# Poisson distribution of two independent random variables

Let X1 and X2 be two independent random variables. Let X1 and Y=X1+X2 have Poisson distributions with means μ1 and μ>μ1, respectively. Find the distribution of X2.

I don't know how to start solving with this kind of problem. I would appreciate if someone could help me. Thanks.

• A good start would be to search our site for related threads. They will reveal simple relationships among sums of Poisson variables as well as various useful techniques for working with them.
– whuber
Commented Apr 3, 2014 at 15:46
• To help you a bit more, whenever you read about the sum of independent random variables, you should think immediately about characteristic functions, or in the discrete case, about probability generating functions. This will give you a short and elegant solution for your problem. By the way, Raikov's Theorem gives an even stronger statement for your specific problem. Commented Apr 3, 2014 at 21:53
• is probability generating function the same as moment generating function? Commented Apr 3, 2014 at 22:43
• user42404 no, they're different, but related -- one generates probability and the other generates moments. Each have their uses, but sometimes you can use either. The wikipedia page for the Poisson lists both. Commented Apr 4, 2014 at 4:38
• The problem with your approach is that you do not know the MGF of X2. I will show you how to do it with the probability generating function (PGF) and you can try to translate. 1) Y has the same distribution as X1 + X2, so they have the same PGF. 2) This PGF is G(z) = exp(lambda(z-1)) 3) The PGF of X1 + X2 is the product of the two PGFs of X1 and X2, namely exp(lambda1(z-1)) H(z). 4) Solve for H(z). 5) Conclude. Commented Apr 4, 2014 at 9:39

Since $Y\sim \mathrm{Poisson}(\mu)$, the probability generating function of $Y$ is given by $$G_Y(z)=e^{\mu(z-1)}$$ and that of $X_1+X_2$ (by independence) as $$G_{X_1+X_2}(z)=G_{X_1}(z)G_{X_2}(z)=e^{\mu_1(z-1)}G_{X_2}(z)\,.$$ But since $Y$ has the same distribution as $X_1+X_2$, it is true that $G_Y=G_{X_1+X_2}$, which means that $$e^{\mu(z-1)}=e^{\mu_1(z-1)}G_{X_2}(z)\,.$$ In consequence, $G_{X_2}(z)=e^{(\mu-\mu_1)(z-1)}$, which we recognize as the probability generating function of a $\mathrm{Poisson}(\mu-\mu_1)$ distributed random variable. So $X_2\sim \mathrm{Poisson}(\mu-\mu_1)$.