## Panjer Recursion II: An Application

My last post was on Panjer recursion, a recursive formula for the probability mass function of the compound distribution $\sum_{j=1}^{N(t)}X_{j}$, where the $X_{j}$ are i.i.d. nonnegative-integer-valued random variables and $N(t)$ is an independent counting process with $(a,b,0)$-class distribution. In today’s brief post, I would like to use Panjer recursion to obtain a recursive formula for the moments of $S$.

For $r\in \mathbb{Z}^{\geq 1}$,

$\begin{array}{lcl}\displaystyle\mathbb{E}\left[S^{r}\right]=\sum_{k=0}^{\infty}k^{r}g_{k}&=&\displaystyle\dfrac{1}{1-af_{0}}\sum_{k=1}^{\infty}k^{r}\sum_{j=1}^{k}\left(a+\dfrac{bj}{k}\right)f_{j}g_{k-j}\\&=&\displaystyle\dfrac{1}{1-af_{0}}\sum_{j=1}^{\infty}\sum_{k=j}^{\infty}\left(ak^{r}+bjk^{r-1}\right)f_{j}g_{k-j}\\&=&\displaystyle\dfrac{1}{1-af_{0}}\sum_{j=1}^{\infty}\sum_{t=0}^{\infty}\left(a(t+j)^{r}+bj(t+j)^{r-1}\right)f_{j}g_{t}\end{array}$

Applying the binomial expansion to $(t+j)^{r}$ and $(t+j)^{r-1}$, we obtain

$\begin{array}{lcl}\displaystyle\sum_{t=0}^{\infty}(t+j)^{r}g_{t}=\sum_{i=0}^{r}{r\choose i}j^{r-i}\mathbb{E}[S^{i}],\indent \sum_{t=0}^{\infty}j(t+j)^{r-1}=\sum_{i=0}^{r-1}{{r-1}\choose i}j^{r-i}\mathbb{E}[S^{i}]\end{array}$

Substituting these identities in, we have

$\begin{array}{lcl} \displaystyle\mathbb{E}\left[S^{r}\right]&=&\displaystyle\dfrac{1}{1-af_{0}}\sum_{j=1}^{\infty}\left[af_{j}\sum_{i=0}^{r}{r\choose i}j^{r-i}\mathbb{E}[S^{i}]+bf_{j}\sum_{i=0}^{r-1}{{r-1}\choose i}j^{r-i}\mathbb{E}[S^{i}]\right]\\&=&\displaystyle\dfrac{1}{1-af_{0}}\left[\sum_{i=0}^{r}{r\choose i}\mathbb{E}[S^{i}]\left(\sum_{j=1}^{\infty}aj^{r-i}f_{j}\right)+\sum_{i=0}^{r-1}{{r-1}\choose i}\mathbb{E}[S^{i}]\left(\sum_{j=1}^{\infty}bj^{r-i}f_{j}\right)\right]\\&=&\displaystyle\dfrac{1}{1-af_{0}}\left[\sum_{i=0}^{r-1}\left[a{r\choose i}+b{{r-1}\choose i}\right]\mathbb{E}[S^{i}]\mathbb{E}[X^{r-i}]+\mathbb{E}[S^{r}]\sum_{j=1}^{\infty}af_{j}\right]\\&=&\displaystyle\dfrac{1}{1-af_{0}}\left[\sum_{i=0}^{r-1}\left[a{r\choose i}+b{{r-1}\choose i}\right]\mathbb{E}[S^{i}]\mathbb{E}[X^{r-i}]+a(1-f_{0})\mathbb{E}[S^{r}\right]\end{array}$

Moving the second term to the LHS and then dividing both sides by $1-a$, we obtain

$\displaystyle\mathbb{E}[S^{r}]=\dfrac{1}{1-a}\sum_{i=0}^{r-1}\left[a{r\choose i}+b{{r-1}\choose i}\right]\mathbb{E}[S^{i}]\mathbb{E}[X^{r-i}]$