Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a question concerning the aforementioned topic :)

So, with $f_X(t)={\lambda} e^{-\lambda t}$, we get: $$\phi_X(t)=\int_{0}^{\infty}e^{tX}\lambda e^{-\lambda X}dX =\lambda \int_{0}^{\infty}e^{(t-\lambda)X}dX =\lambda \frac{1}{t-\lambda}\left[ e^{(t-\lambda)X}\right]_0^\infty =\frac{\lambda}{t-\lambda}[0-(1)]$$ but only, if $(t-\lambda)<0$, else the integral does not converge. But how do we know that $(t-\lambda)<0$? Or do we not know it at all, and can only give the best approximation? Or (of course) am I doing something wrong? :)

Yours, Marie!

share|cite|improve this question
The MGF is defined only for $t<\lambda$, here. The integral, of course, diverges otherwise. – David Mitra Apr 23 '12 at 22:16
up vote 4 down vote accepted

You did nothing wrong. The moment generating function of $X$ simply isn't defined, as your work shows, for $t\ge\lambda$.

share|cite|improve this answer
Notice that this is more than what we need for the usual use of the MGF : obtaining the moments of the random variable. Eventually, provided that all the $n-th$ moments are defined, one has $\frac{\partial^n}{\partial t^n} \phi_X(t)|_{t=0} = E(X^n)$. To be able to use this, you really just need $\phi_X(t)$ to be defined in a neighborhood of $t=0$. – Martin Van der Linden Nov 14 '13 at 18:32

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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