Question

4. Conditioning on the average driving speed (in 10 m/s) Θ-θ, the time until a claim occurs (in years), T. for the driver follows an exponential distribution with mean 1/ . Best Insurance Company covers a pool of policyholders. The average driving speed of these policyholders. Э, follows a gamma distribution with mean 2 and variance 1. For a randomly selected policyholder (a) Calculate the probability that the next claim will arrive in 3 months. (b) Determine the average arrival time of the next claim. (c) Calculate Var(T).

Answer 1

$(a)~The~conditional~pdf~of~T~given~\theta:\\ f(t|\theta)=\theta\exp(-t\theta);~t>0\\ =0~otherwise\\ The~pdf~of~\theta~is\\ f(\theta)=\frac{\lambda}{\Gamma(p)}\exp(-\lambda \theta)(\lambda \theta)^{p-1},~~\theta>0\\ =0~otherwise\\ where,~E(\theta)=\frac{p}{\lambda}=2,~Var(\theta)=\frac{p}{\lambda^2}=1.\\ Then,~\lambda=2,~p=4.\\ Then~the~pdf~of~\theta~is\\ f(\theta)=\frac{2^4}{3!}\exp(-2\theta)\theta^3=\frac{8}{3}\exp(-2\theta)\theta^3,~~\theta>0\\ =0~otherwise\\ The~pdf~of~T~is\\ f(t)=\int_0^\infty f(t|\theta)f(\theta)d\theta=\frac{8}{3}\int_{0}^\infty \exp(-(t+2)\theta)\theta^4d\theta\\ =(8/3)\frac{4!}{(t+2)^5}=\frac{64}{(t+2)^5};~t>0\\ =0~otherwise\\ Probability~that~the~next~claim~will~arrive~in~3~months\\ =P(T\geq 3)=\int_3^\infty \frac{64}{(t+2)^5}dt=64\times\frac{1}{4*5^4}= 0.0256\\ (b)~E(T)=64\int_0^\infty \frac{t}{(t+2)^5}dt=64\left[\int_0^\infty \frac{1}{(t+2)^4}dt -2\int_0^\infty \frac{1}{(t+2)^5}dt\right ]\\ =64*(1/(3*2^3)-2/(4*2^4))=0.6667$

$(c)~E(T^2)=64\int_0^\infty \frac{t^2}{(t+2)^5}dt=64\left[\int_0^\infty \frac{1}{(t+2)^3}dt -4\int_0^\infty \frac{1}{(t+2)^4}dt+4\int_0^\infty \frac{1}{(t+2)^5}dt\right ]\\ =64*(1/(2*2^2)-4/(3*2^3)+4/(4*2^4))=1.3333\\ Var(T)=E(T^2)-E^2(T)=1.3333-0.6667^2=0.8888$