# Probability of random variable $X$ less than $max(Y_i)$

For $\lambda > \mu$, consider independent random variables (i) $X \sim Pois(\mu)$ and (ii) $Y_i \sim Pois(\lambda),\;i\in{1,2}$.

Question : Can we upper-bound the following? $$\mathbb{P}\big(X\geq\max(Y_1, Y_2)\big)$$

• It can be shown that $P(X>Y_i) \leq e^{ -(\sqrt{\lambda}-\sqrt{\mu})^2}$ – Vivek Bagaria Oct 22 '16 at 3:54
• Note : Even though the $X, Y_1, Y_2$ are independent, the event $X>Y_1$ and $X>Y_2$ are not independent. – Vivek Bagaria Oct 26 '16 at 17:20
• This question deals with the distribution of $\max\{Y_1,Y_2\}$: math.stackexchange.com/questions/1113990/… – usul Oct 27 '16 at 5:10

We have $\Pr[\max(Y_1,Y_2) \leq k] = \Pr[Pois(\lambda) \leq k]^2 = e^{-2\lambda}(\sum_{j=0}^k\frac{\lambda^j}{j!})^2$.

Therefore, $\Pr[X \geq \max(Y_1,Y_2)] = \sum_{k \in \mathbb N} \Pr[X = k] \cdot \Pr[\max(Y_1,Y_2) \leq k] = e^{-(2\lambda+\mu)} \bigg (\sum_{k = 0}^\infty\frac{\mu^k}{k!} \cdot (\sum_{j=0}^k \frac{\lambda^j}{j!})^2 \bigg)$.

From this, if you have some concrete values of $\lambda$ and $\mu$, you can try to see which terms are important, and which are negligible.

I guess the important terms are $k < O(\lambda)$.