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

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

<p>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.</p>

<p>I guess the important terms are $k &lt; O(\lambda)$.</p>

CC BY-SA 3.0
2016-11-05T16:04:55.900Z
2016-11-05T16:04:55.900Z
Igor Shinkar
<p>For $\lambda &gt; \mu$, consider independent random variables <strong>(i)</strong>  $X \sim Pois(\mu)$ and <strong>(ii)</strong> $Y_i \sim Pois(\lambda),\;i\in{1,2}$.</p>

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

CC BY-SA 3.0
2016-10-22T03:54:05.710Z
It can be shown that $P(X>Y_i) \leq e^{ -(\sqrt{\lambda}-\sqrt{\mu})^2}$
Vivek Bagaria
CC BY-SA 3.0
2016-10-26T17:20:49.187Z
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
CC BY-SA 3.0
2016-10-27T05:10:25.010Z
This question deals with the distribution of $\max\{Y_1,Y_2\}$: http://math.stackexchange.com/questions/1113990/distribution-of-the-maximum-of-independent-poisson-random-variables
usul
<reference-request><pr.probability><randomness><upper-bounds>
Probability of random variable $X$ less than $max(Y_i)$
Vivek Bagaria
# Probability of random variable $X$ less than $max(Y_i)$

(Asked by: username_0 on 2016-10-22)

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?
$$\mathbb{P}\big(X\geq\max(Y\_1, Y\_2)\big)$$

**Tags:** reference-request, pr.probability, randomness, upper-bounds
> **Comments:**
> **username_0**: It can be shown that $P(X>Y_i) \leq e^{ -(\sqrt{\lambda}-\sqrt{\mu})^2}$  
> **username_0**: Note : Even though the $X, Y_1, Y_2$ are independent, the event $X>Y_1$ and $X>Y_2$ are not independent.  
> **username_1**: This question deals with the distribution of $\max\{Y_1,Y_2\}$: http://math.stackexchange.com/questions/1113990/distribution-of-the-maximum-of-independent-poisson-random-variables  

---

## 1 answers

### Answer by username_2 on 2016-11-05. Score: 1

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 $k < O(\lambda)$.


---

