# Moments and exponential distribution

Stuck on how to solve this, can't seem to find the answers online at all

a. The random variable $X$ has an exponential distribution with parameter and probability density function $f_X(x) = \theta e^{-\theta x}$ , where $x>= 0$.

Obtain the moment generating function $m_X(t) = E[e^{tX}]$ of $X$. For what values of t is this defined?

Suppose that $X_1, . . . ,X_n$ are mutually independent and identically distributed random variables each having an exponential distribution with parameter $\theta$. Let

$S_n = X_1 + · · · + X_n$.

Write down, together with a brief statement of what results you have used, the moment generating function of $S_{n}$ .

b. Suppose that

$y=\frac{\Theta S_{n}}{\sqrt{n}}-\sqrt{n}$

has moment generating function mY (t). Show that

$m_{y}(t)=e^{-t\sqrt{n}}\left ( 1-\frac{t}{\sqrt{n}} \right )^{n}$

Obtain $logmY (t)$ and deduce what happens as n -> $\infty$. What do you conclude about the distribution of $Y$ for large $n$? By expanding $mX(t)$ as a power series in $t$, or otherwise, deduce that $E[Xr] = r!/$$\theta$$^{r}$ for r = 1, 2, 3, . . ..

• Have you tried looking at the definition of the moment generating function? Its an extremely straight forward calculation: en.wikipedia.org/wiki/Moment-generating_function – user25658 Sep 4 '13 at 5:19
• Also, if you not, the above pdf $f_X(x)$ is an exponential distribution so the wikipedia page I provided above actually tells you what the moment generating function is. – user25658 Sep 4 '13 at 5:27
• thanks, that was surprisingly easier with the tips. Also I'm sorry for asking things that are second nature to you, I know I'm not very good but I'm not giving up – Kei Cheung Sep 4 '13 at 6:52
• Can you edit question (b)? Is $\Theta$ is the same as $\theta$ ? – Drew75 Sep 10 '13 at 12:10

Consider adding the self-study tag for homework questions.

The first part of your question, how to find the moment of an exponential, is generally well-explained online, and I'll let you look for that.

For the second question, if the $X_n$ are iid, the moment of the sum $S_n$ simplifies considerably because you can separate then in the expected values.

$M(S_n)=E(e^{tS_n})=E(e^{t(X_1+X_2+...)})=E(e^{tX_1}e^{tX_2}...)$

However the variables are iid, so this this the product of the expected values. The moment becomes the convolution of all the moments. Note this only works because the $X_n$'s are independent.

$M(S_n)=E(e^{tX_1})E(e^{tX_2})...=M(X_1)M(X_2)...$

So you just take the product and you're done.

So to solve this problem you need to calculate

$$m_X(t)=E[e^{tx}]=\int_0^\infty e^{tx}f_X(x)dx=\,...\,$$