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FRM Exam Part I · Random Variables

Expected Value and Variance of a Random Variable

Updated 11 October 2026 · Fact-checked

Expected value is the probability-weighted average of a random variable's outcomes: E(X) = Σ x·P(x). Variance is the probability-weighted average squared distance from that mean: E(X²) − [E(X)]². Standard deviation is its square root. For Y = a + bX, E(Y) = a + bE(X) and Var(Y) = b²Var(X).

Understand Expected Value and Variance

A random variable assigns a number to each outcome of an uncertain event, such as a portfolio's one-day profit. A discrete random variable takes a list of values, each with a probability. The probabilities must add up to 1.

The expected value (mean) is the long-run average. You multiply each outcome by its probability and add. It is a weighted average, not necessarily a value the variable can actually take. A fair die has a mean of 3.5, yet you can never roll 3.5.

The variance measures spread. Take each outcome's distance from the mean, square it, weight it by its probability and add. Squaring stops positive and negative deviations from cancelling. The standard deviation is the square root of variance. It is in the same units as X, so it is the number risk managers quote as volatility.

The linear transformation rules matter most for the exam. Adding a constant shifts the mean but does not change the spread. Multiplying by a constant scales the mean by b and the standard deviation by |b|, so variance scales by b². Expectation is linear, which means E(X + Y) = E(X) + E(Y) always, even when X and Y are dependent. Variance of a sum is different, and it needs covariance. That is covered in a related topic.

Key formulas to remember

Expected value (discrete)
E(X) = Σ xᵢ · P(xᵢ)
Probabilities must sum to 1. For a continuous variable the sum becomes an integral of x·f(x).
Variance (definition)
Var(X) = σ² = E[(X − μ)²] = Σ (xᵢ − μ)² · P(xᵢ)
μ = E(X). Always zero or positive.
Variance (shortcut)
Var(X) = E(X²) − [E(X)]²
Faster for most calculations. E(X²) = Σ xᵢ² · P(xᵢ). Do not confuse it with [E(X)]².
Standard deviation
σ = √Var(X)
Same units as X.
Linear transformation: mean
E(a + bX) = a + b·E(X)
a and b are constants.
Linear transformation: variance
Var(a + bX) = b² · Var(X)
The constant a drops out. Standard deviation of a + bX is |b| · σ.
Expectation of a sum
E(X + Y) = E(X) + E(Y)
Holds whether or not X and Y are independent.
Expectation of a function
E[g(X)] = Σ g(xᵢ) · P(xᵢ)
In general E[g(X)] ≠ g(E[X]). For example, E(X²) ≠ [E(X)]².

How to solve Expected Value and Variance questions

Use this routine for any question on mean, variance or a transformed variable.

  1. 1List every outcome x with its probability P(x). Check that the probabilities add to 1.
  2. 2Compute E(X) = Σ x·P(x). Write the value down.
  3. 3Compute E(X²) = Σ x²·P(x), squaring each outcome before weighting.
  4. 4Compute Var(X) = E(X²) − [E(X)]². Take the square root if the question asks for standard deviation.
  5. 5If the question gives Y = a + bX, do not recompute from scratch. Use E(Y) = a + bE(X) and Var(Y) = b²Var(X).
  6. 6Check the answer: variance cannot be negative, and the mean must lie between the smallest and largest outcome.
  7. 7Check what is asked: mean, variance or standard deviation, and whether the answer needs units such as USD.

Quickest way: Shortcut with the shortcut formula and transformation rules

When to use it: Use it for any discrete table with three to five outcomes, or whenever a question asks about a + bX.

  1. Compute E(X) and E(X²) in a single pass down the table, using a calculator's memory or a scratch column.
  2. Subtract [E(X)]² from E(X²) for variance.
  3. For a + bX, transform the mean and variance directly. Ignore a for variance and square b.
  4. If outcomes are large, subtract a convenient constant c from every outcome first. The variance does not change, and you add c back to the mean.
  5. Eliminate options that are negative variances or that use b instead of b².

Common mistakes in Expected Value and Variance

  • Writing Var(a + bX) = a + b·Var(X) or b·Var(X).

    Students copy the mean rule, which is linear, and apply it to variance.

    Fix: Remember that variance is in squared units. The constant a vanishes and b is squared: Var(a + bX) = b²Var(X).

