程序代写案例-ACTL 3003/5106

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The University of New South Wales
ACTL 3003/5106 Insurance Risk Models
Final Exam|Session 2, 2014
SOLUTIONS
1
Question 1
Denote by th
e aggregate loss
S =
15X
i=1
IiBi +
15X
i=1
Ii+15Bi+15:
where Iiis the indicator of making a claim.
q1 := Pr(I1 = 1) = = Pr(I15 = 1) = 0:005;
q2 := Pr(I16 = 1) = = Pr(I30 = 1) = 0:01;
andB1; B2; ; B30 are i.i.d. Gamma distributed random variables, i.e. Gamma
(; ) with = 2 and = 0:002.
We rst calculate the expectation and variance of S:
E[S] =
15X
i=1
E[Bi]Pr(Ii = 1) +
15X
i=1
E[Bi+15]Pr(Ii+15 = 1)
= 15 2
0:002
0:005 + 15 2
0:002
0:01
= 225;
and
Var[S] =
15X
i=1

q1(1 q1)(E[Bi])2 + q1Var(Bi)

+
15X
i=1

q2(1 q2)(E[Bi+15])2 + q2Var(Bi+15)

= 15

0:005 (1 0:005)

2
0:002
2
+ 0:005 2
0:0022

+15

0:01 (1 0:01)

2
0:002
2
+ 0:01 2
0:0022

= 335625:
With the retained proportion , the net premium collected by the insurer is
(1 + 1)E[S] (1 + 2)(1 )E[S] = (1 2 + + 2)E[S]:
So the probability is calculated as follows
Pr ( S (1 2 + + 2)E[S])
= Pr

S E[S]p
Var[S]
[(1 2)= + 2]E[S]p
Var[S]
!
Pr

Z [(1 2)= + 2]E[S]p
Var[S]
!
:
2
To make the above probability no less than 0:4325 (use the standard normal
table), we should set
[(1 2)= + 2]E[S]p
Var[S]
0:17
Therefore,
2 1
2 0:17
p
Var[S]=E[S]
= 0:2163:
Question 2
a. For column II, the maximum value is 8.5, which is not the minimum value
in its row. So 8.5 is not a saddle point.
For column I, the maximum value is
maxf5:5; Y g =
(
5:5; if Y < 5:5;
Y; if Y > 5:5:
However, the rst case 5:5 is not the minimum value in its row. For the
second case, only when 5:5 < Y < 6:5, Y is the minimum value of its row
and hence a saddle point.
For column III, the maximum value is
maxf6:5; Xg =
(
6:5; if X < 5:5;
X; if X > 5:5:
However, the second case X > 5:5 is not the minimum value in its row.
For the second case, only when Y > 6:5, 6:5 is the minimum value of its
row and hence a saddle point.
So the event that there exists two saddle points can be denoted by
f5:5 < Y < 6:5g \ (fX > 5:5g \ fY > 6:5g) = ;;
Therefore, its probability is zero.
b. Similarly, the event that there exists at least one saddle points can be
denoted by
f5:5 < Y < 6:5g [ (fX > 5:5g \ fY > 6:5g):
3
Therefore, the probability of this event can be calculated as
Pr
f5:5 < Y < 6:5g [ (fX > 5:5g \ fY > 6:5g)
= Pr
f5:5 < Y < 6:5g) + Pr(fX > 5:5g \ fY > 6:5g
Pr(f5:5 < Y < 6:5g \ fX > 5:5g \ fY > 6:5g
= Pr
f5:5 < Y < 6:5g) + Pr(fX > 5:5g \ fY > 6:5g
= Pr
f5:5 < Y < 6:5g) + Pr(fX > 5:5g) PrfY > 6:5g
= Pr

Y = 6) + Pr(X = 6; 7; 8) PrY = 7; 8
=

8
6

0:256 0:752 + 3
9


8
7

0:257 0:75 +

8
8

0:258

= 0:003972:
4
Question 3
a. The moment generating functions of S1 and S2 are given by
mS1(t) = expf1[mX1(t) 1]g;
and
mS2(t) = expf2[mX2(t) 1]g;
The moment generating function of SA and SB can be calculated as
mSA(t) = qmS1(t) + (1 q)mS2(t):
and
mSB (t) = expf1[mX1(qt) 1] + 2[mX2((1 q)t) 1]g:
b. From (a), the moment generating function of SB can be written as
mSB (t) = expf[mX(t) 1]g;
where = 1 + 2 and mX(t) is the moment generating function corre-
sponding to
p(x) =
1

