# Limit of sequence of random variables

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 $$\mathcal{L}^2$$

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

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

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

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.

• Why $X_n$ is a martingale? and I don't understand why this shows that $X_n$ can't converge in $L^1$ or $L^2$. Aug 24, 2021 at 14:34
• @TairGalili, I added more details. Aug 24, 2021 at 14:53
• Thanks, but why the fact that $X_n$ converge a.s to 0 implies that it can't converge in $L^1$ or $L^2$? Aug 24, 2021 at 15:40
• @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 $(X_n)$? Aug 24, 2021 at 15:41
• 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 $A\in\sigma(X_1,\dots,X_n)$ : $\int_{A}X_{n+1}dP=\int_{A}X_{n}dP$ Aug 25, 2021 at 7:58