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### Transcript of Aggregate Loss Models Chapter 9 - University of farhadi/ASPER/Aggregate Loss...

• Aggregate Loss ModelsChapter 9

Note. Here is a proof for E[X1+ +XN |N = n] = E[X1+ +Xn] = nE(X1) : We prove this

for E[S|N = 2] = 2E(X1). First of all let be the common support of the random variables

{X1, ..., Xn} (these variables are identically distributed and so have identical density function

and so have identical support).

Next Step. Note that on the event {N = 2} the random variable S becomes X1 + X2 so it

takes on all values s = x1 + x2 where x1 and x2 . So the support of the conditional

random variable S|{N = 2} is the set = {s = x1 + x2 : x1 , x2 }.

Next Step. For s we have

P{S = s |N = 2} = P{S = x1 + x2 |N = 2} = P{X1 +X2 = x1 + x2 |N = 2}

=P (X1 +X2 = x1 + x2 , N = 2)

P (N = 2)=

P (X1 +X2 = x1 + x2)P (N = 2)

P (N = 2)= P (X1+X2 = x1+x2)

Next Step.

E(S|N = 2) =s

s P (S = s|N = 2) =x1

x2

(x1 + x2)P (X1 +X2 = x1 + x2)

= E(X1 +X2) = E(X1) + E(X2) = 2E(X1)

Theorem. Let {X1, X2, ...} be an independent and identically distributed sample from a claim.

If the number of claims N is independent of the claim amount , then

E(S) = E(N)E(X1)

Proof.

E(S) = E(N

j=1Xj)

=

k=0 P (N = k)E(N

j=1Xj |N = k)

=

k=0 P (N = k)E(k

j=1Xj) independence is used here - check it out

=

k=0 P (N = k) k E(X1)

=(

k=0 k P (N = k))E(X1) = E(N)E(X1)

1

• Theorem. Let {X1, ..., Xn} be an independent and identically distributed sample from a claim.

If the number of claims N is independent of the claim amount , then

(i) E(N

i=1Xi|N) = N E(X1)

(ii) Var(N

i=1Xi|N) = N Var(X1)

(iii) Var(N

i=1Xi) = E(N)Var(X1) + Var(N)E(X1)2 compound variance formula

Proof of (i). For each fixed value n of N we have

E(Ni=1

Xi|N)(n) = E(Ni=1

Xi|N = n) = E(n

i=1

Xi) = nE(X1)

Since this is true for all n we have proved the claim of part (i) .

Proof of (ii). For each fixed value n of N we have

Var(Ni=1

Xi|N)(n) = Var(Ni=1

Xi|N = n) = Var(n

i=1

Xi) = nVar(X1)

Since this is true for all n we have proved the claim of part (ii) .

Proof of (iii).

Var(S) = E[Var(S|N)] + Var[E(S|N)]

= E[Var(N

i=1 |N)] + Var(E[N

i=1 |N ])

= E[N Var(X1)] + Var(N E(X1))

= E(N)Var(X1) + E(X1)2Var(N)

Note. When the primary distribution is Poisson(), then the compound variance formula re-

duces to

Var(S) = E(X2) compound variance for Poisson primary

2

• Example. Let N Poisson(2) and let {X1 , X2 , ...} be an independent identically distributed

sequence with common distribution N(15 , 2 = 5) . Calculate E(N

i=1Xi) and Var(N

i=1Xi)

Solution. We have

E(N) = 2 , Var(N) = 2

Then from the formulas above, we will have:

E(Ni=1

Xi) = E(N)E(X1) = (2)(15) = 30

Var(Ni=1

Xi) = E(N)Var(X1) + Var(N)E(X1)2 = (2)(5) + (2)(15)2 = 460

3

• Aggregate loss: total amount of loss in one period.

Individual risk modelCollective risk modelIndividual risk model: is a model where we study the aggregate loss S = X1 + +Xn

where the losses {X1, ..., Xn} are (stochastically) independent.

Collective risk model: is a model where we study the aggregate loss S = X1 + +XN

where the losses {X1, X2, ...} are (stochastically) independent and identically distributed

(i.i.d.) and where N is a discrete random variable (called frequency) indicating the number

of losses, and that the Xis are independent of N . The distribution of N is called the

primary distribution and the common distribution of Xis is called the secondary

distribution. The collective model is a special case of compound distribution in which we

add up a random number of identically distributed random variables.

In an individual risk model, n is the number of insureds and Xi is the claim size for the

individual i. So under this model we can have P (Xi = 0) > 0 , i.e. there is positive chance

that the insured i will not have any claim in the period of study. But in the collective model,

N represents the variable number of claims and Xi denotes the i-th amount of the claim

that has occurred. So , under the collective model we have P (Xi = 0) = 0.

Convention: For N = 0 we set S = 0.

Example . Let S have a Poisson frequency distribution with parameter = 5. The

individual claim amount has the following distribution:

x fX(x)

100 0.8

500 0.16

1000 0.04

Calculate the probability that aggregate claims will be exactly 600.

Solution.

Case 1. N = 2 in which we have

4

• X1 = 100 , X2 = 500X1 = 500 , X2 = 100with probability

P (S = 600 , N = 2) = P (N = 2 , X1 = 100 , X2 = 500) + P (N = 2 , X1 = 500 , X2 =

100) = 2(e552

2!

