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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.

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  • $\begingroup$ I'm accustomed to the word "expectation" and "expected value" rather than "expectancy", except in actuarial problems. Maybe that last word is used here because that's what this is? $\endgroup$ – Michael Hardy Mar 18 '18 at 15:35
  • $\begingroup$ You consider the case where the number of claims has a Poisson distribution, and you say "Let $m$ be the expected number of claims." But later you consider the expression $m\mu$ and say that $\mu$ is the expected value of the Poisson distribution. But if $m$ is the expected number of claims, and the number of claims has a Poisson distribution, then the expected value of the Poisson distribution is $m. \qquad$ $\endgroup$ – Michael Hardy Mar 18 '18 at 15:42
  • $\begingroup$ Most basic probability texts have a table (sometimes appendix, sometimes inside back cover) that shows population means and variances for various families of distributions. These books also how derivations for mean and variance of several distributions. In addition moment generating functions can be used. You might start by looking at Wikipedia pages for the dist'ns of interest to you. // There are several versions of the negative binomial distribution family, so you have to be sure you get the one that matches your text. $\endgroup$ – BruceET Mar 19 '18 at 7:12

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