹ E [X] \geq 0

4. How to recognize it

Think of this when…

Question asks for an 'average', 'expected', or 'fair price' quantity
A payoff or reward is defined outcome by outcome
You are given a pmf or pdf explicitly
You need $E [X^{2}]$ or another moment, not just $E [X]$

5. Worked example

Worked example

A fair six-sided die is rolled once. Find $E [X]$ and $E [X^{2}]$ .

1
Each face has probability $1/6$ .
$P (X = k) = \frac{1}{6}, k = 1, \dots, 6$
2
Sum $k$ times its probability.
$E [X] = \frac{1}{6} (1 + 2 + 3 + 4 + 5 + 6) = \frac{21}{6} = 3.5$
3
For $E [X^{2}]$ , sum $k^{2}$ times its probability.
$E [X^{2}] = \frac{1}{6} (1 + 4 + 9 + 16 + 25 + 36) = \frac{91}{6} \approx 15.17$
Answer: $E [X] = 3.5$ , $E [X^{2}] = 91/6 \approx 15.17$

6. Common mistakes

Confusing $E [X]$ with the most likely outcome (the mode)
Forgetting that $E [X]$ can be a value $X$ never actually takes
Computing $E [g (X)]$ by plugging $g (E [X])$ instead of averaging $g$ over the distribution — true only when $g$ is linear
Dropping terms with probability zero contribution instead of confirming they are truly zero

7. Quick check

A biased coin shows heads with probability 0.3, paying $10 on heads and $0 on tails. What is $E [payoff]$ ?

For $X$ uniform on $[0, 2]$ , what is $E [X]$ ?

8. Practice

15 QuantSolve questions use this concept.

Practice this concept

9. Related methods

Linearity of Expectation

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