[math]X[/math] and [math]Y[/math] denote the values of two stocks at the end of a five-year period. [math]X[/math] is uniformly distributed on the interval (0, 12). Given [math]X = x[/math], [math]Y[/math] is uniformly distributed on the interval [math](0, x)[/math].

Calculate [math]\operatorname{Cov}(X, Y)[/math] according to this model.

• 0
• 4
• 6
• 12
• 24

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: C

The joint pdf of X and Y is

[[math]] f(x,y) = f_2(y | x) f_1(x) = (1/x)(1/12), \, 0 \lt y \lt x, 0 \lt x \lt 12. [[/math]]

Therefore,

[[math]] \operatorname{E}[X] = \int_0^{12}\int_0^x x \frac{1}{12x} dydx = \int_0^{12} \frac{y}{12} \Big |_0^x dx = \int_0^{12} \frac{x}{12} dx = \frac{x^2}{24} \Big |_0^{12} = 6 [[/math]]

[[math]] \operatorname{E}[Y] = \int_0^{12}\int_0^x y \frac{1}{12x} dydx = \int_0^{12} \frac{y}{24x} \Big |_0^x dx = \int_0^{12} \frac{x}{24} dx = \frac{x^2}{48} \Big |_0^{12} = \frac{144}{48} = 3 [[/math]]

[[math]] \operatorname{E}[XY] = \int_0^{12}\int_0^x \frac{y}{12} dydx = \int_0^{12} \left [\frac{y^2}{24} \right ]_0^x dx = \int_0^{12} \frac{x^2}{24} dx = \frac{x^3}{72} \Big |_0^{12} = \frac{(12)^2}{72} = 24. [[/math]]

[[math]] \operatorname{Cov}(X,Y) = \operatorname{E}[XY] – \operatorname{E}[X]\operatorname{E}[Y] = 24 − (3)(6) = 24 – 18 = 6 . [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.