2005
EXAM FALL C SOLUTIONS
Question #1
Key: D
S (300) = 3 /10 (there are three observations greater than 300)
H (300) = ln[ S (300)] = ln(0.3) = 1.204 .
Question #2
Key: A
E ( X | ) = Var ( X | ) =
= v = E ( ) = ; a = Var ( ) = 2 ; k = v / a = 1/
n
n
Z=
=
n + 1/ n + 1
+
1
0.15 =
(1) +
=
+1
+1
+1
2
1
4 +
0.20 =
(2) +
=
2 + 1
2 + 1
2 + 1
From the first equation,
0.15 + 0.15 = + and so = 0.15 0.85
Then the second equation becomes
0.4 + 0.2 = 4 + 0.15 0.85
0.05 = 2.75 ; = 0.01818
Question #3
Key: E
0.75 =
1
1
; 0.25 =
1 + (100 / )
1 + (500 / )
(100 / ) = 1/ 3; (500 / ) = 3
Taking the ratio of these two equalities produces 5 = 9 . From the second equality,
9 = [(500 / )2 ] = 5 ; (500 / ) 2 = 5; = 223.61
Question #4
Key: B
f ( x) = a + bx + cx 2 + dx 3
0 = f (0) = a
2 = f (2) = a + 2b + 4c + 8d
1 = f '(0) = b
24 = f ''(2) = 2c + 12d ; c = 12 6d
Insert the values for a, b, and c into the second equation to obtain
2 = 2 + 4(12 6d ) + 8d ; 48 = 16d ; d = 3
Then c = 6 and f ( x) = x + 6 x 2 3x3 ; f (1) = 4
Question #5
Key: E
Begin with
y
s
r
350
2
10
500
2
8
1000 1200 1500
1
1
1
5
2
1
8 641
Then S1 (1250) =
= 0.24
10 8 5 2
The likelihood function is
2
2
2
L( ) = 1e 350 / 1e500 / e500 / 1e 1000 / e 1000 / 1e 1200 / 1e1500 /
7 7900 /
= e
l ( ) = 7 ln
7900
; l '( ) =
S 2 (1250) = e 1250(7) / 7900 = 0.33
The absolute difference is 0.09.
7
+
7900
2
= 0; = 7900 / 7
Question #6
Key: E
f ( x) = S '( x) =
4 x 4
( 2 + x 2 )3
L( ) = f (2) f (4) S (4) =
4
4(2) 4
4(4) 4
128 12
=2
( 2 + 22 )3 ( 2 + 42 )3 ( 2 + 42 ) 2 ( + 4)3 ( 2 + 16)5
l ( ) = ln128 + 12 ln 3ln( 2 + 4) 5ln( 2 + 16)
12
6
10
2
= 0;12( 4 + 20 2 + 64) 6( 4 + 16 2 ) 10( 4 + 4 2 ) = 0
l '( ) = 2
+ 4 + 16
0 = 4 4 + 104 2 + 768 = 4 26 2 192
26 262 + 4(192)
=
= 32; = 5.657
2
2
Question #7
Key: A
2x
0
2
E( X | ) = x
dx =
2
2x
4 2 2 4 2 2
; Var ( X | ) = x 2 2 dx
=
=
0
3
9
2
9
18
1
= (2 / 3) E ( ) = (2 / 3) 4 4 d = 8 /15
0
1
EVPV = v = (1/18) E ( 2 ) = (1/18) 4 5 d = 1/ 27
0
VHM = a = (2 / 3) Var ( ) = (4 / 9) 4 / 6 (4 / 5) 2 = 8 / 675
2
1/ 27
1
= 25 / 8; Z =
= 8 / 33
8 / 675
1 + 25 / 8
Estimate is (8 / 33)(0.1) + (25 / 33)(8 /15) = 0.428.
k=
Question #8
Key: D
From the Poisson(4) distribution the probabilities at 0, 1, and 2 are 0.0183, 0.0733, and 0.1463.
The cumulative probabilities are 0.0183, 0.0916, and 0.2381. Because 0.0916 < 0.13 < 0.2381
the simulated number of claims is 2. Claim amounts are simulated from solving
u = 1 e x /1000 for x = 1000 ln(1 u ) . The two simulated amounts are 51.29 and 2995.73 for a
total of 3047.02
Question #9
Key: B
It may be easiest to show this by graphing the density functions. For the first function the three
components are each constant. One is of height 1/20 from 0 to 2 (representing the empirical
probability of 1/10 at 1, one is height 1/20 from 1 to 3 and one is height 8/20 from 2 to 4. The
following figure shows each of them and their sum, the kernel density estimator.
