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Suppose that $\{X_n\}_{n=1}^\infty$ is a sequence of random variables such that: $$X_1=\lambda\ , \ X_{n+1}\sim\text{Poi}(X_n)$$

(First, we draw $X_n$, if $X_n=k$ then $X_{n+1}\sim Poi(k)$

I am trying to find out whether or not the sequence converges:

  1. almost surely
  2. in $\mathcal{L}^1$
  3. in $\mathcal{L}^2$

Well, by definition, the only thing I can tell about this sequence is that:

$$P(X_{n+1}=k)=\sum_{m=0}^\infty P(X_{n+1}=k|X_n=m)P(X_n=m)=\sum_{m=0}^\infty \frac{e^{-m}m^k}{k!}P(X_n=m)$$

and I don't know how to proceed from here... Am I in the right way? How can I proceed?

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  • $\begingroup$ Are you familiar with the martingale convergence theorems? $\endgroup$ Aug 24, 2021 at 13:38
  • $\begingroup$ @user6247850 Yes. $\endgroup$ Aug 24, 2021 at 14:25

1 Answer 1

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Since the mean of $\operatorname{Poi}(\mu)$ is $\mu$, we have

$$ \mathbf{E}[X_{n} \mid X_1,\ldots,X_{n-1}] = X_{n-1} $$

and hence $(X_n)$ is a non-negative martingale. So by the martingale convergence theorem, $(X_n)$ converges a.s. Then by the generalized Borel–Cantelli lemma,

\begin{align*} \mathbf{P}(X_n = 0 \text{ i.o.}) &= \mathbf{P}\Biggl( \sum_{n=1}^{\infty} \mathbf{P}(X_n = 0 \mid X_1,\ldots,X_{n-1}) = \infty \Biggr) \\ &= \mathbf{P}\Biggl( \sum_{n=1}^{\infty} e^{-X_{n-1}} = \infty \Biggr) \\ &= 1, \end{align*}

where the last line follows from the fact that $X_n$ converges a.s. Together this and the observation that $X_{n+k} = 0$ for all $k \geq 0$ whenever $X_n = 0$ (this is because $\operatorname{Poi}(0) = \delta_0$), we find that $(X_n)$ is eventually constant with the value $0$ almost surely. Therefore

$$ \lim_{n\to\infty} X_n = 0 \quad \text{a.s.} $$

This then shows that $(X_n)_{n\geq 1}$ cannot converge in $L^1$ or $L^2$.


Alternatively, given that a.s.-convergence of $(X_n)$ has been established, we may directly prove that $X_{\infty} := \lim X_n = 0$ a.s. Indeed, by the law of iterated expectation, for each $s \geq 0$ we have

$$ \mathbf{E}[e^{-sX_n}] = \mathbf{E}[\mathbf{E}[e^{-sX_{n}}\mid X_{n-1}]] = \mathbf{E}[e^{X_{n-1}(e^{-s}-1)}] = \mathbf{E}[e^{-f(s)X_{n-1}}], $$

where $ f(s) = 1 - e^{-s} $ and the moment generating function formula for the Poisson distribution is utilized in the second step. From this recurrence formula, we get

$$ \mathbf{E}[e^{-sX_n}] = e^{-f^{\circ n}(s)\lambda}. $$

However, by noting that $0 < f(s) < s$ whenever $s > 0$, it is not hard to check that $f^{\circ n}(s) \to 0$ as $n \to \infty$ for any $s \geq 0$. By this and the dominated convergence altogether,

$$ \mathbf{E}[e^{-sX_{\infty}}] = \mathbf{E}[\lim_{n\to\infty} e^{-sX_n}] = \lim_{n\to\infty} \mathbf{E}[e^{-sX_n}] = 1.$$

This implies that $e^{-s X_{\infty}} = 1$ a.s. and hence $X_{\infty} = 0$ a.s. as desired.

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    $\begingroup$ Why $X_n$ is a martingale? and I don't understand why this shows that $X_n$ can't converge in $L^1$ or $L^2$. $\endgroup$ Aug 24, 2021 at 14:34
  • $\begingroup$ @TairGalili, I added more details. $\endgroup$ Aug 24, 2021 at 14:53
  • $\begingroup$ Thanks, but why the fact that $X_n$ converge a.s to 0 implies that it can't converge in $L^1$ or $L^2$? $\endgroup$ Aug 24, 2021 at 15:40
  • $\begingroup$ @TairGalili, Can you study the behavior of $\mathbf{E}[|X_n-X_{\infty}|]$ and $\mathbf{E}[|X_n-X_{\infty}|^2]$, where $X_{\infty}=\lim X_n$ is the a.s. limit of $(X_n)$? $\endgroup$ Aug 24, 2021 at 15:41
  • $\begingroup$ Hmm, I do have one last question... Why the fact that $E[\text{Poi}(\mu)]=\mu$ implies that $\mathbf{E}[X_{n} \mid X_1,\ldots,X_{n-1}] = X_{n-1}$. Intuitively, it seems right, but how can I formally prove that for every $A\in\sigma(X_1,\dots,X_n)$ : $\int_{A}X_{n+1}dP=\int_{A}X_{n}dP$ $\endgroup$ Aug 25, 2021 at 7:58

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