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?


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.

Additional Update:

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.

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    $\begingroup$ Hint: $\bar{X}_n$ has a Gamma distribution $\endgroup$ – StijnDeVuyst Nov 23 '19 at 21:46
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    $\begingroup$ 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 $n\to \infty$? $\endgroup$ – Glen_b -Reinstate Monica Nov 23 '19 at 22:16
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    $\begingroup$ 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.) $\endgroup$ – whuber Nov 23 '19 at 22:24
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    $\begingroup$ @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. $\endgroup$ – Glen_b -Reinstate Monica Nov 24 '19 at 2:50
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    $\begingroup$ No, it's incredibly easy to deal with. What's $\mu_Y$? What's $\sigma_Y$? Write down a standardized $\bar{Y}$ ... $\endgroup$ – Glen_b -Reinstate Monica Nov 24 '19 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

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  • $\begingroup$ What is $o(\frac{1}{n})$? Sorry, I am not familiar with this notation. Additionally, how do you get that equality? $\endgroup$ – Ron Snow Nov 24 '19 at 1:49
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    $\begingroup$ @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 $x$ $\endgroup$ – Henry Nov 24 '19 at 2:03
  • $\begingroup$ 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 $o(\frac{1}{n})$? $\endgroup$ – Ron Snow Nov 24 '19 at 3:17

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