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
f1
f2
f3
f
0
1
2
3
4
The triangular one is similar. For the triangle from 0 to 2, the area must be 1/10. With a base of
2, the height is 1/10. the same holds for the second triangle. The third has height 8/10. When
added they look as follows;
0.9
0.8
0.7
0.6
f1
0.5
f2
0.4
f3
0.3
f
0.2
0.1
0
0
1
2
3
4
The question asks about cumulative probabilities. From 0 to 1 the first is linear and the second is
quadratic, but by x = 1 both have accumulated 0.05 of probability. Because the cumulative
distribution functions are the same at 1 and the density functions are identical from 1 to 2, the
distribution functions must be identical from 1 to 2.
Question #10
Key: D and E
For the Poisson distribution, the mean, , is estimated as 230/1000 = 0.23.
# of Days
Poisson
Probability
0.794533
0.182743
0.021015
0.001709
0
1
2
3 or more
Total
Expected # of
Workers
794.53
182.74
21.02
1.71
Observed # of
Workers
818
153
25
4
1000
2
0.69
4.84
0.75
3.07
9.35
The 2 distribution has 2 degrees of freedom because there are four categories and the Poisson
parameter is estimated (d.f. = 4 1 1 = 2).
The critical values for a chi-square test with two degrees of freedom are shown in the following
table.
Significance Level
10%
5%
2.5%
1%
Critical Value
4.61
5.99
7.38
9.21
9.35 is greater than 9.21 so the null hypothesis is rejected at the 1% significance level.
Question #11
Key: D
25(480 472.73) 2 + 30(466.67 472.73) 2
= 2423.03 where 480 = 12,000/25,
2 1
466.67 = 14,000/30, and 472.73 = 26,000/55.
55
k = 2423.03 / 254 = 9.54; Z =
= 0.852
55 + 9.54
EVPV = v =
Question #12
Key: C
1
2
0 < x <1
1.6 x,
f ''( x) =
S = (1.6 x) 2 dx + (2.4 0.8 x) 2 dx = 2.56
0
1
1.6 0.8( x 1) = 2.4 0.8 x, 1 < x < 3
Question #13
Key: C
Relative risk = e 1 2
which has partial derivatives e0.2 at 1 = 0.05 and 2 = 0.15
Using the delta method, the variance of the relative risk is
0.2 7e 0.4
1
0.2
0.2 2 1 e
e
e
= 0.000469
=
10, 000
1 3 e 0.2 10, 000
Std dev = 0.0217
upper limit = e 0.2 + 1.96 ( 0.0217 )
(
)
= 0.8613
Alternatively, consider the quantity 1 + 2 . The variance is
2 1 1
1
7
= 0.0007 . The lower limit for this quantity is
(1 1)
=
10, 000
1 3 1 10, 000
0.2 1.96 0.0007 = 0.1481 and the upper limit for the relative risk is e 0.1481 = 0.8623 .
Question #14
Key: C
ln 5000
The quantity of interest is P = Pr( X 5000) =
. The point estimate is
6.84
ln 5000
= (1.125) = 0.87 .
1.49
For the delta method:
P (1.125)
P 1.125 (1.125)
1 z2 / 2
=
= 0.1422;
=
= 0.1600 where ( z ) =
e
.
1.49
1.49
2
Then the variance of P is estimated as (0.1422) 2 0.0444 + (0.16) 2 0.0222 = 0.001466 and the
lower limit is PL = 0.87 1.96 0.001466 = 0.79496 .
Question #15
Key: A
Pr( = 0.1| X 1 = 1) =
Pr( X 1 = 1| = 0.1) Pr( = 0.1)
Pr( X 1 = 1| = 0.1) Pr( = 0.1) + Pr( X 1 = 1| = 0.3) Pr( = 0.3)
0.1(0.8)
4
=
0.1(0.8) + 0.3(0.2) 7
Then,
E ( X 2 | = 0.1) = 0(0.2) + 1(0.1) + 2(0.7) 1.5
=
E = ( X 2 | = 0.3) = 0(0.6) + 1(0.3) + 2(0.1) = 0.5
4
3
E ( X 2 | X 1 = 1) = (1.5) + (0.5) = 1.071
7
7
Question #16
Key: D
The requirement is that
F (1500) S (1500)
0.01F (1500) 1.96
N
2
P
P( N P)
0.0001 2 3.8416
N
N3
NP
38, 416.
