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}]

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