EXAM P SAMPLE SOLUTIONS

105. Solution: A
The calculation requires integrating over the indicated region.
2x1 1 2 8 1 1 1
2 4 2 2 4 2 2 2 4 4 5
x
( ) ∫∫ ∫ ∫ ( ) ∫
E X = x y dy dx = x y dx = x 4x− x dx = 4xdx
= x
3 3 3 5
0 x
0 0 0
x
2x 1
1 2 8 1 1 1
2 8 3 8 3 3 56 4 56 5
x
( ) ∫∫ ( 8 )
x ∫ ∫ ∫
E Y = xy dy dx = xy dy dx = x x − x dx = x dx = x
3 9 9 9 45
0 0 0 0
x
2x
1 1
2 2 2 3 2 3 3 5
1 2x8 18
8 56 56 28
E( XY) = ∫∫ x y dydx= x y dx x ( 8x
x ) dx x d
0 x 3 ∫
= − =
09
∫09 ∫ x = =
0 9 54 27
28 ⎛56 ⎞⎛4⎞ Cov ( XY , ) = E( XY) − E( X) EY ( ) = − ⎜ ⎟⎜ ⎟=
0.04
27 ⎝45 ⎠⎝5⎠ --------------------------------------------------------------------------------------------------------
106. Solution: C
The joint pdf of X and Y is f(x,y) = f 2(y|x) f 1(x)
= (1/x)(1/12), 0 < y < x, 0 < x < 12 .
Therefore,
12 x
12
1 y x
E[X] = ∫∫x⋅ dydx= 12x ∫12 0
12 2
x x 12
dx= ∫ dx=
12 24 0
= 6
0 0 0 0
12 12 2 x 12
2
12
x
y ⎡ y ⎤ x x 144
E[Y] = dydx = ⎢ ⎥ dx = dx = = = 3
E[XY] =
∫∫ ∫ ∫
12x 0 0 0⎣24x⎦024 0 48 0 48
12 12 2 x 12 2 3
12
3
x
y ⎡y ⎤ x x (12)
dydx = ⎢ ⎥ dx = dx = =
∫∫ ∫ ∫
12 0
0 0 0⎣24⎦ 24 72 72
0 0
Cov(X,Y) = E[XY] – E[X]E[Y] = 24 − (3)(6) = 24 – 18 = 6 .
x
Page 45 of 55
= 24
0
=
0
4
5
=
56
45