Martingale convergence theorem for Poisson.

Question

Let $\{A_n\}_n$ be random variables such that $A_0=1$ and given $\{A_j, j=0,\dots , n-1\}, A_n \sim Poisson(A_{n-1}).$
It is straightforward that $\mathbb{E}[A_n| A_1, \dots , A_{n-1}]=A_{n-1}$, and that $\{A_j\}$ is a nonnegative martingale. By martingale convergence theorem, we get $\lim{A_n}=a < \infty.$ Is there anything I can say for $P(A_n =k| A_j)=\frac{A_{n-1}^k}{k!}e^{-A_{n-1}}$? I'm kind of lost, any hint would be appreciated. Thanks in advance.

Answer 1

Let's examine the generating function of $a$. Evidently $$ \Bbb E[s^{A_n}]=\Bbb E[e^{-A_{n-1}(1-s)}], $$ so letting $n\to\infty$ we obtain $$ \Bbb E[s^a]=\Bbb E[e^{-a(1-s)}]. $$ Now let $s\to 0+$: $$ \Bbb P[a=0] = \Bbb E[e^{-a}]=\Bbb P[a=0]+\Bbb E [e^{-a}1_{\{a\ge 1\}}]. $$ Thus $\Bbb E[e^{-a}1_{\{a\ge 1\}}]=0$, forcing $\Bbb P[a=0]=1$.

Answer 2

I think I've solved it, but i'm not sure about a certain part of my proof.
Anyways, since $A_n$ given the previous $A_j$ is $Poisson(A_{n-1}) \Rightarrow \mathbb{E}\{A_n^2 | A_1, \dots, A_{n-1}\} = A_{n-1}^2 + A_{n-1} (*)$
now taking limits as $n \rightarrow \infty$ in $(*)$ we get that $a^2=a^2+a \Rightarrow a=0$.
My question is the following: is it legit to take such limit in the conditional expectation with respect to such a $\sigma$-algebra?
Intuitively, I think it should be OK, since as $n$ increases to $N$ (large), the conditional expectation of our interest is calculated given $A_1, \dots A_{N-1},$ thing that is required in order for $A_N$ to keep it's Poisson distribution (since A_N is Poisson onlygiven the previous values of $A_j$).