= Solution
The <exponential distribution> with rate $\beta>0$ has <moment-generating function> $\beta/(\beta-t)$ for $t<\beta$. Shifting it by $\alpha$ multiplies the <moment-generating function> by $e^{\alpha t}$. Therefore the <shifted-exponential Poisson mixture> has <probability generating function>
$$
\boxed{G_N(z)=e^{\alpha(z-1)}\frac{\beta}{\beta+1-z}.}
$$
The <probability generating function> converges and is analytic for $|z|<\beta+1$, and in particular is finite for real $0\le z<\beta+1$.
Write $p=\beta/(\beta+1)$ and $q=1/(\beta+1)$, so $p+q=1$. The factor $e^{\alpha(z-1)}$ is the <probability generating function> of $V\sim\operatorname{Poisson}(\alpha)$, while $p/(1-qz)$ is that of a <geometric distribution> on $\{0,1,\ldots\}$, with $\mathbb P(W=j)=pq^j$. Choose $V$ and $W$ independently. The product rule for the <probability generating function> of a sum of independent counts then proves $N\overset d=V+W$. The support-zero convention for $W$ is essential here.
The <convolution of independent random variables> gives the finite sum
$$
\boxed{p_n=e^{-\alpha}\frac{\beta}{\beta+1}\sum_{k=0}^{n}\frac{\alpha^k}{k!}(\beta+1)^{-(n-k)},\qquad n\ge0.}
$$
Taking the logarithmic derivative of the <probability generating function> gives
$$
\frac{G_N'(z)}{G_N(z)}=\alpha+\frac1{\beta+1-z},
\qquad
(\beta+1-z)G_N'(z)=\{1+\alpha(\beta+1)-\alpha z\}G_N(z).
$$
Compare coefficients of $z^{n-1}$, for $n\ge2$. The left side is $(\beta+1)np_n-(n-1)p_{n-1}$, and the right side is $\{1+\alpha(\beta+1)\}p_{n-1}-\alpha p_{n-2}$. Hence the <shifted-exponential Poisson count recursion> is
$$
\boxed{p_n=\frac{n+\alpha(\beta+1)}{n(\beta+1)}p_{n-1}-\frac{\alpha}{n(\beta+1)}p_{n-2},\qquad n\ge2.}
$$
The constant and linear coefficients supply the starting values
$$
p_0=e^{-\alpha}\frac{\beta}{\beta+1},\qquad
p_1=\left(\alpha+\frac1{\beta+1}\right)p_0.
$$
These two values determine every subsequent <probability> by the recursion. The negative second term does not mean that the distribution has negative probabilities: the convolution formula exhibits every $p_n$ as a sum of nonnegative quantities.
At $\alpha=0$ the independent <Poisson distribution> component is identically zero and the count has a <geometric distribution>. Both the finite sum and the recursion reduce to $p_n=q p_{n-1}$. Consequently the <Panjer claim-count class> parameters are
$$
\boxed{a=\frac1{\beta+1},\qquad b=0.}
$$
Although the shifted model initially has positive $\alpha$, this zero-shift boundary case is well defined.
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