# The probability of maximum of two independent Poisson-distributed variables

I have two Poisson distributions with parameters $\lambda_1$ and $\lambda_2$ and two independent variables $A$ and $B$ from these two distributions. I know that

$$\begin{equation*} \mathsf P(\min(A,B)>K)=\mathsf P(A>K) \ast \mathsf P(B>K) \end{equation*}$$

I'm just wondering what $\mathsf P(\max(A,B)>K)$ is. Thanks.

If the maximum of two numbers is greater than some constant, then either one or the other of the numbers is greater than it;   they are not both less than or equal to it.

If the random variables are independently Poisson distributed then:

\begin{align} \mathsf P(\max(A,B)>K) & =\mathsf P(A>K \cup B>K) \\[1ex] & = 1 - \mathsf P(A\leq K)\mathsf P(B\leq K) \\[1ex] & = 1 - \sum_{j=0}^K\frac{\lambda_1^j e^{-\lambda_1}}{j!}\;\sum_{j=0}^K\frac{\lambda_2^j e^{-\lambda_2}}{j!} \\[1ex] & = 1 - \frac{\Gamma(K+1,\lambda_1)\Gamma(K+1,\lambda_2)}{K!^2} \end{align}

$\Gamma(\cdot,\cdot)$ is the Incomplete Gamma Function

• Thank you. Is this approach also true for more than two variables? – Taban May 11 '15 at 10:53
• Both of the Answers are applicable to several independent random variables. This one for several independent Poisson random variables. – BruceET May 12 '15 at 3:13

Let $W = \max(A, B)$, then $P(W > k) = 1 - P(W \le k) = 1 - P(A \le k)P(B \le k).$

• I think $W$ became $X$ and $K$ became $k$. +1 anyway. – Nicolas May 11 '15 at 9:08
• Right. Phone rang while proofing. – BruceET May 11 '15 at 9:09