Question

Please provide additional steps.

Xn be a sequence of independent random vari Example 38.1. Let Xi,..., ables that are all Poisson(A). Then, E(X1-λ, Var(X)-λ, and MX1 (t) = eA(e*-1). Let us compute the mgt of Zn-(X1 + . . . + Xn-nλ)WnX: ηλ nex According to the Taylor-Macl.aurin expansion, etv1smaller terms. Thus, Mz, (t) = exp {-tvAn + tvAn + 2 + smaller terms} n艹eta

Answer 1

$=e^{-t\sqrt{n \lambda}}(E(e^{\frac{t}{\sqrt{n \lambda}}X_1}))^n$ ( since X_i are iid)

$=e^{-t\sqrt{n \lambda}}(M_{X_1}(\frac{t}{\sqrt{n \lambda}}))^n$

Now, MGF of $Poisson(\lambda)$

$M_X(t)=E(e^{tX})= \sum_{x=0}^{\infty}e^{tx}\frac{e^{- \lambda} \lambda^x}{x!}= e^{- \lambda}\sum_{x=0}^{\infty}\frac{(\lambda e^t)^x}{x!} = e^{-\lambda} e^{\lambda e^t}$ ( From the Taylor series expansion)

So,

$M_{Z_n}(t)=e^{-t\sqrt{n \lambda}}(e^{-\lambda + \lambda e^{\frac{t}{\sqrt{n \lambda}}}})^n\\\\ =e^{-t\sqrt{n \lambda}}e^{ -n\lambda +n\lambda e^{\frac{t}{\sqrt{n \lambda}}} }$

According to the Taylor series expansion given in the question,

$=e^{-t\sqrt{n \lambda}}e^{ -n\lambda +n\lambda (1+\frac{t}{\sqrt{n\lambda }}+\frac{t^2}{2n \lambda} +\frac{t^3}{6(n\lambda)^{\frac{3}{2}}}+......)}$

$=e^{-t\sqrt{n \lambda}}e^{ -n\lambda +n\lambda+t\sqrt{n \lambda}+\frac{t^2}{2} +\frac{t^3}{6(n\lambda)^{\frac{1}{2}}}+......)}$

$=e^{ (\frac{t^2}{2} +\frac{t^3}{6(n\lambda)^{\frac{1}{2}}}+......)}$

Now it is know that as n tends to infinity $\frac{1}{n^r}$ goes to 0 for r>0

$n \to \infty =e^{\frac{t^2}{2}}$