FINA 3222A Assignment 2
Due: 11:59 PM, October 17, 2022
To be submitted through Blackboard
Instructions
• If working is not specifically requested, a correct answer will receive full marks. However,
an incorrect answer could still receive partial credits if working is provided. An incorrect
answer with no working will receive zero marks.
• Your final numerical answers should be given to 4 decimal places, e.g., 1.2345 or 0.00004567.
• Lateness Penalty: 20% of the total marks of the assignment deducted for every 8-hour
period late.
Questions
Question 1. [11 pts] Let N be the logarithmic rv with pmf
pk =
βk
,
k(1 + β)k ln(1 + β)
k = 1, 2, ...
with parameter β = 10 and p0 = 0. It is also given that E [N ] =
β
β (1+β− ln(1+β)
)
. Consider its zero-modified version N M with pM
0 = 0.1.
ln(1+β)
β
ln(1+β)
and Var(N ) =
(a) [4 pts] Find P N M > 2 .
(b) [3 pts] Find E N M . (c) [4 pts] Find Var N M .
Solution. (a) Since
pM
k =
1 − pM
10k
0
pk = 0.9pk = 0.9
, (1 mark)
1 − p0
k · 11k ln 11
k = 1, 2, . . . ,
it follows that
pM
1 = 0.9
10
102
= 0.3412 (1 mark) and pM
=
0.9
= 0.1551. (1 mark)
2
11 ln 11
2 · 112 ln 11
1
Thus,
M
M
P N M > 2 = 1 − pM
0 − p1 − p2 = 1 − 0.1 − 0.3412 − 0.1551 = 0.4037. (1 mark)
(b) Since pM
k = 0.9pk for k = 1, 2, . . ., we have
∞
M X
E N
=
kpM
k (1 mark)
=
k=0
∞
X
kpM
k
k=1
= 0.9
∞
X
kpk (1 mark)
k=1
= 0.9E [N ] = 0.9
10
= 3.7533 (1 mark) .
ln 11
(c) Similarly,
h
E
N
M 2
i
=
∞
X
k 2 pM
k (1 mark)
k=1
= 0.9E N 2 (1 mark)
= 0.9
10(11 − ln1011 )
+
ln 11
10
ln 11
2 !
= 41.2862, (1 mark)
and
Var N M = 41.2862 − (3.7533)2 = 27.1989. (1 mark)
2
Question 2. [12 pts] The density function of the ground-up loss rv X is given by
x
1
1
q+
(1 − q) x e− 100 , x > 0,
fX (x) =
100
100
where 0 < q < 1. With an ordinary deductible of 100, the expected value of amount paid per
payment is known to be 125. Determine the expected value of amount paid per payment if the
deductible level is adjusted to 200.
Solution.
First, we shall determine the value of q using the information that E [YP ] = E [X − d|X > d] = 125
when d = 100. The survival function of X is
Z ∞
fX (y)dy
F X (x) =
x
Z ∞
Z
1
q
(1 − q) ∞ − 1 y
y
− 100
=
e
dy +
ye 100 dy (1 mark)
100 x
1002 x
1
x − 1 x
=qe− 100 x + (1 − q) 1 +
e 100
100
1
x − 1 x
e 100 , x > 0. (1 mark)
=e− 100 x + (1 − q)
100
For y ≥ 0, since YL = (X − d)+ , the survival function of YL is
y+d
F YL (y) = F X (y + d) = e− 100 + (1 − q)
y + d − y+d
e 100 ,
100
y > 0. (1 mark)
This implies
∞
y + d − y+d
− y+d
100
100
+ (1 − q)
E [YL ] =
e
dy (1 mark)
e
100
0
Z ∞
Z ∞
x
x −x
− 100
e
=
dx + (1 − q)
e 100 dx (1 mark)
100
d
d
Z
d
d
=100e− 100 + (1 − q) (d + 100) e− 100 , (1 mark)
and
E [YP ] =
E [YL ]
(1 mark)
F YL (0)
d
=
d
100e− 100 + (1 − q) (d + 100) e− 100
d
d
d − 100
e− 100 + (1 − q) 100
e
100 + (1 − q) (d + 100)
=
. (1 mark)
d
1 + (1 − q) 100
When d = 100,
125 = E [YP ] =
(1 mark)
300 − 200q
, (1 mark)
2−q
3
which implies
2
q = . (1 mark)
3
Hence, when d = 200,
E [YP ] =
100 + 31 (200 + 100)
= 120. (1 mark)
1 + 13 200
100
Question 3. [5 pts] The cdf of a ground-up loss is given by
x
3
FX (x) = 1 − e−( 100 ) ,
x ≥ 0.
The number of losses NL has a negative binomial distribution with r = 2 and β = 1.5. Suppose
that an ordinary deductible of 50 is applied to each individual loss. Determine the distribution of
