Moment Generating function

Question

The random variable $X$ has the density function $f(x)=e^{-x}I_{(0,\infty)}(x)$. Then the MGF will be $$M_X(t)=\int_{-\infty}^{\infty}e^{tx}e^{-x}I_{(0,\infty)}(x) =(l-t)^{-1} \textsf{ provided that t<1 }$$ My question is , why $ t $ should be $<1$ what if the condition that $t\neq1$, in other words , why should the expected value of $e^{tX}$ exist for every value of $t$ in some open interval containing $0$ , why MGF could not be a discontinuous function ?

Answer 1

Indeed the MGF is a particular case of a Theorem relationed with Fourier Transform. The idea behin is the concept of convergence in distribution for a sequence of random variables. That definition generalizes the MGF, and it works by mean of defining the set of continuity of a distribution function.

I hope it helped you.