N P
For the five answer choices, the left hand side is 34,364, 15,000, 27,125, 39,243, and 37,688.
Only answer D meets the condition.
Question #17
Key: D
^
^
s4
= H ( y 4 ) H ( y 3 ) = 0.5691 0.4128 = 0.1563 .
r4
s4
^^
^^
= V [ H ( y 4 )] V [ H ( y 3 )] = 0.014448 0.009565 = 0.004883 .
2
r4
( s4 / r4 ) 2
0.15632
=
= 5.
Therefore, s4 =
s4 / r42
0.004833
Question #18
Key: A
ln f ( x) = ln 2 ln( + x)
ln f ( x) 1
2
=
+x
2
ln f ( x)
1
2
= 2 +
2
( + x) 2
2 ln f ( x)
1
2
1
2
1
2
1
E
dx = 2 +
= 2 + 2 = 2
= 2 + 0
2
4
3
3( + x) 0
3
( + x)
3
I ( ) =
n
3 2
; Var =
3 2
n
Question #19
Key: B
= E[ E ( X | )] = E ( ) = 1(0.9) + 10(0.09) + 20(0.01) = 2
EVPV = v = E[Var ( X | )] = E ( ) = 2
VHM = a = Var[ E ( X | )] = Var ( ) = 1(0.9) + 100(0.09) + 400(0.01) 22 = 9.9
Z=
1
= 0.83193; 11.983 = 0.83193 x + 0.16807(2); x = 14
1 + 2 / 9.9
Question #20
Key: A
The given interval for H can be written as 0.775 1.96 0.063 and therefore the estimated
variance of H is 0.063. To apply the delta method,
dS
= e H ; Var ( S ) (e H ) 2 Var ( H ) = (e0.775 ) 2 (0.063) = 0.134 .
S = e H ;
dH
The point estimate of S is e 0.775 = 0.4607 and the confidence interval is
0.4607 1.96 0.0134 = 0.2269 or (0.23, 0.69).
Question #21
Key: B
The first step is to trend the year 1 data by 1.21 and the year 2 data by 1.1. The observations are
now 24.2, 48.4, 60.5, 33, 44, 99, and 132.
The first two sample moments are 63.014 and 5262.64. The equations to solve are
2
63.014 = e + 0.5 ; 4.14336 = + 0.5 2
5262.64 = e 2 + 2 ; 8.56839 = 2 + 2 2 .
Taking four times the first equation and subtracting the second gives 2 and therefore
4(4.14336) 8.56839
=
= 4.00 .
2
2
Question #22
Key: A
= x = 12 / 60 = 0.2, EVPV = v = x = 0.2
10(0.4 0.2) 2 + 20(0.25 0.2) 2 + 30(0.1 0.2) 2 (3 1)(0.2)
= 0.009545
102 + 202 + 302
60
60
10
= 0.323
k = 20.9524; Z =
10 + 20.9524
VHM = a =
Question #23
Key: E
By elimination, (A) is incorrect because f ''(3) = 1.833 0 , (B) is incorrect because
f ''(0) = 2 0 , (C) is incorrect because f ''(0) = 1 0 , and (D) is incorrect because
f ''(0) = 1 0 . Therefore (E) must be correct. Also, this function does meet all the
requirements:
f ''(0) = 0; f 0 (1) = f1 (1) = 2; f 0' (1) = f1' (1) = 0; f 0'' (1) = f1'' (1) = 3; f (3) = 6; f ''(3) = 0
Question #24
Key: B and C
For males, c j = 1 and for females, c j = e0.27 = 1.31 . Then,
H (20) =
1.31
1
1
1
+
+
= 0.6965 and S female (20) = ( e 0.6965 ) = 0.402 .
3 + 2(1.31) 2 + 2(1.31) 2 + 1.31
.
