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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – whuber
    Commented Apr 3, 2014 at 15:46
  • 2
    $\begingroup$ 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. $\endgroup$
    – binkyhorse
    Commented Apr 3, 2014 at 21:53
  • $\begingroup$ is probability generating function the same as moment generating function? $\endgroup$
    – user42404
    Commented Apr 3, 2014 at 22:43
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    $\begingroup$ 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. $\endgroup$
    – Glen_b
    Commented Apr 4, 2014 at 4:38
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    $\begingroup$ 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. $\endgroup$
    – binkyhorse
    Commented Apr 4, 2014 at 9:39

1 Answer 1


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 $X_2\sim \mathrm{Poisson}(\mu-\mu_1)$.

  • $\begingroup$ Nice. After that I'll just have to inetgrate it to get the cdf. Thanks! Im not really familiar with these pgfs, instead im familiar with their mgfs. Some textbooks don't discuss pgf. $\endgroup$
    – user42404
    Commented Apr 5, 2014 at 0:48
  • $\begingroup$ You are welcome. Note that no more integration is required - by showing that the PDF of X_2 is the PDF of a Poisson random variable, we already showed that the distribution of X_2 is indeed Poisson (and we also found its parameter). By the way, the same idea can be used to prove the central limit theorem (there, with characteristic functions replacing the PGF): find out the characteristic function of the limit and see that it's the same as that of a standard normal random variable. Then argue that by uniqueness of the characteristic function, the limit must have a normal distribution. $\endgroup$
    – binkyhorse
    Commented Apr 5, 2014 at 20:19

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