  • Using [E(X)]² in place of E(X²).

    The two look alike, and squaring after averaging is easier than squaring before.

    Fix: Square each outcome first, then weight by probability. The difference E(X²) − [E(X)]² is exactly the variance.

  • Forgetting to take the square root when asked for standard deviation.

    Students stop at the variance they just computed.

    Fix: Read the last line of the question. Standard deviation is √variance, and a wrong option is often the variance itself.

  • Dropping the absolute value for a negative multiplier.

    Students write σ(bX) = b·σ, which gives a negative standard deviation when b < 0.

    Fix: Use |b|·σ. Variance already uses b², so it stays positive.

  • Dividing by n or n − 1 when the probabilities are given.

    It is mixed up with sample variance.

    Fix: With a probability table you are computing a population-style moment. Weight by probabilities and do not divide by anything.

  • Using P(x) values that do not add up to 1 without noticing.

    Students rush into the arithmetic.

    Fix: Add the probabilities first. If one is missing, find it as 1 minus the sum of the others.

Worked examples

Example 1

A bond portfolio's one-year profit X (in USD million) has this distribution: −4 with probability 0.10, 0 with probability 0.30, 5 with probability 0.40, 10 with probability 0.20. Find the expected value, variance and standard deviation of X.

Show the solution
  1. Check the probabilities: 0.10 + 0.30 + 0.40 + 0.20 = 1.00.
  2. E(X) = (−4)(0.10) + 0(0.30) + 5(0.40) + 10(0.20) = −0.4 + 0 + 2.0 + 2.0 = 3.6.
  3. E(X²) = 16(0.10) + 0(0.30) + 25(0.40) + 100(0.20) = 1.6 + 0 + 10.0 + 20.0 = 31.6.
  4. Var(X) = 31.6 − 3.6² = 31.6 − 12.96 = 18.64.
  5. σ = √18.64 ≈ 4.32.

Answer: E(X) = USD 3.6 million, Var(X) = 18.64, standard deviation ≈ USD 4.32 million.

Example 2

A trader's daily P&L X has mean USD 20,000 and standard deviation USD 50,000. The desk converts it to Y = 3X − 10,000 (a leveraged position less a fixed cost). Find the mean and standard deviation of Y.

Show the solution
  1. Here a = −10,000 and b = 3.
  2. E(Y) = a + b·E(X) = −10,000 + 3(20,000) = −10,000 + 60,000 = 50,000.
  3. Var(X) = 50,000² = 2,500,000,000.
  4. Var(Y) = b²·Var(X) = 9 × 2,500,000,000 = 22,500,000,000.
  5. σ(Y) = √22,500,000,000 = 150,000. This equals |b|·σ(X) = 3 × 50,000, and the constant −10,000 has no effect.

Answer: E(Y) = USD 50,000 and standard deviation of Y = USD 150,000.

Exam tips

  • Questions often give a table and ask for variance, with the mean as a trap option. Compute E(X²) − [E(X)]² and check which quantity the question wants.
  • For a + bX questions, expect distractors that use b instead of b², add a to the variance, or leave out the square root. Rule these out by inspection.
  • Keep full decimals in E(X) until the end. Rounding the mean early can shift the variance enough to pick the wrong option.
  • Remember E(X + Y) = E(X) + E(Y) needs no independence, but Var(X + Y) = Var(X) + Var(Y) does (or zero covariance).
  • Use the calculator's statistics mode or memory keys to accumulate Σ xP and Σ x²P without retyping.

Practice questions from Random Variables

Expected Value and Variance in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

Expected Value and Variance: frequently asked questions

What is the difference between E(X²) and [E(X)]²?

E(X²) squares each outcome and then averages. [E(X)]² averages first and then squares. Their difference is the variance, so E(X²) is never smaller than [E(X)]².

Why does adding a constant not change variance?

Adding a constant moves every outcome and the mean by the same amount. Each deviation from the mean stays the same, so the spread is unchanged.

Does E(X + Y) = E(X) + E(Y) need independence?

No. Expectation is linear and this holds for any two random variables. Variance of a sum does depend on covariance, so independence matters there.

How do I find the expected value of a function such as X²?

Apply the function to each outcome and weight by its probability: E[g(X)] = Σ g(x)·P(x). Do not apply the function to E(X) instead, because in general E[g(X)] ≠ g(E[X]).