# 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}$$M _{\bar{X_{n}}}\left ( \sqrt{n}t \right) By knowing MGF of \bar{X_{n}}, I will be able to plug it back in (1) to derive MGF of Y_n. (b) What is the limiting distribution of Y_n , as n \space \rightarrow \infty ? I believe after finding the MGF of Y_n I will be able to know the distribution of Y_n , meaning it will allow me to find the pdf by using MGF? Then set n \space \rightarrow \infty to find a distribution that match the result? ## 1 Answer First, if X_1, X_2, \ldots, X_n are IID random variables with common distribution X and moment generating function M_X(t) = \operatorname{E}[e^{tX}], then$$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? • I tried to plug in$\infty$for n, but the MGF for$Y_n$is approaching$\infty$instead of$e^{t^2/2}$. I searched for your other responses: math.stackexchange.com/questions/1268881/… and observed similar method. I know that$\sum_{i=1}^n X_i$is a Gamma random variable so does$\bar X_n $, but with n approaching to$\infty$Gamma function will converge to 1, and leaving just$e^{-t\sqrt{n}}$, which look the closest to exponential distribution? I really appreciate your explanation, I learned a lot from them. – lydias May 7 '17 at 21:57 • actually by central limit theorem, the distribution will be approaching normal as n approaches$\infty$– lydias May 8 '17 at 2:58 • Hi Heropup, would you be willing to show how you used Taylor expansion to arrive at$e^{\frac{t^2}{2}}$? I understand that as n approaches$\infty$, MGF of$Y_n$=$e^{\frac{-t}{\sqrt{n}}}e^{-nIn(1-t/\sqrt{n})}$by Taylor expansion it will converge to$e^{\frac{t^2}{2}}$, which shows the limit distribution is converge to standard Normal with MGF:$e^{\mu t +\sigma^2t^2/2}$in this case$\mu = 0 $and$\sigma=1\$ by standard normal. – lydias May 9 '17 at 4:11
• heropup: Context. You might be more successful than I have been, to understand what is going on... – Did May 9 '17 at 6:34