91Ó°ÊÓ

The accident parameter is $\mathrm{λ}$is Poisson distributed with mean $\mathrm{λ}$.

$\mathrm{λ}$is a random variable with distributed Gamma $(s,\mathrm{Î\pm })$.

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 $\mathrm{λ}$,

$f_{\mathrm{λ}}\left(l\right)=\frac{{\mathrm{Î\pm }}^s}{\mathrm{Γ}\left(s\right)}{l}^{s-1}{e}^{-\mathrm{Î\pm }l}$

Now, with N = n

We need o find the conditional density of $\mathrm{λ}$.

Then using the Bayesian formula,

$f_{\mathrm{λ}âˆ\pounds N}(lâˆ\pounds n)=\frac{P(N=nâˆ\pounds \mathrm{λ}=l)f_{\mathrm{λ}}\left(l\right)}{P(N=n)}$

Now define

$P(N=n)=p_n$

we have

$\frac{P(N=nâˆ\pounds \mathrm{λ}=l)f_{\mathrm{λ}}\left(l\right)}{P(N=n)}=\frac{1}{p_n}·\frac{l^n}{n!}{e}^{-l}·\frac{{\mathrm{Î\pm }}^s}{\mathrm{Γ}\left(s\right)}{l}^{s-1}{e}^{-\mathrm{Î\pm }l}$

such that

localid="1647241920552" $f_{\mathrm{λ}âˆ\pounds N}(lâˆ\pounds n)=\frac{a^s}{n!·p_n·\mathrm{Γ}\left(s\right)}{l}^{n+s-1}{e}^{-l(\mathrm{Î\pm }+1)}$

with parameters $n+sand\mathrm{Î\pm }+1$,

$\mathrm{λ}/N=n$has Gamma distribution.

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

localid="1647241966218" $E(\mathrm{λ}âˆ\pounds N=n)=\frac{n+s}{\mathrm{Î\pm }+1}$

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