Question #25
Key: C
5
5
j =1
j =1
l ( , ) = ln f ( x j ) = ln + ( 1) ln x j ln ( x j / ) . Under the null hypothesis it is
5
l (2, ) = ln 2 + ln x j 2 ln ( x j / ) 2 . Inserting the maximizing value of 816.7 for gives
j =1
35.28 . The likelihood ratio test statistic is 2( 33.05 + 35.28) = 4.46 . There is one degree of
freedom. At a 5% significance level the critical value is 3.84 and at a 2.5% significance level it
is 5.02.
Question #26
Key: C
It is given that n = 4, v = 8, and Z = 0.4. Then, 0.4 =
4
8
4+
a
which solves for a = 4/3. For the
covariance,
Cov( X i , X j ) = E ( X i X j ) E ( X i ) E ( X j )
= E[ E ( X i X j | )] E[ E ( X i | )]E[ E ( X j | )]
= E[ ( ) 2 ] E[ ( )]2 = Var[ ( )] = a = 4 / 3.
Question #27
Key: A
u
0.6217
0.9941
0.8686
0.0485
z
0.31
2.52
1.12
1.66
x
5.8325
7.49
6.44
4.355
lognormal
341.21
1790.05
626.41
77.87
Average
with deductible
241.21
1690.05
526.41
0
614.42
The value of z is obtained by inversion from the standard normal table. That is, u = Pr( Z z ) .
The value of x is obtained from x = 0.75 z + 5.6 . The lognormal value is obtained by
exponentiating x and the final column applies the deductible.
Question #28
Key: B
MSE = E[( X 2 2 ) 2 ] = E ( X 4 2 X 2 2 + 4 )
= 24 4 2(2 2 ) 2 + 4 = 21 4
Question #29
Key: C
157(0) + 66(1) + 19(2) + 4(3) + 2(4)
= 0.5 is the maximum likelihood
248
estimate of the geometric parameter as well as the method of moments estimate of the Poisson
parameter . Then, P = (1 + 0.5) 1 = 0.6667 and Q = e 0.5 = 0.6065 . The absolute difference is
0.0602.
The sample mean of
Question #30
Key: D
5000(0) + 2100(1) + 750(2) + 100(3) + 50(4)
= 0.5125 and
8000
5000(0.5125) 2 + 2100(0.4875) 2 + 750(1.4875) 2 + 100(2.4875) 2 + 50(3.4875)2
s2 =
= 0.5874 .
7999
Then, = v = x = 0.5125 and a = s 2 x = 0.0749 . The credibility factor is
1
Z=
= 0.1275 and the estimate is 0.1275(1) + 0.8725(0.5125) = 0.5747.
1 + 0.5125 / 0.0749
x=
Question #31
Key: B
s = Fn (3000) = 4 / 8 = 0.5 because for the p-p plot the denominator is n+1.
t = F (3000) = 1 e 3000 / 3300 = 0.59711 . For the difference plot, D uses a denominator of n and so
D = 4 / 7 0.59711 = 0.02568 and the answer is 0.5 0.59711 + 0.02568 = 0.071.
Question #32
Key: B
(q | 2, 2) f (2 | q ) f (2 | q) (q) = q(q)(q 2 / 0.039) q 4 . Because
0.5
0.2
q 4 dq = 0.006186 ,
(q | 2, 2) = q 4 / 0.006186 . Given q, the expected number of claims is
E ( N | q) = 0(0.1) + 1(0.9 q ) + 2q = 0.9 + q . The Bayesian estimate is
0.5
E ( N | 2, 2) = (0.9 + q)
0.2
q4
dq = 1.319.
0.006186
Question #33
Key: E
200 + 110 310 + x
0.689 = F500 (1500) = 0.5 F500 (1000) + 0.5 F500 (2000) = 0.5
+
x = 69
500
500
310 + 69 379 + y
0.839 = F500 (3500) = 0.5F500 (2000) + 0.5 F500 (5000) = 0.5
+
y = 81
500
500
Question #34
Key: A
A is false because the test works best when the expected number of observations is about the
same from interval to interval. B is true (Loss Models, 427-8), C is true (Loss Models, 428), and
D is true (Loss Models, 430).
Question #35
Key: E
2
2
n 0 1 + Y ; Y = = 10, 000 ; Y = 2 = 108
Y
1.96
n
0.1
2
108
1
1 + 108 2 = 384.16(1 + )
Because is needed, but not given, the answer cannot be determined from the information
given.