Question

2. An insurance company supposes that each person has a accident parameter and that the yearly number of accidents of someone whose accident parameter is is Poisson distributed with mean 1. They also suppose that the parameter value of a newly insured person can be assumed to be the value of an exponential random variable with parameter s. (a) If a newly insured person has n accidents in her first year, find the conditional density of her accident parameter. (b) Determine the expected number of accidents that she will have in the following year.

Answer 1

Let the random variable X denotes the number of accidents. According to the question, X follows a Poisson distribution with parameter 2, that is X 2- Poisson (2). e 2* ,x>0. х! Thus, P(X =x|2)= Also, the parameter value 2 follows the exponential distribution with parameter s, that is, 2- Exponential (s). Thus, f(a) = se , s >0. -s. (a) The conditional density of the accident parameter a given a newly insured person has n accidents is obtained as shown below: P(X =no1) P(X =n) Sar-n(2)X = n) = P(X=n\2)f(2) P(X =n) Now, P(X = n2)f (2) is calculated as shown below: %3D X se s2 п! P(X =n|2) f(2)=e"2" -(1+s)2 2"s n!

The P(X=n) is obtained by integrating P(X=n2)f(2) as shown below: P(X =n)=[P(X=n|2)S(2)da (1+s) 2"s -je* %3D п! Г(n+1)s n+1 (1+ s)** n! (1+s)| n+1 (1+ s)** Thus, P(X =n) = (1+ s)* * n+1 Thus, the required probability is calculated as shown below: t(2)X = n) = P(X =n|2)f(a) P(X =n) Sдк Sar-, (4) X = n) = %3D -(1+s)2 2"s n! (1+s)*I n+1 n+1 (1+s)2 2" (1+s)** n!

(1+s) 2" (1+s) n+1 Thus, far (4|X =n)= which is the gamma distribution with 2x=n n! parameters (n+1, s+1). (++) 2" (1+s)** n+1 Hence, the conditional probability is fax-n (2)X = n) = -,2>0. n! (b) The mean of the random variable X- Gamma (a, B) is E(X)=. So, the expected number of accidents is the mean of the random variable 1 X =n~ Gamma (n+1, s +1). n+1 Hence, the expected value is E(X)= s+1