Let $X$ be Poisson distributed and $\tau \in (0, 1)$. Are there upper bounds of $\min \{n \in \mathbb N : \mathbb P [X \ge n] \le \tau\}$?

Question

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!

Answer 1

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 $\lambda / \tau$.

Answer 2

A very simple upper bound is given by Markov's inequality :

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