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Describe the way an expected value of a function is obtained for a discrete bivariate random variable.

FRM® Part I · Quantitative Analysis · Multivariate Random Variables · FRM1.QA.4.c

The idea

For a discrete bivariate random variable, the expected value of any function g(X, Y) is: Expected value = Σ g(x, y) × f(x, y). The sum runs over all cells of the joint matrix. You evaluate the function in each cell, weight it by that cell's probability and add. This covers E(X), where g is just x, as well as E(X²), E(XY), which is needed for covariance, and nonlinear payoffs such as max(X − Y, 0). The classic mistake is to plug the means into the function, g(E(X), E(Y)). That is generally correct only when g is linear. Likewise, E(XY) equals E(X) × E(Y) only when the covariance is zero. For E(X) alone, you can also use the marginal probabilities: E(X) = Σ x × fX(x). Always weight by probabilities, never by equal shares.

Formulas

Expected value of a function of a bivariate variable

E[g(X, Y)] = Σ g(x, y) × f(x, y)

g(x, y)
the function evaluated in each cell
f(x, y)
the joint probability of that cell

The sum runs over every cell of the joint matrix. Plugging the means into g is right only when g is linear.

A practice question

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

For a discrete bivariate random variable (X, Y) with joint pmf f(x, y), how is the expected value of a function g(X, Y) obtained?

  1. Evaluate g at the expected values, g(E[X], E[Y]).Why not: Applies the function to the means; this is generally different from E[g(X, Y)] when g is nonlinear.
  2. Compute g(x, y) for every cell, weight it by that cell's probability f(x, y), and add over all cells. - correct
  3. Average g(x, y) over all cells with equal weights, ignoring the probabilities.Why not: Uses equal weights rather than the joint probabilities, so likely and unlikely pairs count the same.
  4. Multiply E[X] by E[Y] and then apply g to the product.Why not: Treats E[XY] as E[X]E[Y] (valid only without dependence) and then applies g in the wrong place.

E[g(X, Y)] = sum over x and y of g(x, y) f(x, y). The probabilities weight the values of the function itself, not the inputs, which is why E[g(X, Y)] is generally not g(E[X], E[Y]).

A harder one

X takes values 1, 2, 3 and Y takes values 1, 2 with joint pmf: P(1,1) = 0.10, P(1,2) = 0.20, P(2,1) = 0.25, P(2,2) = 0.15, P(3,1) = 0.20, P(3,2) = 0.10. What is E[XY]?

  1. 2.80 - correct
  2. 2.90Why not: Multiplies E[X] = 2.00 by E[Y] = 1.45, which equals E[XY] only if X and Y are uncorrelated.
  3. 3.00Why not: Averages x times y over the six cells with equal weights instead of using the joint probabilities.
  4. 3.45Why not: Adds the two means, which is E[X + Y], not E[XY].

E[XY] = 1x1x0.10 + 1x2x0.20 + 2x1x0.25 + 2x2x0.15 + 3x1x0.20 + 3x2x0.10 = 0.10 + 0.40 + 0.50 + 0.60 + 0.60 + 0.60 = 2.80.

The hardest tier

Strategy A earns X in {0, 2, 4} and strategy B earns Y in {1, 3} (USD million). The joint pmf is: P(0,1) = 0.15, P(0,3) = 0.10, P(2,1) = 0.20, P(2,3) = 0.25, P(4,1) = 0.25, P(4,3) = 0.05. A contract pays max(X - Y, 0). What is its expected payoff?

  1. 0.30Why not: Applies the payoff to the expected values, max(E[X] - E[Y], 0) = max(2.10 - 1.80, 0); this ignores that the payoff is convex.
  2. 1.70Why not: Computes E|X - Y|, which also counts the cells where Y exceeds X; the payoff is zero there.
  3. 0.83Why not: Averages the six payoffs with equal weights instead of the joint probabilities.
  4. 1.00 - correct

Nonzero payoffs: (2,1) pays 1 x 0.20 = 0.20; (4,1) pays 3 x 0.25 = 0.75; (4,3) pays 1 x 0.05 = 0.05. Total = 1.00. Using the means gives 0.30, which is wrong because E[max(.)] differs from max(E[.]).

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