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Define the mathematical expectation of a random variable and use it in calculations.

FRM® Part I · Quantitative Analysis · Random Variables · FRM1.QA.2.b

The idea

The expected value E[X] of a random variable is its probability-weighted average. For a discrete variable, multiply each possible value by its probability and add: Expected value = Σ x × p(x). For a continuous variable it is the integral of x × f(x) over all values, where f(x) is the density. It is the long-run average, not necessarily a value the variable can take, and it is not in general the median or the mode. Expectation is linear, whether or not X and Y are independent, so the expected value of aX + bY + c is: Linearity = a × E[X] + b × E[Y] + c. It is not generally multiplicative, so the expected value of the product XY is: Product = E[X] × E[Y] + Cov(X, Y). And E[g(X)] is usually not g(E[X]) when g is nonlinear, as with option payoffs. The law of total expectation says the overall mean is the probability-weighted average of the conditional means. Most errors come from applying a nonlinear function to the mean instead of averaging the function.

Formulas

Expected value of a discrete random variable

E[X] = Σ x × p(x)

x
each possible value
p(x)
the probability of that value

The probabilities must sum to 1. E[g(X)] is not g(E[X]) when g is nonlinear, as with an option payoff.

Expected value of a product

E[XY] = E[X] × E[Y] + Cov(X, Y)

X, Y
the two random variables
Cov(X, Y)
their covariance

Always true. E[XY] equals E[X] × E[Y] only when the covariance is zero, unlike linearity, which holds whatever the dependence.

A practice question

From the question bank, with the reason each wrong option is wrong.

Which expression gives the expected value of a discrete random variable X with probability mass function p(x)?

  1. The simple average of the possible values of x, ignoring p(x).Why not: Treats all outcomes as equally likely instead of weighting by probability.
  2. The sum over all possible x of x times p(x). - correct
  3. The sum over all possible x of x² times p(x).Why not: This is the second raw moment E[X²], not the mean.
  4. The value of x with the largest p(x).Why not: This is the mode, the most likely value, not the probability-weighted average.

E[X] is the probability-weighted average: sum of x p(x). It need not equal the most likely value or a simple average of the outcomes.

A harder one

At expiry a stock price S is 80, 100 or 120 with probabilities 0.25, 0.45 and 0.30. A call option with strike 100 pays max(S - 100, 0). What is the expected payoff?

  1. 1.0Why not: Applies the payoff to the expected price, max(E[S] - 100, 0), instead of averaging the payoffs.
  2. 101.0Why not: Reports the expected stock price E[S] rather than the expected option payoff.
  3. 6.0 - correct
  4. 20.0Why not: Takes the payoff in the up state without weighting by its probability of 0.30.

E[S] = 101, but E[g(S)] differs from g(E[S]) for a nonlinear payoff. Expected payoff = 0.25(0) + 0.45(0) + 0.30(20) = 6.0.

The hardest tier

Two random variables have E[X] = 3, E[Y] = 5 and a covariance of 2. What is E[XY]?

  1. 15Why not: Assumes E[XY] = E[X] E[Y], which holds only if the covariance is zero.
  2. 13Why not: Subtracts the covariance instead of adding it: E[X] E[Y] - Cov.
  3. 10Why not: Adds the means and the covariance (3 + 5 + 2) instead of multiplying the means.
  4. 17 - correct

Cov(X,Y) = E[XY] - E[X] E[Y], so E[XY] = Cov + E[X] E[Y] = 2 + 15 = 17.

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