Question

An insurance company supposes that the number of accidents that each of its policyholders
will have in a year is Poisson distributed, with the mean of the Poisson depending on the policyholder. Suppose the Poisson mean of a randomly chosen policyholder follows exponential distribution with mean=1 [The density for exponential distribution with mean=1 is f(y) = e^(−y) ]. Let X be the number of accidents next year for a randomly chosen policyholder. Then Var(X) is?

(The answer is supposed to be 2, but I keep finding 1...)

Answer 1

why it is two? if i recall correctly, the mean is the variance in a poisson distribution

Answer 2

Well I don't know it's from an answer key. I'm trying to get it but I keep getting 1 o.o

Answer 3

maybe i don' t understand this question either. not having a lot of success tonight

Answer 4

I think I'm getting confused because the Mean of the Poisson is itself a random variable (exponential distribution)...

Answer 5

If we say X~Poisson (M) where M is the mean, but M ~ exp(1) ?

Answer 6

Hmm could it be that... E(X) = M? but M is an r.v.... Do we say E(X|M=m) = E[M^k*e^(-M))/k!] by Poisson distribution. By substitution, => E=[(m^k*e^-m)/k!] = m

So E(X|M) = M
By double expectation, E(X)=E(E[X|M)] = E(M) = 1

Similarly, Var(X|M) = 1

And I just realized we have a formula: Var(X) = E[Var(X|Y)] + Var[E(X|Y)]
So, Var X = E(1) + Var(1) = 2?? o_O

Answer 7

sometimes things are easy

Answer 8

Are you saying my reasoning is sensible? Or... hehe I'm not sure what you are implying there :P