)(0.8)(0.16) = 0.0216

Case 2. N = 6 in which we have X1 = = X6 = 100. Then:

P (S = 600 , N = 6) = P (N = 6 , X1 = 100 , ... , X6 = 100) =(e556

6!

)(0.8)6 = 0.0383

P (S = 600) = 0.0216 + 0.0383 = 0.0599

Example . Consider the following collective risk model.

# of claims probability claim size probability

0 15

1 35 2513

150 23

2 15 5023

200 13

Find the variance of the aggregate loss.

5

• Solution.

N

0{

S = 0 Prob. = 15

1

S = 25 Prob. = (35)(13) = 15S = 150 Prob. = (35)(23) = 25

2

S = 100 Prob. = (15)(

23)(

23) =

445

S = 250 Prob. = 2(15)(23)(

13) =

445

S = 400 Prob. = (15)(13)(

13) =

145

E(S) = (0)(15) + (25)(15) + (150)(

25) + (100)(

445) + (250)(

445) + (400)(

145) = 105

E(S2) = (0)2(15) + (25)2(15) + (150)

2(25) + (100)2( 445) + (250)

2( 445) + (400)2( 145) = 19125

Var(S) = E(S2) E(S)2 = 19125 1052 = 8100

Example (Problem 9.55 of the textbook). A population has two classes of drivers. The

number of accidents per individual driver has a geometric distribution. For a driver selected

at random from Class I, the geometric distribution parameter has a uniform distribution over

the interval (0,1). Twenty-five percent of the drivers are in Class I. All drivers in Class II have

expected number of claims 0.25. For a driver selected at random from this population,

determine the probability of exactly two accidents.

Solution.

P(N = 2 |Class I) = 10

P (N = 2|)(1)d = 10

2

(1 + )3d

=

21

(u 1)2

u3du u = 1 +

=

21

u2 2u+ 1u3

du =

21

(1

u 2

u2+

1

u3

)du =

21

(1

u 2u2 + u3

)du

6

• =[ln |u|+ 2u1 + u

2

2

]u=2u=1

= 0.068

P(N = 2 |Class II) = 2

(1 + )3=

(0.25)2

(1.25)3= 0.032

P(N = 2) = P (N = 2|class II)P (class II) + P (N = 2|class I)P (class I) =

(0.068)(0.25) + (0.032)(0.75) = 0.041

Example. There are two groups of good and bad drivers in the city. The good drivers make

up 70% of the population of the drivers. The annual claim amount of each good driver is

uniformly distributed on (0, 3000). The annual claim for each bad driver is exponentially

distributed with mean = 5000. As many as 100 drivers are chosen at random. Find the

mean and variance of the aggregate claim.

Solution. We have S = X1 + +X100 , where each Xi is Uniform(0, 5000) with probability

0.7 and is Exponential( = 5000) with probability 0.3.

E(X) = E(X|good)P (good) + E(X|bad)P (bad) = (1500)(0.7) + (5000)(0.3) = 2550

E(X2|good) = 30000

x21

3000dx =

1

9000

[x3

]30000

= 3000, 000

Using the table , for Exponential with mean the k-th moment is k (k + 1). therefore:

E(X2|bad) = (5000)2(3) = (25, 000, 000)(2) = 50, 000, 000

= (3000, 000)(0.7) + (25, 000, 000)(0.3) = 9, 600, 000

Var(X) = E(X2) E(X)2 = 9, 600, 000 (2550)2 = 3, 097, 500

E(S) = 100E(X) = 255, 000

Var(S) = 100Var(X) = 309, 750, 000

7

• Some statistics of compound distribution

Note 1:

If PX(z) = E[zX ] is the PGF of a random variable X , then

P X(z) = E

[d

dz(zX)

]= E

[X zX1

]P X(z) = E

[d

dz(X zX1)

]= E

[X(X 1) zX2

]and in general, for all ks we have:

P(k)X (z) = E

[X(X 1) (X k + 1) zXk

]By putting z = 1 in both sides of these equalities, we get:

P X(1) = E(X)

P X(1) = E[X(X 1)]

P(k)X (1) = E [X(X 1) (X k + 1)]

Note 2:

For the aggregate model we have:

8

• PS(z) = E(zS) = E(z0)P (N = 0) +

k=1E

[zX1++Xk |N = k

]P (N = k)

= P (N = 0) +

k=1E[zX1++Xk

]P (N = k)

= P (N = 0) +

k=1E[zX1

]kP (N = k) independent and identical distribution

= P (N = 0) +

k=1wk P (N = k) w = E(zX1)

=

k=0wk P (N = k)

= E(wN

)= PN (w) = PN (PX(z))

So

PS(z) = PN (PX(z))

Then the moment generating function of the aggregate S is:

MS(t) = PS(et) = PN [PX(e

t)] = PN [MX(t)]

Then : M S(t) = P N [MX(t)]M X(t)M S(t) = P N [MX(t)] (M X(t))2 + P N [MX(t)]M X(t)E(S) = M S(0) = P

N [MX(0)]M

X(0) = P

N (1)M

X(0) = E(N)E(X)

9

• E(S2) = M S(0) = PN [MX(0)] (M

X(0))

2 + P N [MX(0)]MX(0)

= P N (1) (MX(0))

2 + P N (1)MX(0)

= E[N(N 1)]E(X)2 + E(N)E(X2)

= E(N){E(X2) E(X)2

}+ E(N2)E(X)2

= E(N)Var(X) + E(N2)E(X)2

T