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Express a probability mass function through a probability matrix and explain how this works.

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

The idea

A bivariate discrete random variable (X, Y) is described by its joint probability mass function (pmf) f(x, y), the probability that X takes the value x and Y takes the value y at the same time. A probability matrix lays this out as a table with the values of X down the rows and the values of Y across the columns, so each cell holds the probability that that pair occurs. Two rules make it valid: every cell lies between 0 and 1, and all the cells together sum to 1. Events such as X > Y, or a total X + Y of 3, are found by adding the cells that satisfy them. A frequent mistake is to force each row or column to sum to 1, which describes conditional distributions rather than the joint pmf. When a cell is missing, subtract the sum of the known cells from 1.

A practice question

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

A probability matrix is used to describe the joint probability mass function of two discrete random variables X and Y. Which condition must the entries of a valid probability matrix satisfy?

  1. Each row sums to 1 and each column sums to 1.Why not: Treats the table as a set of conditional distributions or a transition matrix; only the grand total must equal 1.
  2. Every entry is between 0 and 1, and all the entries taken together sum to 1. - correct
  3. The row totals must equal the column totals, so that the matrix is symmetric.Why not: Confuses a joint pmf with a symmetric matrix; X and Y can take different values and need not have equal marginals.
  4. The entries on the main diagonal sum to 1 and the off-diagonal entries are zero.Why not: Describes perfect dependence only; a general joint pmf can put positive probability off the diagonal.

Each cell is P(X = x, Y = y), so each is a probability and the cells together exhaust all possible (x, y) pairs, giving a grand total of 1. Rows summing to 1 would describe conditional distributions, not the joint pmf.

A harder one

The joint pmf of X (rows, values 0 and 1) and Y (columns, values 1, 2 and 3) is given by a probability matrix. Row X = 0 reads 0.10, 0.20, 0.15. Row X = 1 reads 0.25, 0.10 and an unreadable entry for Y = 3. What is P(X = 1, Y = 3)?

  1. 0.20 - correct
  2. 0.80Why not: Reports the sum of the known cells instead of subtracting it from 1.
  3. 0.65Why not: Subtracts only the two known cells in its own row from 1, ignoring row X = 0, which also uses up probability.
  4. 0.55Why not: Uses the whole row X = 1 total (0.55) as if it were the single missing cell.

All cells must sum to 1. Known cells sum to 0.80, so the missing cell is 1 - 0.80 = 0.20.

The hardest tier

Two desks report the number of limit breaches in a month, X for desk 1 and Y for desk 2, each taking values 1, 2 or 3. The joint pmf is: P(1,1) = 0.05, P(1,2) = 0.10, P(1,3) = 0.05, P(2,1) = 0.15, P(2,2) = 0.20, P(2,3) = 0.10, P(3,1) = 0.10, P(3,2) = 0.15, P(3,3) = 0.10, where P(x,y) = P(X = x, Y = y). What is the probability that desk 1 has strictly more breaches than desk 2?

  1. 0.40 - correct
  2. 0.25Why not: Adds the cells where X < Y, the reverse inequality.
  3. 0.35Why not: Adds only the diagonal cells where X = Y, i.e. the ties.
  4. 0.75Why not: Computes P(X >= Y), including the ties, when the question asks for strictly more.

Sum the cells with x > y: P(2,1) + P(3,1) + P(3,2) = 0.15 + 0.10 + 0.15 = 0.40. The three regions X > Y, X = Y and X < Y have probabilities 0.40, 0.35 and 0.25, which sum to 1.

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