A uniform tail bound on Poisson random variable.

Question

Let $N$ be a Poisson random variable with mean $\lambda>0$. Prove that there exists uniform constants $M,c>0$ (do not depend on $\lambda$) such that $\mathbb{P}(N\geq n) \leq \lambda M\, e^{-cn}$.

I tried to prove this with Chernoff style tail bound on $N$. But most of them give a bound on $\mathbb{P}(N \geq \lambda + t)$ where $t\geq 0$. Any help is appreciated.

Answer 1

The statement you are trying to prove is wrong. Take $\lambda = 2n$, $n\ge 1$. Then your inequality reads, $$ \mathbb{P}(N_{2n}\geq n) \leq 2n M\,e^{-cn}\to 0, n\to\infty. $$ However, $$ \mathbb{P}(N_{2n}< n) = \mathbb{P}\Bigl(\frac{N_{2n}}{2n}< \frac12\Bigr) \to 0, n\to\infty, $$ since by LLN, $\frac{N_{2n}}{2n}\to 1$, $n\to\infty$, in probability.

most of them give a bound on $\mathbb{P}(N \geq \lambda + t)$ where $t\geq 0$

for a reason. Here $\mathbb E[N] =\lambda$, and it is sensible to expect that $\mathbb P(N>x)$ is low only for $x>\mathbb E[N]$.