the number of payments NP .
Solution.
From the pgf relationship between NL and NP is given by
GNP (t) = GNL (1 − α + αt) ,
where
50
3
α = F X (50) = e−( 100 ) = 0.8825. (1 mark)
Given that GNL (t) = (1 − 1.5(t − 1))−2 , it follows that
GNP (t) = GNL (1 − α + αt)
= (1 − 1.5(1 − α + αt − 1))−2 (1 mark)
= (1 − 1.5α(t − 1))−2
= (1 − 1.3237 (t − 1))−2 . (1 mark)
Therefore, NP has a negative binomial distribution (1 mark) with r = 2 (0.5 marks) , and
β = 1.3237 (0.5 marks) .
4
Question 4. [6 pts] A towing company provides all towing services to members of the City
Automobile Club. You are given:
Towing distance (in miles)
0-9.99
10-29.99
30+
Towing cost
80
100
160
Probability
50%
40%
10%
• The automobile owner must pay 10% of the cost and the remainder is paid by the City
Automobile Club.
• The number of towing has a Poisson distribution with mean of 1000 per year.
• The number of towings and the costs of individual towings are all mutually independent.
Using the normal approximation for the distribution of aggregate towing costs, calculate the
probability that the City Automobile Club pays more than 90,000 in any given year.
Solution.
Let X be the individual towing cost. Then
E[X] = 80(0.5) + 100(0.4) + 160(0.1) = 96, (1 mark)
and
Var(X) = (80 − 96)2 (0.5) + (100 − 96)2 (0.4) + (160 − 96)2 (0.1) = 544. (1 mark)
P
Let S = N
i=1 Xi , where N ∼ P oisson(1000). Then
E[S] = 96(1000) = 96, 000 (1 mark)
and
Var(S) = 1000(544) + 1000 ∗ 962 = 9, 760, 000. (1 mark)
Using normal approximation, the aggregate towing cost is approximated by
S̃ ∼ N ormal(96, 000, 9, 760, 000).
Then the probability that the City Automobile Club pays more than 90,000 can be approximated
by
P(0.9S̃ > 90, 000) (1 mark)
100, 000 − 96, 000
S̃ − 96, 000
√
=1 − P(S̃ ≤ 100, 000) = 1 − P( √
≤
)
9, 760, 000
9, 760, 000
=1 − Φ(1.2804) = 0.1002. (1 mark)
(Alternatively, one can find the mean and variance for the aggregate costs paid by the club
first, then apply the normal approximation.)
5
Question 5. [7 pts] For a given insurance portfolio, the number of losses N is assumed to have
pmf
0.04,
n = 0,
pn =
3(2+n)(1+n) 4 n
, n = 1, 2, 3, . . .
775
5
(a) [4 pts] Show that N is an (a, b, 1) member. Identify the values of a and b.
(b) [3 pts] Argue that N is a zero-modified negative binomial (r = 3, β = 4) with a probability
mass at 0 of 0.04.
Solution.
(a) We shall prove that
b
pn
=a+
pn−1
n
for n = 2, 3, . . . It follows that
pn
=
pn−1
=
3(2+n)(1+n) 4 n
775
5
3(1+n)(n) 4 n−1
775
5
4
(2 + n) 5
(1 mark)
n 2
4
=
+1
n
5
8
4
= + 5 , (1 mark)
5 n
for n = 2, 3, . . . (1 mark) , which implies that a = 54 (0.5 marks) and b = 85 (0.5 marks) .
(b) If N is a zero-modified negative binomial (r = 3, β = 4) rv with a probability mass at 0 of
0.04, its pmf is such that
p0 = P(N = 0) = 0.04 (0.5 marks)
and
pn = P(N = n)
n 3
1 − 0.04 3 + n − 1
4
1
=
(1.5 marks)
3
3−1
5
5
1 − 15
n
4 1 − 100
4
1
2+n
=
1
2
5
125
1 − 125
n
96
(2 + n)(1 + n) 4
= 100
124
2
5
n
3
4
=
(2 + n)(1 + n)
775
5
(1 mark for showing that the pmf of the zero-modified negative binomial reduces to pn for n ≥ 1.)
for n = 1, 2, ...,which is consistent with the above.
6
Question 6. [11 pts] Suppose the ground-up loss X ∼ Exp(500) (mean is 500). The amount
paid per loss after some policy adjustment is given by

