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FRM Part I · FRM Exam Part I · Multivariate Random Variables

Annual returns on X and Y are bivariate normal with means of 12% and 2%, both standard deviations equal to 10%, and a correlation of 0.5. What is the probability that Y's return exceeds X's return? (Phi(-1) = 0.1587, Phi(-0.707) = 0.2398, Phi(-0.577) = 0.2820)

The difference X - Y is normal with mean 10% and variance 100 + 100 - 2(0.5)(100) = 100, so a standard deviation of 10%. The probability that it is below zero is Phi(-1), about 15.87%. Subtracting the covariance term is essential for a difference.

  1. A15.87%Correct
  2. B23.98%
  3. C28.20%
  4. D84.13%

Explanation

D = X - Y is normal with mean 10% and variance 100 + 100 - 2*0.5*10*10 = 100, so the standard deviation is 10%. P(Y > X) = P(D < 0) = Phi((0-10)/10) = Phi(-1) = 15.87%. Setting correlation to zero gives sigma = 14.14 and 23.98%. Using the sum's variance (300) gives 28.20%. 84.13% is the probability that X exceeds Y.

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