EXAM P SAMPLE SOLUTIONS

2
[ X Y] E⎡( X Y) ⎤ ( E[ X Y]
)
8= Var + =
⎣
+ −
⎦
+
= ⎡
⎣ + + ⎤
⎦−
+
( [ ] [ ] )
2 2
E X 2XY
Y E X E Y
2 2
2
= E⎡ ⎣X ⎤
⎦+ 2E[ XY] + E⎡ ⎣Y ⎤
⎦−
( 5 + 7)
= 27.4 + 2E[ XY]
+ 51.4 −144
= 2E[ XY]
−65.2
[ ] = ( 8 + 65.2) 2 = 36.6
( CC 1, 2 ) = ( ) − =
E XY
Finally, Cov 2.2 36.6 71.72 8.8
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108. Solution: A
The joint density of 1 T2
is given by
> 0
and T
−t1 −t2
f ( t1, t2) = e e
Therefore,
, t1 > 0 , t2
Pr X ≤ x = Pr 2T
+ T ≤ x
[ ] [ ]
1 2
1
x ( x−t2) x ⎡
2 −t1 −t2 −t2 −t1
= ∫∫ e e dt1dt2 =
0 0 ∫ e ⎢−e 0
⎢⎣ 1
( x−t2) ⎤
2
⎥dt2
0 ⎥⎦
1 1 1 1
x − x+ t2 x
− x − t
−t ⎡ ⎤ ⎛ 2
2 2 2 −t
⎞
2 2 2
= ∫ e ⎢1− e ⎥dt2
= e e e dt2
0 ∫ ⎜ − ⎟
0
⎣ ⎦ ⎝ ⎠
⎡
= ⎢− e
⎣
+ e e
⎤
⎥
⎦
=− e + e e + − e
1 1 1 1 1
− x − t2− x − x
− x
−t2 2 2 x −x
2 2 2
2 0 2 1 2
1 1
− x − x
−x
−x = 1− e + 2e
− 2e 2 = 1− 2 e 2 −x
+ e , x>
0
It follows that the density of X is given by
1 1
d ⎡ − x ⎤ − x
2 −x 2 −x
g( x) = ⎢1− 2 e + e ⎥ = e − e
dx ⎣ ⎦
, x > 0
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