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?


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?

  • $\begingroup$ 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. $\endgroup$ – lydias May 7 '17 at 21:57
  • $\begingroup$ actually by central limit theorem, the distribution will be approaching normal as n approaches $\infty$ $\endgroup$ – lydias May 8 '17 at 2:58
  • $\begingroup$ 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. $\endgroup$ – lydias May 9 '17 at 4:11
  • $\begingroup$ heropup: Context. You might be more successful than I have been, to understand what is going on... $\endgroup$ – Did May 9 '17 at 6:34

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