# What is the limiting distribution of $Y_n = \sqrt{n}(\bar{X}_n-1)$ as $n \to \infty$?

Let $$X_1,\cdots,X_n$$ be independently and identically distributed with pdf $$f(x)=e^{-x}, 0 < x < \infty$$. Let $$Y_n = \sqrt{n}(\bar{X}_n-1)$$.

What is the limiting distribution of $$Y_n$$ as $$n \to \infty$$?

My work:

I decided to try an mgf approach. Clearly, $$X_1,\cdots,X_n \sim Exp(1)$$, so $$M_{X_i}(t)=\frac{1}{1-t}, t < 1$$. After a bit of work, I found that

$$M_{Y_n}(t)=[e^{t/\sqrt{n}}(1-\frac{t}{\sqrt{n}})]^{-n}, t < \sqrt{n}$$. This does not appear to resemble a known distribution's mgf. Should I change my approach?

Updated:

I think I'm supposed to solve the problem using an mfg method. Now I am getting this:

$$M_{y_n}(t)=[[1 + \frac{t}{\sqrt{n}} + (\frac{t}{\sqrt{n}})^2\cdot1/2! + \cdots]-[\frac{t}{\sqrt{n}} +(\frac{t}{\sqrt{n}})^2+(\frac{t}{\sqrt{n}})^3\cdot 1/3!+\cdots]]^{-n}$$, which resembles some work that we have done in class, but I am not too sure how I can evaluate this.

Due to Glen_b's comment, I attempted using CLT. Here is my updated "updated work."

Since $$X_i \sim Exp(1), i=1,\cdots,n$$, then $$E(X_i)=1, Var(X_i)=1$$. So,

$$\sqrt{n}(\bar{X}_n-1) \to N(0,1)$$ in distribution by Central Limit Theorem, which matches the answers provided below.

• Hint: $\bar{X}_n$ has a Gamma distribution Nov 23, 2019 at 21:46
• If you're using the mgf approach (NB I have not checked your mgf is correct), what happens to the function in the limit as $n\to \infty$? Nov 23, 2019 at 22:16
• Are you allowed to invoke the Central Limit Theorem? (Even if not, it tells you immediately what the limiting distribution is, which can guide your demonstration.)
– whuber
Nov 23, 2019 at 22:24
• @edison (i) regarding to your discussion with whuber -- what does the CLT say? (ii) On the other hand if (as I had previously assumed) you can't invoke the CLT, you have already expanded $M$ (though as I say I am not checking your work), what you'd need next is to use facts about limits. Nov 24, 2019 at 2:50
• No, it's incredibly easy to deal with. What's $\mu_Y$? What's $\sigma_Y$? Write down a standardized $\bar{Y}$ ... Nov 24, 2019 at 6:19

In your updated version, you will get $$\left[1 - \frac{t^2/2}{n} +o(\frac{1}{n})\right]^{-n} = \left[1 + \frac{t^2/2}{n} +o(\frac{1}{n})\right]^{n}$$
and the limit of that is $$e^{t^2/2}$$, which is the moment generating function of a standard normal distribution with mean $$0$$ and variance $$1$$, much as you might expect from the central limit theorem
• What is $o(\frac{1}{n})$? Sorry, I am not familiar with this notation. Additionally, how do you get that equality? Nov 24, 2019 at 1:49
• @Edison This is Little o notation and here means means that the remainder when divided by $\frac1n$ tends to $0$ as $n$ increases so eventually very much smaller in magnitude than $\frac1n$ or $\frac{t^2/2}{n}$. The equality is similar to saying $\frac{1}{1-x} = 1+x +\cdots$ but here we have $\frac{t^2/2}{n}$ rather than $x$ Nov 24, 2019 at 2:03
• Thank you for linking the Wikipedia article- that was very helpful. I get that $\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n = e^x$, but how come we can neglect $o(\frac{1}{n})$? Also, how did you know to keep the $-\frac{t^2/n}{n}$ term out of $o(\frac{1}{n})$? Nov 24, 2019 at 3:17