Expectation of shifted function of Poisson random variable

Question

Let $\gamma>0$, and suppose $X$ is Poisson distributed random variable with parameter $\gamma$. Suppose $f:\mathbf{N}_0 \rightarrow \mathbf{R}_+$ is such that $\mathbf{E}[f(X)^{1+\epsilon}] < \infty$ for some $\epsilon > 0$. Show that $\mathbf{E}[f(X+k)]<\infty$ for any $k \in \mathbf{N}$.

I've tried the following: $$\mathbf{E}[X(X-1)\dots(X-k+1)f(X)]= \sum_{n=k}^{\infty}n(n-1)\dots(n-k+1)f(n)\frac{\lambda^n}{n!}e^{-\lambda}=\lambda^k \sum_{n=0}^{\infty}f(n+k)\frac{\lambda^n}{n!}e^{-\lambda}=\lambda^k \mathbf{E}(f(X+k))$$ So, now we need to prove that, if $\mathbf{E}[f(X)^{1+\epsilon}]<\infty$ for some $\epsilon>0$ then $\mathbf{E}[X(X-1)\dots(X-k+1)f(X)] < \infty$. I'm not sure how to continue. Help would be much appreciated.

Answer 1

Assuming $\mathbb E[X(X-1)...(X-k+1)f(X)] = \infty$, then $ \mathbb E[X^kf(X)] = \infty$, too.

Note that, then $ \sum_{n=0}^\infty \frac{f(n)n^k\lambda^n}{n!} = \infty$. By D'Alambert it means that limit superior of $ \frac{f(n+1)}{f(n)} (1+\frac{1}{n})^k \frac{\lambda}{n+1} $ is greater or equal to $1$, and it's equivalent to $\limsup$ of $\frac{f(n+1)}{f(n)} \frac{1}{n+1}$ being greater or equal to $\frac{1}{\lambda}$. In particular there is a subsequence $(n_k)$ such that $\frac{f(n_k +1)}{f(n_k)} \frac{1}{n_k + 1} \to \frac{1}{\lambda} + \delta$, where $\delta \ge 0$ (maybe $+\infty$). Again, that means $\frac{f(n_k+1)}{f(n_k)}$ ~ $(1+a_k)n_k$, where $a_k$ is a sequence of non-negative numbers. (By that I meant, there are constants $0<c<C<\infty$ such that $c(1+a_k)n_k \le \frac{f(n_k+1)}{f(n_k)} < C(1+a_k)n_k$. Indeed, it cannot be of order less than $n_k$, because then the limit would be $0$ (the higher bound is not important for us (in fact, it tells us nothing, cause $a_k$ can diverge to $\infty$ very quickly)). But then $\frac{f(n_k+1)^{1+\varepsilon}}{f(n_k)^{1+\varepsilon}}$~$(1+a_k)^{1+\varepsilon}n_k^{1+\varepsilon}$, so in particular $(\frac{f(n_k+1)}{f(n_k)})^{1+\varepsilon} \ge Bn_k^{1+\varepsilon}$.

But looking at $\mathbb E[f(X)^{1+\varepsilon}] = \sum_{n=0}^\infty e^{-\lambda} \frac{\lambda^n f(n)^{1+\varepsilon}}{n!}$, we see that $\sum_{n=0}^\infty f(n)^{1+\varepsilon} \frac{\lambda^n}{n!}$ must be convergent, hence by D'Alambert $\limsup$ of $(\frac{f(n+1)}{f(n)})^{1+\varepsilon} \frac{\lambda}{n+1}$ must be less or equal to $1$. But on subsequence $(n_k)$ it's infinity, a contradiction to assumption $\mathbb E[f(X)^{1+\varepsilon}] < \infty$, hence for any $k \in \mathbb N_0$ we get $\mathbb E[X(X-1)...(X-k+1)f(X)] < \infty$, so $\mathbb E[f(X+k)] < \infty$ for any $k \in \mathbb N_0$, too (as you showed those are equal).

Answer 2

A slightly different approach:

Let $k \in \mathbb N$ and $\varepsilon>0$ be fixed, and let $$P(f, \varepsilon):E(f(X)^{1+\varepsilon})< \infty$$ be true. We want to show that $$Q(f,k): E(f(X+k))<\infty$$ is implied by $P(f, \varepsilon)$.

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Consider the auxiliarry function $$\hat f(n) = \max\left\{f(n), \left(\frac{n!}{\lambda^{n}} \times \frac{1}{n^2}\right)^{\frac{1}{1+ \varepsilon}}\right\}$$

Notice that $P(f, \epsilon) \implies P(\hat{f}, \varepsilon)$ and $Q(\hat{f},k) \implies Q(f,k)$. The second implication is justified by $\hat{f}\ge f$ and the first implication is justified by the fact that $\max(U,V)^{1+\varepsilon} \le (U+V)^{1+\varepsilon}\le 2^{\varepsilon}(U^{1+\varepsilon} + V^{1+\varepsilon})$ * for any two non-negative values $U,V$.

Let us prove that $P(\hat f, \varepsilon) \implies Q(\hat f, k)$.

Formally $e^{\lambda}E(\hat f(X+k)) = \sum_{n=0}^\infty \hat f(n+k)\frac{\lambda^n}{n!}= \sum_{n=k}^\infty \hat f(n)\frac{\lambda^n}{n!}\frac{(n-k+1)\cdots n}{\lambda^{k}}\le \frac{1}{\lambda^k}\sum_{n=k}^\infty \hat f(n)\frac{\lambda^n}{n!}n^k. $

We have $\hat f(n) \ge \left(\frac{n!}{\lambda^{n}} \times \frac{1}{n^2}\right)^{\frac{1}{1+ \varepsilon}} \ge^\text{Stirling} C \left(\frac{n^{n-2}}{(e \lambda)^n}\right)^{\frac{1}{1+ \varepsilon}} \ge C' n^{\frac{k}{\varepsilon}} \implies \frac{(\hat{f})}{C''}^\varepsilon \ge n^{k}$ for some positive constants $C,C',C''$.

Hence $\sum_{n=k}^\infty \hat f(n)\frac{\lambda^n}{n!}n^k \le \sum_{n=k}^\infty \hat f(n)\frac{\lambda^n}{n!}\frac{(\hat{f})^\varepsilon}{C''} = \frac{1}{C''}\sum_{n=k}^\infty \hat f(n)^{1+\varepsilon}\frac{\lambda^n}{n!} \overset{\text{by } P(\hat{f}, \varepsilon) }{<}+\infty$ which finishes the proof.

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* or $\max(U,V)^{1+\varepsilon} = \max(U^{1+\varepsilon},V^{1+\varepsilon}) \le (U^{1+\varepsilon}+V^{1+\varepsilon})$