Course 1 May 2000 Multiple Choice Exams - Society of Actuaries

10. Answer: C
Let T 1 be the time until the next Basic Policy claim, and let T 2 be the time until the next Deluxe
policy claim. Then the joint pdf of T 1 and T 2 is
⎛1 −t1/2 ⎞⎛1 −t2/3 ⎞ 1 −t1/2 −t2/3
f( t1, t2)
= ⎜ e ⎟⎜ e ⎟=
e e , 0 < t 1 < ∞ , 0 < t 2 < ∞ and we need to find
⎝2 ⎠⎝3 ⎠ 6
P[T 2 < T 1 ] =
∞ t1
∞
1 −t1/2 −t2/3 ⎡ 1 −t1/2 −t2
/3⎤
t1
∫∫ e e dt2dt1 = e e dt1
6
∫ ⎢ −
2 ⎥
0 0 0 ⎣ ⎦ 0
∞
∞
⎡1 −t1/2 1 −t1/2 −t1/3 ⎤ ⎡1 −t1/2 1 −5 t1/6⎤
= ⎢ e − e e dt1 = e − e dt1
0⎣2 2 ⎥
⎦
⎢
0⎣2 2 ⎥
⎦
∞
/2 5 /6⎤
− + = 1− =
−t1 − t1
∫ ∫ =
⎢
e e
⎡
⎣
3 3 2
5 ⎥
= 0.4 .
⎦ 5 5
0
11. Answer: D
We are given that the joint pdf of X and Y is f(x,y) = 2(x+y), 0 < y < x < 1 .
Now f x (x) =
x
2
x
∫ (2x + 2 y) dy = ⎡
⎣2xy + y ⎤
⎦ = 2x 2 + x 2 = 3x 2 , 0 < x < 1
0
0
f( x, y) 2( x+ y) 2⎛1
y ⎞
so f(y|x) = ( ) = 3 =
2 3 ⎜ +
2 ⎟
fx x x ⎝ x x ⎠ , 0 < y < x
2⎡
1 y ⎤ 2
f(y|x = 0.10) = [ 10 100 ]
3 ⎢
+ = + y
0.1 0.01⎥
, 0 < y < 0.10
⎣ ⎦ 3
0.05
P[Y < 0.05|X = 0.10] = ∫ 2 ⎡20 100 2 ⎤ 0.05 1 1 5
[ 10 + 100ydy ] = y y
3 ⎢
+ = + =
⎣ 3 3 ⎥
⎦ 3 12 12
= 0.4167 .
0
0
12. Answer: E
⎡ π π ⎤
We are given x = sin(t) + t, y = cos(t) – t, t ∈ ⎢
− ,
⎣ 2 2 ⎥
. We want to find the slope of the tangent
⎦
line at (x,y) = (0,1) . Therefore, note 0 = sin(t) + t fi sin(t) = - t fi t = 0 . Then
dy dy dx −sin(0) −1 1
dx/dt = cos(t) + 1, dy/dt = - sin(t) – 1. Then
( xy , ) = (0,1)
= /
t=
0= =− .
dx dt dt cos(0) + 1 2
The equation of the tangent line is given by y – 1 = (-1/2) (x – 0) or y = (–1/2)x + 1 .
13. Answer: E
r(p) = (2000 – 10p)p = 2000p – 10p 2 . The graph of this function hits the x axis twice (at p = 0
and p = 200). The derivative r¢(p) = 2000 – 20p implies that the graph in E is better than the
graph in D .
Course 1 5 May 2000