Answer to Question #339374 in Statistics and Probability for ken

Question

Let X, Y be independent and identically distributed random variables, each having Poisson distribution with rate parameter λ. Find the probability generating function of W = X + Y and hence the mean and variance of W

Answer 1

$X$ has a Poisson distribution. It means that $X$ takes values from the set of nonnegative integers $0,1,2,...$ and $P(X=k)=\frac{\lambda^ke^{-\lambda}}{k!}$, where $k=0,1,2,...$. For $Y$ we have: $P(Y=n)=\frac{\lambda^ne^{-\lambda}}{n!}$, $n=0,1,2,...$. A random variable $W=X+Y$ has a Poisson distribution with parameter $2\lambda$. It is a well-known fact. It can be also checked directly: $P(W=n)=P(X+Y=n)=\sum_{k=0}^nP(X=k,Y=n-k)=\sum_{k=0}^n\frac{\lambda^ke^{-\lambda}}{k!}\frac{\lambda^{n-k}e^{-\lambda}}{(n-k)!}=\sum_{k=0}^n\frac{\lambda^ne^{-2\lambda}}{k!(n-k)!}=\frac{e^{-2\lambda}}{n!}\sum_{k=0}^n\frac{n!\lambda^n}{k!(n-k)!}=\frac{(2\lambda)^n}{n!}e^{-2\lambda}$

We used a binomial formula. Remind that the probability generating function for $W$ has the form:

$G(z)=E[z^W]=\sum_{x=0}^{+\infty}p(x)z^x=\sum_{x=0}^{+\infty}\frac{(2\lambda)^x}{x!}e^{-2\lambda}z^x=e^{-2\lambda}\sum_{x=0}^{+\infty}\frac{(2\lambda z)^x}{x!}=e^{2\lambda(z-1)}$

We used a Taylor series expansion for $e^{2\lambda z}$. The latter is a known formula. Substitute $z=e^t$. We receive: $M(t)=E[e^{tW}]=e^{2\lambda(e^t-1)}$. This is a moment-generating function. Using the known formulae, we have: $E[W]=M'(0)=(2\lambda e^te^{2\lambda(e^t-1)})|_{t=0}=2\lambda$.

$E[W^2]=M''(0)=(2\lambda e^te^{2\lambda(e^t-1)}+4\lambda^2 e^{2t}e^{2\lambda(e^t-1)})|_{t=0}=2\lambda+4\lambda^2$.

$Var[W]=E[W^2]-(E[W])^2=2\lambda+4\lambda^2-4\lambda^2=2\lambda$

Answer: The probability generating function of $W$ is: $G(z)=E[z^W]=e^{2\lambda(z-1)}$. The mean is: $E[W]=2\lambda$, $Var[W]=2\lambda$.