I am trying to do the following:
Let $S$ be the total claim size when the number of claims follow a Negative Binomial Distribution.
How can I derive a formula for the $E(S)$ - expectancy and $V(S)$ - variance for a Negative Binomial distribution for the total claim size?
I found the one for the Poison distribution and its seems that it is related to the conditional expectancy formula.
Where the formula for the total claim size for the expectancy and variance for a Poisson distribution is given by:
Let $m$ be the expected number of claim.
Then,
$E(S)=m\mu$ because the mean of the Poisson distribution is $\mu$,
$$V(S)=m\alpha_2$$
where for a Poisson distribution $\alpha_2=\mu+\mu^2$ and $\alpha_1=\mu$ finding by applying derivative to the mgf of the Poisson distribution at $t=0$
My thought is that:
For a Negative Binomial distribution with parameters $p$ and $k$ with Mean $\frac{k(1-p)}{p}$ and Variance $\frac{k(1-p)}{p^(2)}$
The formula for the Expectancy and Variance for the total claim size follows the same logic and it is given by:
$$E(S)=m\frac{k(1-p)}{p}$$
where $\alpha_2$ is given by: $\frac{k^2p^2-2k^2p+k^2-kp+k}{p^2}$, and I calculated this by finding the second derivative of the moment generating function of the Negative Binomial distribution at $t=0.$
$$V(S)=m\frac{k^2p^2-2k^2p+k^2-kp+k}{p^2}$$
Then as both distributions have different formulas for expectancy and variance of the total claim size I do not understand for which values of $p$ and $k$ of the Negative Binomial dist. can we obtain different results for $E(S)$ from that of the homogeneous Poisson case.
Can anyone help me on this please?
Thanks.