X ≤ 100,
 0,
X − 100,
100 < X ≤ 1000,
YL =

0.8X + 100, X > 1000.
(a) [5 pts] Find the survival function F YL (y) of amount paid per loss YL for all y ∈ R.
(b) [1 pt] Determine the value of k ∈ R such that
YL = (X − 100)+ + k(X − 1000)+ .
(c) [5 pts] Calculate the loss elimination ratio.
Solution. (a)
F YL (y) = P {YL > y}

y < 0,
 1,
P {X − 100 > y} ,
0 ≤ y < 900,
=

P {0.8X + 100 > y} , y ≥ 900,

y < 0,
 1,
F X (y + 100),
0 ≤ y < 900,
=

F X (1.25y − 125), y ≥ 900,

y < 0, (1 mark)
 1, y+100
− 500
(1 mark) ,
0 ≤ y < 900 (1 mark) ,
e
=
 − 1.25y−125
500
e
(1 mark) , y ≥ 900 (1 mark) .
(b) It is easy to see from the expression of YL that
YL = (X − 100)+ − 0.2(X − 1000)+ ,
i.e., k = −0.2. (1 mark)
(c) Since
E[YL ] = E([X − 100)+ ] − 0.2E [(X − 1000)+ ]
Z ∞
Z ∞
x
x
− 500
=
e
dx (1 mark) − 0.2
e− 500 dx (1 mark)
100
1000
−0.2
−2
= 500e
− 0.2 · 500e
= 395.8318, (1 mark)
the loss elimination ratio
LER =
E[X − YL ]
E[X]
7
E[YL ]
(1 mark)
E[X]
395.8318
=1−
= 0.2083. (1 mark)
500
=1−
8
Formula Sheet
• Exponential distribution with parameter θ > 0. For x > 0 and k = 1, 2, . . .,
x
1
f (x) = e−x/θ , F (x) = 1 − e− θ , E X k = θk k!.
θ
• Gamma distribution with parameters α, θ > 0. For x > 0 and k = 1, 2, . . .,
f (x) =
where Γ(α) =
R∞
0
(x/θ)α e−x/θ
, E X k = θk (α + k − 1) · · · α,
xΓ(α)
xα−1 e−x dx.
• Pareto distribution with parameters α, θ > 0. For x > 0,
α
k
θk k!
αθα
θ
f (x) =
,
F
(x)
=
1
−
X
.
,
E
=
(x + θ)α+1
x+θ
(α − 1) · · · (α − k)
• Beta distribution with parameters (a, b, 1). For x ∈ (0, 1),
Γ(a + b) a
1
u (1 − u)b−1 , 0 < x < θ,
Γ(a)Γ(b)
x
k
θ Γ(a + b)Γ(a + k)
E Xk =
, k > −a
Γ(a)Γ(a + b + k)
f (x) =
u = x/θ,
• Poisson distribution with parameter λ > 0. For k = 0, 1, 2 . . . ,
pk =
λk e−λ
, E [N ] = Var(N ) = λ, G(t) = eλ(t−1) .
k!
• Geometric distribution with parameter β > 0. For k = 0, 1, 2 . . . ,
pk =
βk
, E [N ] = β, Var(N ) = β(1 + β), G(t) = [1 − β(t − 1)]−1 .
(1 + β)k+1
• Binomial distribution with parameters m ∈ N and q ∈ (0, 1). For k = 0, 1, . . . , m,
m k
pk =
q (1 − q)m−k , E [N ] = mq, Var(N ) = mq(1 − q), G(t) = [1 + q(t − 1)]m .
k
• Negative binomial distribution with parameters β, r > 0. For k = 0, 1, 2 . . . ,
r(r + 1) · · · (r + k − 1)β k
pk =
, E [N ] = rβ, Var(N ) = rβ(1 + β), G(t) = (1 + β − βt)−r .
r+k
k!(1 + β)
9