# Sum of two Poissons (not as easy as it looks!)

• Mar 1st 2009, 11:13 PM
Amanda1990
Sum of two Poissons (not as easy as it looks!)
I can show that the sum of two Poisson processes of rates a and b is a Poisson process of rate a+b.

Now, find the probability that the arrival of the combined (a+b) Poisson process comes from the process of rate a. I know that the answer must surely be a/(a+b), but how do we actually go about formally proving it?
• Mar 2nd 2009, 12:53 AM
Laurent
Quote:

Originally Posted by Amanda1990
I can show that the sum of two Poisson processes of rates a and b is a Poisson process of rate a+b.

Now, find the probability that the arrival of the combined (a+b) Poisson process comes from the process of rate a. I know that the answer must surely be a/(a+b), but how do we actually go about formally proving it?

I'm not sure what exactly you mean by "the arrival" of a Poisson process, but I'm pretty sure what you need to prove eventually is that if $X,Y$ are independent, such that the distribution of $X$ is exponential with parameter $a$ and the distribution of $Y$ is exponential with parameter $b$, then $P(X\leq Y)=\frac{a}{a+b}$. I have seen this done on the forum a few times; for instance you can write, using independence, $P(X\leq Y)=\int_0^\infty P(Y\geq x) a e^{-a x} dx= \int_0^\infty e^{-bx} a e^{-a x} dx$, etc.