TITLE: How to show tightness of a Poission$(\lambda)$ family, $\lambda \in \Lambda$ implies $\Lambda$ is bounded.
QUESTION [6 upvotes]: I'm still learning how to properly work with the tightness of a sequence. The question is as in the title:
Assume that a family Poisson$(\lambda)$, $\lambda \in \Lambda$ is tight. Show that $\Lambda$ must be bounded.
What I have so far I have from the definition that we must have $\lim_{M \rightarrow \infty}\inf_\lambda\mathbb{P}_\lambda([-M,M]) = 1$. Writing out the definition for the probability measure I have
\begin{align*}
\mathbb{P}_\lambda([-M,M]) = \mathbb{P}_\lambda([0,M]) &= \sum_{k=0}^M \frac{e^{-\lambda}\lambda^k}{k!} \\
&= e^{-\lambda}\sum_{k=0}^M \frac{\lambda^k}{k!}
\end{align*}
From here we can see that as $M \rightarrow \infty$ we have that $e^{-\lambda}\sum_{k=0}^M \frac{\lambda^k}{k!} \rightarrow e^{-\lambda}e^\lambda = 1$, but from what I can see this holds for arbitrary values of $\lambda$. What am I missing?

REPLY [2 votes]: A simple proof is as follows. For every $\lambda\gt0$, $\mathbb P_\lambda([\lambda,+\infty))\gt0$, and, by the central limit theorem, $\mathbb P_\lambda([\lambda,+\infty))\to\frac12$ when $\lambda\to\infty$, hence there exists some positive $c$ such that $\mathbb P_\lambda([\lambda,+\infty))\geqslant 2c$ for every $\lambda\geqslant1$.
Turning things the other way around, this means that if there exists $M\geqslant1$ such that $\inf\limits_{\lambda\in\Lambda}\mathbb P_\lambda([-M,M])\geqslant1-c$, then no $\lambda\geqslant M$ is in $\Lambda$, that is, $\Lambda\subseteq[0,M]$.