# Show that a sequence is bounded.

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.

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.
• That first equality is wrong. It converges to $0$ as $l\to\infty.$ Mar 17, 2023 at 14:57
• Ah, you meant $\sum_0^\infty$ in the first equality, not $\sum_0^l.$ Mar 17, 2023 at 15:02