IAI Actuarial Core Principles · Actuarial Statistics · Jointly distributed random variables
X and Y have joint density f(x,y) = 2 for 0 < y < x < 1, and zero elsewhere. What is E(Y | X = 0.6)?
E(Y | X = 0.6) is 0.3, because given X = x, Y is uniform on (0, x) and so has mean x/2.
- A0.2
- B0.3Correct
- C0.4
- D0.5
- 0.6
Explanation
Given X = x, the density of Y is f(x,y)/f_X(x) = 2/(2x) = 1/x on (0,x), which is uniform on (0,x). The mean is x/2 = 0.3. Using 0.5 would wrongly treat Y as uniform on (0,1).
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