Answer to Question #338484 in Statistics and Probability for Calvin

Question

Let X, Y be a 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

Suppose that random variables $X$ and $Y$ have Poisson distribution with parameter $\lambda$. It means that their densities have the following form: $P(X=k)=\frac{\lambda^ke^{-\lambda}}{k!},$ $P(Y=n)=\frac{\lambda^ne^{-\lambda}}{n!}$ , where $k,n=0,1,2,...$. Compute the density for $W$: $P(W=n)=\sum_{l=0}^nP(X=l)P(Y=n-l)=\sum_{l=0}^n\frac{\lambda^{n}e^{-2\lambda}}{l!(n-l)!}=\frac{1}{n!}e^{-2\lambda}\sum_{l=0}^nC_n^l\lambda^n$, where $C_n^l=\frac{n!}{l!(n-l)!}$. We use the binomial formula and get: $P(W=n)=\frac{2\lambda e^{-2\lambda}}{n!}$. Thus, $W$ also has a Poisson distribution with parameter: $\mu=2\lambda$. It is a known fact. We remind that the probability generating function is defined by: $G(z)=E[z^W]=\sum_{k=0}^{+\infty}P(W=k)z^k=\sum_{k=0}^{+\infty}\frac{\mu^ke^{-\mu}}{k!}z^k=e^{-\mu}\sum_{k=0}^{+\infty}\frac{(\mu z)^k}{k!}=e^{-\mu}e^{\mu z}=e^{\mu(z-1)}$

We used the Taylor series for $e^{\mu z}$. We received that the probability generating function has the form: $G(z)=e^{\mu(z-1)}$. It is also a known fact. After setting: $z=e^t$ we receive: $M(t)=e^{\mu(e^t-1)}$. It is a moment-generating function. We remind a known fact that $M^{(n)}(0)=E[W^n]$. I.e., $n$-th derivative ($n=0,1,2,...$ ) of $M(t)$ provides the expectation $E[W^n].$ We receive: $M'(t)=\mu e^te^{\mu(e^t-1)}$, $M''(t)=\mu e^{t} e^{\mu(e^t-1)}+(\mu e^t)^2e^{\mu(e^t-1)}$.

Thus, $E[W]=M'(0)=\mu$, $E[W^2]=M''(0)=\mu+\mu^2$. The variance is: $Var[W]=E[W^2]-(E[W)]^2=\mu+\mu^2-\mu^2=\mu$.

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