Moment generating function of sample mean and limiting distribution 
Problem: 
  Let $\bar{X_{n}}$ be the mean of a random sample of size n from an exponential distribution with the following density fctn:
  $$f(x) = e^{-x}, 0<x<\infty \space, 0 \space otherwise$$
  $Y_n=\sqrt{n}(\bar{X_{n}}-1)$
(a) What is the MGF of $Y_n?$ 

M(t) of exponential is $\frac{1}{1-t}$ and M(t) of $\bar{X_n}$ is $M_X(\frac{t}{n})^n$. How do I derive MGF for $\bar{X_{n}}$ with given condition?
From a hint, the first step to obtain MGF of $Y_n$ is getting 
(1) $M _{Y_{n}}(t)$= $e^{-\sqrt{n}t}$$M _{\bar{X_{n}}}\left ( \sqrt{n}t \right)$
By knowing MGF of $\bar{X_{n}}$, I will be able to plug it back in (1) to derive MGF of $Y_n$.

(b) What is the limiting distribution of $Y_n$ , as $n \space \rightarrow \infty$ ?

I believe after finding  the MGF of $Y_n$ I will be able to know the distribution of $Y_n$ , meaning it will allow me to find the pdf by using MGF? Then set $n \space \rightarrow \infty$ to find a distribution that match the result? 
 A: First, if $X_1, X_2, \ldots, X_n$ are IID random variables with common distribution $X$ and moment generating function $M_X(t) = \operatorname{E}[e^{tX}]$, then $$S_n = \sum_{i=1}^n X_i$$ has moment generating function $$M_{S_n}(t) = \operatorname{E}[e^{tS_n}] = \operatorname{E}[e^{t\sum_{i=1}^n X_i}] = \operatorname{E}\left[\prod_{i=1}^n e^{tX_i}\right] \overset{\text{ind}}{=} \prod_{i=1}^n \operatorname{E}[e^{tX_i}] = \prod_{i=1}^n M_X(t) = (M_X(t))^n.$$  That is to say, the MGF of a sum of $n$ IID random variables is equal to the MGF of one such random variable raised to the $n^{\rm th}$ power.
It follows from this that if $X \sim \operatorname{Exponential}(\lambda)$, where $\lambda$ is a rate parameter, then $$M_X(t) = \frac{\lambda}{\lambda - t}.$$  Your problem is the special case $\lambda = 1$.  Therefore, $$M_{S_n}(t) = \left(\frac{\lambda}{\lambda - t}\right)^n,$$ where we have defined $S_n$ as the sample total as described above.  How does this help us with the MGF of $Y_n$?  Well, we know $$Y_n = \sqrt{n}(\bar X_n - 1).$$ We also know that $$\bar X_n = S_n/n,$$ that is to say, the sample mean is simply the sample total divided by the sample size $n$.  Thus, $$M_{Y_n}(t) = \operatorname{E}[e^{t(\sqrt{n}(\bar X_n - 1))}] = \operatorname{E}[e^{(t/\sqrt{n})S_n - t\sqrt{n}}] = \operatorname{E}[e^{(t/\sqrt{n}) S_n}] \operatorname{E}[e^{-t\sqrt{n}}] = e^{-t\sqrt{n}} M_{S_n}(t/\sqrt{n}).$$  The first equality is the definition of MGF.  The second follows from simple algebraic manipulation of the exponent.  The third is the result of factoring out the non-random term $e^{-t\sqrt{n}}$, which is not a function of any random variable, and the fourth is the definition of MGF again.  Finally, we use the earlier result for $M_{S_n}(t)$ to get $$M_{Y_n}(t) = e^{-t\sqrt{n}} \left(\frac{1}{1 - t/\sqrt{n}}\right)^n.$$
To find the limiting distribution of $Y_n$, you need to evaluate $$\lim_{n \to \infty} M_{Y_n}(t).$$  This should give you $e^{t^2/2}$, but I have left the proof as an exercise.  What familiar distribution has such an MGF?  Why does this make sense?
