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FRM Part I · FRM Exam Part I · Fundamentals of Probability

A loss random variable L (in USD millions) has a density f(x) = c(4 - x) for 0 ≤ x ≤ 4 and zero elsewhere. For f to be a valid probability density function, what must c equal, and what is the resulting P(L ≤ 2)?

Setting the integral of c(4 - x) from 0 to 4 equal to one gives 8c = 1, so c = 1/8. Then the probability that L is at most 2 is (1/8)(8 - 2) = 0.75, since the density is larger at low losses.

  1. Ac = 1/8; P(L ≤ 2) = 0.75Correct
  2. Bc = 1/4; P(L ≤ 2) = 0.50
  3. Cc = 1/8; P(L ≤ 2) = 0.50
  4. Dc = 1/4; P(L ≤ 2) = 0.75

Explanation

The integral of c(4-x) over 0 to 4 is c(16 - 8) = 8c = 1, so c = 1/8. Then P(L ≤ 2) = (1/8)(4·2 - 2) = (1/8)(6) = 0.75. Check: P(L>2) = (1/8)∫2..4(4-x)dx = (1/8)(2) = 0.25. Choosing c = 1/4 forgets the 1/2 factor in the integral of x.

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