Question

Romeo and Juliet decide to start a new business venture: JR Insurance Ltd. to insure people against damages from car accidents. They assume that car accidents resulting in claims occur according to a nonhomogeneous Poisson process at rate ? ? = 2? per month. As almost any other insurance policy; there is a stochastic delay between the time each accident occurs and the payment is made. These claims not yet paid are said to be “outstanding claims.” Suppose that these durations between the accident and payment are independent and identically distributed random variables, uniformly distributed between 0 and 5 months. Also suppose that the amount of the claims are independent and identically distributed random variables, having an exponential distribution with mean $10.000. Find the followings.

i. Expected number of accidents occurring in the first 4 months.

ii. Probability that 3 accidents occur during the first month and no

accident occurs during the second month.

iii. Let T1 be the time of the first accident. Find P(T1 > 2 months).

iv. Probability that the total amount of all accident claims during the first

8 months exceeds $200,000. (Hint: use normal approximation.)

v. The mean number of outstanding claims at time 6.

Answer 1

i) Let N(t) be the number of accidents in the interval [0, t]. The stochastic process {N(t) : t ≥ 0} is a nonhomogeneous Poisson process with arrival rate λ(t) = 2t,

The expected number of accidents in the interval [0, t] is

E[N(t)] ≡ m(t) = λ(s) ds.

In our case, we want E[N(4)] ≡ m(4) = 2s ds = 16.

ii) These are independent Poisson random variables,

Probability that 3 accidents occur during the first month

m(1) − m(0) = m(1) = 1

prob ( 3 accidents occur during the first month) =

=0.06131

no accident occurs during the second month

m(2) − m(1) = 4 − 1 = 3.

prob ( 3 accidents occur during the first month) =

0.049787

iii. Let T1 be the time of the first accident. Find P(T1 > 2 months).

Let T1 be the time of the first accident.

For any t ≥ 0, P(T1 > t) = P(N(t) = 0) =

=

m(t) = λ(s) ds.

=​​​​​​​2s. ds

= =4

P(T1 > t) = and P(T1 > 1) = =0.0183156

iv. Probability that the total amount of all accident claims during the first 8 months exceeds $200,000.

Mean =X(t) =

(Yi) , t ≥ 0,

where Yi is the random dollar value of the claim associated with the i th accident.

In general, E[X(t)] = m(t)E[Y1], where m(t) = E[N(t)]

Here then E[N(8)] = = 64 and

E[X(8)] = 100E[Y1] = 64 × 64 = 4096

Variance,

SD(X(8)) = V ar(X(8)) = = 200000

P(X(8) > 14, 000) = P (X(8) − E[X(8)]/SD(X(10)) > (200000 − 4096) /200000

=0.979

by z table probability = 0.3365