EXAM P SAMPLE SOLUTIONS

54. Solution: B
Let Y denote the claim payment made by the insurance company.
Then
⎧0
with probability 0.94
⎪
Y = ⎨Max
( 0, x−1
) with probability 0.04
⎪
⎩14
with probability 0.02
and
EY = 0.94 0 + 0.04 0.5003
15
− x /2
x− 1 e dx+
0.02 14
[ ] ( )( ) ( )( ) ( ) ( )(
( )
15 15
⎡ −x/2 −x/2
xe dx e dx⎤
∫1 ∫1
= 0.020012
⎢⎣ − + 0.28
⎥⎦
15 15
−x/2 15 −x/2 −x/2
= 0.28 + ( 0.020012) ⎡− 2xe | 1 + 2 e dx− e dx⎤
⎢⎣ ∫1 ∫1
⎥⎦
15
−7.5 −0.5 −x/
2
= 0.28 + ( 0.020012) ⎡− 30e + 2e
+ e dx⎤
⎢⎣ ∫1
⎥⎦
−7.5 −0.5 −x/
2 15
= 0.28 + ( 0.020012) ⎡
⎣− 30e + 2e −2e
| ⎤ 1 ⎦
= 0.28 + 0.020012
−7.5 −0.5 −7.5 −0.5
− 30e + 2e − 2e + 2e
∫
( )( )
( )( −7.5 e
−0.5
e )
= 0.28 + 0.020012 − 32 + 4
( )( )
= 0.28 + 0.020012 2.408
= 0.328 (in thousands)
It follows that the expected claim payment is 328 .
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55. Solution: C
k
The pdf of x is given by f(x) = 4
(1 + x)
, 0 < x < ∞ . To find k, note
∞
k k 1
1 = ∫ dx =− 4 3
(1 + x) 3 (1 + x)
0
k = 3
∞
0
k
=
3
∞
3x
It then follows that E[x] = ∫ dx and substituting u = 1 + x, du = dx, we see
4
(1 + x)
∞
∞
0
3 4
−2 u
−3
u
3( u −1)
− − ⎡ ⎤ ⎡1 1⎤
E[x] = ∫ du = 3 ( u − u ) du = 3⎢ − = 3 −
4
u ∫
−2 −3 ⎥ ⎢
⎣2 3⎥
= 3/2 – 1 = ½ .
1
⎣ ⎦ ⎦
1 1
1
Page 23 of 55
∞
)