Skip to content

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

X takes values 1 or 2 and Y takes values 10 or 20 with joint probabilities: P(1,10)=0.30, P(1,20)=0.20, P(2,10)=0.10, P(2,20)=0.40. What is Cov(X, Y)?

The covariance is 1. E[XY] is 25 from the joint table, while E[X] is 1.5 and E[Y] is 16, so the product of the means is 24. Covariance equals E[XY] minus E[X]E[Y], which is 25 minus 24, or 1.

  1. A0
  2. B1Correct
  3. C2
  4. D25

Explanation

E[XY]=1×10×0.3+1×20×0.2+2×10×0.1+2×20×0.4=3+4+2+16=25. E[X]=1(0.5)+2(0.5)=1.5 and E[Y]=10(0.4)+20(0.6)=16. Cov = 25 − 1.5×16 = 1. A value of 0 would wrongly imply independence, and 25 is just E[XY].

Did you get it right without looking?

One question tells you little. A timed set on Multivariate Random Variables shows your real accuracy, how long you take and where you lose marks.

Practise Multivariate Random VariablesFree diagnostic mock

More Multivariate Random Variables questions