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Let $X$ be a random variable whose distribution is Poisson with parameter $\lambda>0$. Let $$ A_\tau := \{n \in \mathbb N : \mathbb P [X \ge n] \le \tau\} \quad \forall \tau \in (0, 1). $$

Are there some upper bounds of $\min A_\tau$ in terms of $\tau$ and $\lambda$? Thank you so much for your elaboration!

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  • $\begingroup$ Perhaps you could try something like a one-tailed version of Chebyshev's inequality to give loose bounds $\endgroup$
    – Henry
    Commented Dec 12, 2022 at 1:24

2 Answers 2

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A somewhat more complicated upper bound than @Adtiya Dhawan's answer. But the technique is standard, it is the Cramer-Chernoff technique.

We have that $$ P(X \ge n) \le \inf_{\eta > 0} \exp(-n \eta) E[\exp(\eta X)] = \inf_{\eta > 0} \exp(-n \eta + \lambda(\exp(\eta)-1)) = e^{-\lambda}\left(\frac{e \lambda}{n}\right)^n.$$

Hence, $$\min A_{\tau} \le \inf\left\{n : \left(\frac{e \lambda}{n}\right)^n \le \tau e^{\lambda}\right\}.$$ This would be better than $\lambda / \tau$.

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    $\begingroup$ Using Stirling's approximation, your bound is equivalent to $e^{-\lambda}\frac{\lambda^n}{n!}\cdot \sqrt{2\pi n}=P(X=n)\cdot \sqrt{2\pi n}$. This leaves open the possibility that $\limsup P(X\ge n)/P(X=n)=+\infty$. In actuality, you can prove $\lim_n P(X\ge n)/P(X=n)=1$. $\endgroup$
    – Mike Earnest
    Commented Dec 12, 2022 at 17:08
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    $\begingroup$ $$ \begin{align} P[X\ge n] &=\frac{e^{-\lambda}\lambda^n}{n!}\sum_{k\ge 0}\frac{\lambda^k}{(n+1)(n+2)\cdots (n+k)} \\&\le \frac{e^{-\lambda}\lambda^n}{n!}\sum_{k\ge 0}\frac{\lambda^k}{n^k} \\&=P[X=n]\cdot \frac1{1-\lambda/n} \end{align} $$ $\endgroup$
    – Mike Earnest
    Commented Dec 12, 2022 at 17:10
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A very simple upper bound is given by Markov's inequality :

$$\mathbb{P}(X \geq n) \leq \frac{\mathbb{E}(X)}{n} = \frac{\lambda}{n} \leq \tau \ \ \text{if} \ n \geq \frac{\lambda}{\tau} \ \text{so that} \ A_{\tau} \leq \frac{\lambda}{\tau}$$

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    $\begingroup$ While correct I feel like it's important to note that Markov's inequality often yields a trivial upper bound (for example, there are cases when it simply guarantees the probability is less than or equal to some constant $a\ge1$ which is always the case.. $\endgroup$
    – Alborz
    Commented Dec 12, 2022 at 2:56

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