FRM Part I · FRM Exam Part I · Multivariate Random Variables
The joint distribution of X and Y has three equally likely outcomes: (X,Y) = (1,2), (2,4), (3,3). What is the covariance of X and Y (population, using the probabilities given)?
The covariance is 1/3. The means are 2 for X and 3 for Y. Multiplying deviations gives 1, 0 and 0 across the three equally likely outcomes, which sum to 1 and average to 1/3. Equivalently, E[XY] of 19/3 minus 2 times 3 equals 1/3.
- A1/3
- B2/3Correct
- C1
- D0
Explanation
E[X]=2, E[Y]=3. E[XY]=(2+8+9)/3=19/3. Cov = 19/3 - 6 = 1/3. Check via deviations: (-1)(-1)=1, (0)(1)=0, (1)(0)=0, sum 1, divided by 3 = 1/3.
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