X has a Poisson distribution. It means that X takes values from the set of nonnegative integers 0,1,2,... and P(X=k)=k!λke−λ, where k=0,1,2,.... For Y we have: P(Y=n)=n!λne−λ, n=0,1,2,.... A random variable W=X+Y has a Poisson distribution with parameter 2λ. It is a well-known fact. It can be also checked directly: P(W=n)=P(X+Y=n)=∑k=0nP(X=k,Y=n−k)=∑k=0nk!λke−λ(n−k)!λn−ke−λ=∑k=0nk!(n−k)!λne−2λ=n!e−2λ∑k=0nk!(n−k)!n!λn=n!(2λ)ne−2λ
We used a binomial formula. Remind that the probability generating function for W has the form:
G(z)=E[zW]=∑x=0+∞p(x)zx=∑x=0+∞x!(2λ)xe−2λzx=e−2λ∑x=0+∞x!(2λz)x=e2λ(z−1)
We used a Taylor series expansion for e2λz. The latter is a known formula. Substitute z=et. We receive: M(t)=E[etW]=e2λ(et−1). This is a moment-generating function. Using the known formulae, we have: E[W]=M′(0)=(2λete2λ(et−1))∣t=0=2λ.
E[W2]=M′′(0)=(2λete2λ(et−1)+4λ2e2te2λ(et−1))∣t=0=2λ+4λ2.
Var[W]=E[W2]−(E[W])2=2λ+4λ2−4λ2=2λ
Answer: The probability generating function of W is: G(z)=E[zW]=e2λ(z−1). The mean is: E[W]=2λ, Var[W]=2λ.
Comments