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

The accident parameter is $\lambda$is Poisson distributed with mean $\lambda$.

$\lambda$is a random variable with distributed Gamma $(s,\alpha )$.

The newly insured person has n accidents in her first year.

Let N be the random variable that marks the number of accidents of some person in a certain year.

According to the statement, the density function of $\lambda$,

$f_{\lambda }\left(l\right)=\frac{{\alpha }^s}{\Gamma \left(s\right)}{l}^{s-1}{e}^{-\alpha l}$

Now, with N = n

We need o find the conditional density of $\lambda$.

Then using the Bayesian formula,

$f_{\lambda \mid N}(l\mid n)=\frac{P(N=n\mid \lambda =l)f_{\lambda }\left(l\right)}{P(N=n)}$

Now define

$P(N=n)=p_n$

we have

$\frac{P(N=n\mid \lambda =l)f_{\lambda }\left(l\right)}{P(N=n)}=\frac{1}{p_n}·\frac{l^n}{n!}{e}^{-l}·\frac{{\alpha }^s}{\Gamma \left(s\right)}{l}^{s-1}{e}^{-\alpha l}$

such that

localid="1647241920552" $f_{\lambda \mid N}(l\mid n)=\frac{a^s}{n!·p_n·\Gamma \left(s\right)}{l}^{n+s-1}{e}^{-l(\alpha +1)}$

with parameters $n+sand\alpha +1$,

$\lambda /N=n$has Gamma distribution.

Moreover, the expected number of accidents that she will have in the following year,

localid="1647241966218" $E(\lambda \mid N=n)=\frac{n+s}{\alpha +1}$

Where the formula for the expected value of gamma distribution has been used.