## Related questions with answers

$\begin{array} { l } { \text { Let } X _ { n } \sim \text { Pois } ( n ) \text { for all positive integers } n \text { . Use MGFs to show that the distribution of } } \\ { \text { the standardized version of } X _ { n } \text { converges to a Normal distribution as } n \rightarrow \infty , \text { without } } \\ { \text { invoking the CLT. } } \end{array}$

Solution

VerifiedWe are given that $X_n \sim \text{Pois}(n)$. Since $E(X_n) = n$ and $\operatorname{Var}(X_n) = n$, we define standardized variables

$\begin{align*} Y_n := \frac{X_n - n}{\sqrt{n}} \end{align*}$

Let's find the MGF of $Y_n$. We have that

$\begin{align*} M_{Y_n}(t) = E[e^{tY_n}] = E[e^{t\frac{X_n - n}{\sqrt{n}}}] = e^{-t\sqrt{n}}M_{X_n}(\frac{t}{\sqrt{n}}) \end{align*}$

Since we know MGF of Poisson distribution, we have that

$\begin{align*} e^{-t\sqrt{n}}M_{X_n}(\frac{t}{\sqrt{n}}) = e^{-t\sqrt{n}} e^{n(e^{t/\sqrt{n}}-1)} \end{align*}$

We are interested in what happens with $M_{Y_n}(t)$ as $n \rightarrow \infty$. Let's consider the expression that is in exponential function. We have that

$\begin{align*} \lim_{n \rightarrow \infty} \left(n(e^{t/\sqrt{n}}-1) -t\sqrt{n} \right) &= \lim_{x \rightarrow \infty} \left(x(e^{t/\sqrt{x}}-1) -t\sqrt{x} \right) = [\text{use } x = \frac{1}{y^2}] \\ &= \lim_{y \rightarrow 0} \left(\frac{1}{y^2}(e^{ty}-1) -t\frac{1}{y} \right) \\ &= \lim_{y \rightarrow 0} \frac{e^{ty}-1-ty}{y^2} = [\text{L'Hopital}] \\ &= \frac{t}{2} \lim_{y \rightarrow 0} \frac{e^{ty}-1}{y} = - \frac{t^2}{2} \end{align*}$

Hence, we have that

$\begin{align*} M_{Y_n}(t) = e^{-t\sqrt{n}} e^{n(e^{t/\sqrt{n}}-1)} \rightarrow e^{- \frac{t^2}{2}} \end{align*}$

as $n \rightarrow \infty$. But the limit MGF is exactly the MGF of Standard Normal Distribution. Hence

$\begin{align*} Y_n \rightarrow \mathcal{N}(0,1) \end{align*}$

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