Question #339374

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


1
Expert's answer
2022-05-10T23:31:40-0400

XX has a Poisson distribution. It means that XX takes values from the set of nonnegative integers 0,1,2,...0,1,2,... and P(X=k)=λkeλk!P(X=k)=\frac{\lambda^ke^{-\lambda}}{k!}, where k=0,1,2,...k=0,1,2,.... For YY we have: P(Y=n)=λneλn!P(Y=n)=\frac{\lambda^ne^{-\lambda}}{n!}, n=0,1,2,...n=0,1,2,.... A random variable W=X+YW=X+Y has a Poisson distribution with parameter 2λ2\lambda. It is a well-known fact. It can be also checked directly: P(W=n)=P(X+Y=n)=k=0nP(X=k,Y=nk)=k=0nλkeλk!λnkeλ(nk)!=k=0nλne2λk!(nk)!=e2λn!k=0nn!λnk!(nk)!=(2λ)nn!e2λP(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 WW has the form:

G(z)=E[zW]=x=0+p(x)zx=x=0+(2λ)xx!e2λzx=e2λx=0+(2λz)xx!=e2λ(z1)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 e2λze^{2\lambda z}. The latter is a known formula. Substitute z=etz=e^t. We receive: M(t)=E[etW]=e2λ(et1)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)=(2λete2λ(et1))t=0=2λE[W]=M'(0)=(2\lambda e^te^{2\lambda(e^t-1)})|_{t=0}=2\lambda.

E[W2]=M(0)=(2λete2λ(et1)+4λ2e2te2λ(et1))t=0=2λ+4λ2E[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[W2](E[W])2=2λ+4λ24λ2=2λVar[W]=E[W^2]-(E[W])^2=2\lambda+4\lambda^2-4\lambda^2=2\lambda

Answer: The probability generating function of WW is: G(z)=E[zW]=e2λ(z1)G(z)=E[z^W]=e^{2\lambda(z-1)}. The mean is: E[W]=2λE[W]=2\lambda, Var[W]=2λVar[W]=2\lambda.


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