p1(
x
q
) +
2

p2(
x
1 q ):
So SB follows the compound Poisson distribution with and p(x). How-
ever, mSA(t) can not be written as the general parametric form of the
mgf of a compound Poisson distribution and hence SA is not a compound
Poisson distribution.
c. Although SA is not a compound Poisson distribution, we can apply the
law of total probability rst
fSA(s) = fSA(sjI = 1)Pr(I = 1) + fSA(sjI = 0)Pr(I = 0)
= qfS1(s) + (1 q)fS2(s):
Then using Panjer's algorithm to fSi(s) gives
fS1(0) = e
1(01) = 0:36788;
fS1(1) =
1
1
1 1 1
6
fS1(0) = 0:061313;
fS1(2) =
1
2
(1 1 1
6
fS1(1) + 1 2
1
6
fS1(0)) = 0:066423;
and
fS2(0) = e
2( 171) = 0:18009;
fS2(1) =
1
1
1 1 1
7
fS1(0) = 0:025727;
fS2(2) =
1
2
(1 1 1
7
fS1(1) + 1 2
1
7
fS1(0)) = 0:027565;
5
Therefore,
fSA(2) = 0:5fS1(2) + 0:5fS2(2) = 0:046994:
d. For SA, the probability mass functions at s = 0; 1; 2 are given by
fSA(0) = 0:5fS1(0) + 0:5fS2(0) = 0:27399;
fSA(1) = 0:5fS1(1) + 0:5fS2(1) = 0:04352;
fSA(2) = 0:5fS1(2) + 0:5fS2(2) = 0:046994:
Then the distribution functions at s = 0; 1; 2 are
FSA(0) = 0:27399;
FSA(1) = 0:27399 + 0:04352 = 0:31751;
FSA(2) = 0:31751 + 0:046994 = 0:36450:
Using the recursive formulas gives
E[I0] = E[SA]
= qE[S1] + (1 q)E[S2]
= 0:5 1 1
6
(1 + 2 + 6) + 0:5 2 1
7
(0 + 1 + 2 + 6)
= 4:75;
and
E[I1] = E[I0] (1 FSA(0)) = 4:75 (1 0:27399) = 4:02399;
E[I2] = E[I1] (1 FSA(1)) = 4:02399 (1 0:31751) = 3:3415;
E[I2:5] = E[I2] (2:5 2)(1 FSA(2)) = 3:3415 0:5(1 0:36450) = 3:02375;
Question 4.
The development factors are
7;982
7;638 = 1:04504,
7;638+8;028
5;496+5;162 = 1:46988
1 1f = 1 11:045041:46988 = 0:3490.
Hence, emerging liability for 2012 is 10; 024 0:85 0:349 = 2; 974.
As the reported claims are 6,434, the ultimate liability is 2; 974 + 6; 434 =
9; 408.
6
As a result, reserve for 2012=9,408-2,944=6,464.
Question 5.
(a) (i) The Bayesian point estimator under quadratic loss is the posterior mean:
E(j(y1; ; yn)) =
2
1 + (
Pn
i=1 yi)
2
2
21 + n
2
2
= (1 Z)+ Z(
Pn
i=1 yi
n
);
where Z =
n22
21+n
2
2
is the credibility factor.
(ii) Z =
n22
21+n
2
2
= 0:8889
Credibility premium=Zy + (1 Z) = 244:9708
Premium= 869:4 (1 + 25%) = 306:21
(iii) Decrease 22 will decrease the credibility factor and hence the credibility
premium moves towards prior mean.
(b)
0:4 = Z =
4
4 + 8a
) a = 4
3
Cov(X1; X3) = E(X1X3) E(X1)E(X3)
= E[E(X1X3j)] E[E(X1j)]E[E(X3j)]
= E[(E(X1j))2] E[(E(X1j))]2
= V ar(E(X1j)) = a = 4
3
Question 6.
^ = x = 12=60 = 0:2
v^ = x = 0:2
a^ =
10(0:4 0:2)2 + 20(0:25 0:2)2 + 30(0:1 0:2)2 (3 1)0:2
60 (102 + 202 + 302)=60 = 0:009545
k =
0:2
a^
= 20:9524
Z =
10
10 + 20:9524
= 0:323
Question 7. (a) A (b) B (c) B
End of Paper
7

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