Consider a simple random sample in which $ X_i \sim \exp(1) $, and consider the standardized variable $Z_n = (\bar{X_n} -1)\sqrt{n}$, in which $\bar{X_n}$ represents the sample mean.
(i) Determine the mgf of $X_i$ and $\bar{X_n}$.
(ii) Without using the CLT, show that $\lim_{n \to \infty} E(e^{tZ_n}) = e^{t^2/2}$, and conclude that $Z_n$ converges in distribution to the standard normal distribution.
Attempt
(i) The mgf of $ X_i $ is $ M_{X_i}(t) = \frac{1}{1 – t}$, for $t < 1$. I won't show the calculations for this. Using properties of MGFs, we have that
$$ M_{ \bar{X_n} }(t) = M_{\frac{X_1 + \cdots + X_n}{n}}(t) = M_{\frac{X_1}{n}}(t) M_{\frac{X_2}{n}}(t) \cdots M_{\frac{X_n}{n}}(t) = M_{X_1}(t/n) M_{X_2}(t/n) \cdots M_{X_n}(t/n) $$
$$ M_{ \bar{X_n} }(t) = \Bigl[ M_{X_1}(t/n) \Bigr]^n = \Bigl[ \frac{1}{1 – t/n} \Bigr]^n = \frac{1}{ (1 – t/n)^n} = (1 – \frac{t}{n})^{-n} $$
We recognize this mgf as the mgf of a Gamma$(n, n)$ distribution, which implies that $\bar{X_n} \sim $ Gamma$(n, n)$.
(ii) First of all, we observe that $E(e^{tZ_n})$ is, by definition, the mgf of $Z_n$. To calculate this mgf, we need the pdf of $Z_n$, which we obtain as follows
$$ P(Z_n \leq x) = P(\bar{X_n}\sqrt{n} – \sqrt{n} \leq x) = P( \bar{X_n} \leq \frac{x}{\sqrt{n}} + 1) = F_{\bar{X_n}}(\frac{x}{\sqrt{n}} + 1) $$
By differentiating, we obtain
$$ f_{Z_n}(x) = f_{\bar{X_n}}(\frac{x}{\sqrt{n}} + 1)\frac{1}{\sqrt{n}} = \frac{n^n}{\Gamma(n)} \Bigl( \frac{x}{\sqrt{n}} + 1 \Bigr)^{n-1} e^{-n(\frac{x}{\sqrt{n}} + 1)} \frac{1}{\sqrt{n}}, ~~ x > 0. $$
Now, finally, we can calculate $E(e^{tZ_n})$, the mgf of Z_n.
$$ E(e^{tZ_n}) = \int_{0}^{\infty}e^{tx}f_{Z_n}(x) = \int_{0}^{\infty}e^{tx} \frac{n^n}{\Gamma(n)} \Bigl( \frac{x}{\sqrt{n}} + 1 \Bigr)^{n-1} e^{-n(\frac{x}{\sqrt{n}} + 1)} \frac{1}{\sqrt{n}} dx $$
This is where I'm currently stuck. This integral seems crazy and it makes me think I may be doing something wrong, or at least it makes me think there must be some easier path to solving this problem.
Edit:
Thanks to Clement's comment, I realized that I can calculate $ E(e^{tZ_n}) $ using the mgf of $\bar{X_n}$, which I had already calculated. Then the calculation of the desired limit, which is not a trivial one, has been shown in the answer to the question. This question can be closed.
Best Answer
$\newcommand{\e}{\operatorname E}$ \begin{align} M_{Z_n}(t) = {} & \e(e^{tZ_n} ) = \e\left( e^{t(\overline X_n -1)\sqrt n} \right) \\[8pt] = {} & e^{-t\sqrt n}\e\left(e^{t(X_1+\cdots +X_n)/\sqrt n} \right) = e^{-t\sqrt n} \left( \e\left(e^{tX_1/\sqrt n} \right) \right)^n \\[8pt] = {} & e^{-t\sqrt n} \big( M_{X_1}(t/\sqrt n) \big)^n = e^{-t\sqrt n} \left( \frac 1 {1 - \frac t {\sqrt n}} \right)^n \\[8pt] = {} & \exp \left( -t\sqrt n - n \log\left( 1 - \tfrac t {\sqrt n} \right) \right) \end{align} This logarithm has a power series: \begin{align} & -t\sqrt n - n \log\left( 1 - \tfrac t {\sqrt n} \right) \\[8pt] = {} & -t\sqrt n - n\left( \tfrac{-t}{\sqrt n} - \tfrac{(-t/\sqrt n)^2} 2 + \tfrac{(-t/\sqrt n)^3} 3 - \tfrac {(-t/\sqrt n)^4} 4 + \cdots \right) \\[8pt] = {} & \frac{t^2} 2 + (\text{terms that approach 0 as } n\to\infty). \end{align}
Ultimately the goal is to show that the sequence of CDFs converges to the CDF of the standard normal. What is shown here is that the sequence of MGFs converges to the MGF of the standard normal. A "continuity theorem," often omitted from statements of exercises like this one, say that if the sequence of MGFs converges, then so does the sequence of CDFs.