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, . . ..

  • 2
    $\begingroup$ 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 $\endgroup$ – user25658 Sep 4 '13 at 5:19
  • $\begingroup$ 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. $\endgroup$ – user25658 Sep 4 '13 at 5:27
  • $\begingroup$ 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 $\endgroup$ – Kei Cheung Sep 4 '13 at 6:52
  • $\begingroup$ Can you edit question (b)? Is $\Theta$ is the same as $\theta$ ? $\endgroup$ – 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.


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.


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


So to solve this problem you need to calculate

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.