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Let $f: \mathbb{R}_+ \to \mathbb{R}$ be a Lipschitz continuous function, i.e. there exits some $C > 0$ such that for all $x,y \in \mathbb{R}_+$, we have $$ |f(x) - f(y)| \leq C|x-y|. $$ If $N \sim Poi(\lambda)$, $\lambda>0$, we can consider the random variable $f(N)$. Assume that $\mathbb{E}[f(N)] = 0$, i.e. $e^{-\lambda}\sum_{n=0}^\infty \frac{\lambda^n}{n!}f(n) = 0$.

Now, we consider the sequence $$ a_l := \frac{l!}{\lambda^{l+1}} \sum_{n=0}^l\frac{\lambda^n}{n!}f(n). $$ The goal is to show that $a_l$ is a bounded sequence.

My thoughts: I was able to prove an equivalent representation of $a_l$ given by $$ a_l = \frac{\mathbb{E}[f(N)1_{\{N \leq l\}}]}{\lambda \mathbb{P}[N = l]}. $$ Since $$ a_l := \underbrace{\frac{l!}{\lambda^{l+1}}}_{\to \infty} \underbrace{{\sum_{n=0}^l\frac{\lambda^n}{n!}f(n)}}_{\to 0} $$ it would suffice that the partial sums on the RHS converge fast enough to 0. However, I am not sure how to proceed. It is not clear to me on how I can apply the Lipschitz condition in order to show boundedness. Some help or guidance into the right direction would be really appreciated.

Thanks in advance!

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1 Answer 1

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Since $\mathbb{E} f(N)=0$, we have $$ \sum_{n=0}^l\frac{\lambda^n}{n!}f(n)=0 $$ and thus $$ -a_l=\frac{l!}{\lambda^{l+1}}\sum_{n=l+1}^{\infty}\frac{\lambda^n}{n!}f(n) $$

From Lipschitz, assuming $|f(0)|=a$, you immediately get that $|f(n)|\le Cn+a\le C_1 n$ for $n\ge 1$ for $C_1=C+a$. Hence $$ \begin{align}|a_l|&\le \frac{l!}{\lambda^{l+1}}\sum_{n=l+1}^{\infty}\frac{\lambda^n}{n!}|f(n)|\\ &\le \frac{l!}{\lambda^{l+1}}\sum_{n=l+1}^\infty\frac{\lambda^n}{n!}C_1 n\\ &\le C_1 l!\sum_{k=0}^\infty\frac{\lambda^k}{(k+l)!}\\ &=C_1\left[1+\frac{\lambda}{l+1}+\frac{\lambda^2}{(l+1)(l+2)}+\dots\right]\\ &\le C_1\left[1+\frac{\lambda}{1!}+\frac{\lambda^2}{2!}+\dots\right]\\&\le C_1 e^{\lambda} \end{align}$$ which is uniformly bounded.

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    $\begingroup$ Thanks for the great answer! $\endgroup$ Mar 17, 2023 at 13:53
  • $\begingroup$ That first equality is wrong. It converges to $0$ as $l\to\infty.$ $\endgroup$ Mar 17, 2023 at 14:57
  • $\begingroup$ Ah, you meant $\sum_0^\infty$ in the first equality, not $\sum_0^l.$ $\endgroup$ Mar 17, 2023 at 15:02

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