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!

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    $\begingroup$ The MGF is defined only for $t<\lambda$, here. The integral, of course, diverges otherwise. $\endgroup$ – David Mitra Apr 23 '12 at 22:16

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

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    $\begingroup$ 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 $t=0$. $\endgroup$ – Martin Van der Linden Nov 14 '13 at 18:32

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