# Central Limit Theorem; Exponential Distribution

I'm trying to prove the Central Limit Theorem for the exponential distribution and I'm running into problems. This is what I've done so far:

Given $S = X_1+X_2+...+X_n$ where each $X_i$ is an independent and identically distributed random variable with probability density function equal to $f(x) = \alpha e^{-\alpha x}$ for $x > 0$, show that as $n \rightarrow \infty$, the distribution of $\frac{S-n\mu}{\sqrt{n}\sigma}$ approaches the standard normal distribution, i.e., $N(0,1)$.

My strategy was to use moment generating functions (since if you can show that the mgf approaches that of the standard normal r.v., i.e., $e^{\frac{t^2}2}$, then you have proven the result by the uniqueness of mgf's):

$M_x(t)=\int_0^{\infty}e^{tx} \alpha e^{-\alpha x} dx = \frac{\alpha}{\alpha-t}$ for $t<\alpha$.

We therefore have:

$Ms(t)=M_{X_1} \cdot M_{X_2} \cdot ...\cdot M_{X_n} = [M_{x}]^n$

Thus we are examining the limiting behavior of

$\frac{[M_{x}]^n-n\mu}{\sqrt{n}\sigma} = \frac{[M_{x}]^n-\frac{n}{\alpha}}{\frac{\sqrt{n}}{\alpha}} = \frac{[\frac{\alpha}{\alpha-t}]^n-\frac{n}{\alpha}}{\frac{\sqrt{n}}{\alpha}}$ since $\mu = \sigma = \frac{1}{\alpha}$ for the exponential distribution.

After this point I don't know what to do. Could someone help me calculate the limit as n approaches infinity? It should end up being $e^{\frac{t^2}2}$. Thanks

• You need to revise the way to express the MGF of $(S-a)/b$ in terms of the MGF of $S$... No idea where the formula $((M_X)^n-n\mu)/(\sqrt{n}\sigma)$ is coming from. – Did Apr 15 '16 at 20:29
• @Did I explained in the question...I simply put the derived formula for $M_S$ into the expression for the distribution I mentioned in the second paragraph ...the formula for $M_S$ was derived from a property of moment-generating functions, that the mgf of a sum is the product of the mgfs – Gabriel Apr 15 '16 at 22:15
• Indeed $M_S=(M_X)^n$ but $M_{(S-a)/b}$ is certainly not $(M_S-a)/b$. – Did Apr 16 '16 at 6:39

Write $S$ in the form $S=\sum_{i=1}Y_i,$ where we define $$Y_i:= {X_i-\mu\over \sqrt n\sigma}.$$ The $Y$'s are are iid, so the MGF of $S$ is the $n$th power of the MGF of a single $Y$: $$M_S(t)=( M_Y(t) ) ^n.$$ As for the MGF of $Y$, we have \begin{align} M_Y(t)&=E\exp t( \frac{X-\mu}{\sqrt n\sigma} )\\ & =\exp(-\frac{t\mu}{\sqrt n\sigma})E \exp\frac{tX}{\sqrt n\sigma}\\ & =\exp(-\frac{t\mu}{\sqrt n\sigma})M_X(\frac{t}{\sqrt n\sigma}). \end{align} Plug in the MGF for the exponential with $\mu=\sigma=\frac1\alpha$ and this simplifies to $$M_Y(t) = {\exp(-\frac t{\sqrt n})\over 1-{t\over\sqrt n}}.$$ Raise this to the $n$th power, and take the limit as $n\to\infty$. (Your best bet is to log everything and use the power series expansion for $\log(1-x)$ for small $x$; in log space the limit will be $t^2/2$.)
• Why did you define $Y_i$ as $\frac{X_I-\mu}{\sqrt{n}\sigma}$ and not as $\frac{X_I-n\mu}{\sqrt{n}\sigma}$? – Gabriel Apr 17 '16 at 20:00
• If you sum $X_i-\mu\over\sqrt n\sigma$ over all $i$ you get $n$ copies of $\mu$. If you define $Y_i$ with $n\mu$ you get $n^2\mu$, which is wrong. – grand_chat Apr 18 '16 at 2:02