Mgf of double exponential RV

Question

In class the other day we were talking about a double exponential RV $X$ with a pdf $f(x)=\frac{1}{2}e^{-|x|}$ for $-\infty<x<\infty$. The professor noted that the mgf was $M(t)=\frac{1}{1-t^2}$ for $|t|<1$. However, I'm trying to verify this now and I keep getting $M(t)=\frac{1}{2\pm 2t}$. Am I incorrect? I'm using the definition of mgf, where $M(t)=E[e^{tx}]$.

Answer 1

$$\begin{align*} M_X(t) &= \operatorname{E}[e^{tX}] = \int_{x=-\infty}^\infty e^{tX} \cdot \frac{1}{2} e^{-|x|} \, dx \\ &= \frac{1}{2} \int_{x=-\infty}^0 e^{tx} e^{-(-x)} \, dx + \frac{1}{2} \int_{x=0}^\infty e^{tx} e^{-x} \, dx \\ &= \frac{1}{2} \int_{x=-\infty}^0 e^{x(1+t)} \, dx + \frac{1}{2} \int_{x=0}^\infty e^{x(t-1)} \, dx \\ &= \frac{1}{2} \left[ \frac{e^{x(1+t)}}{1+t} \right]_{x=-\infty}^0 + \frac{1}{2} \left[ \frac{e^{x(t-1)}}{t-1} \right]_{x=0}^\infty \\ &= \frac{1}{2}\left( \frac{1}{1+t} - 0 + 0 - \frac{1}{t-1} \right), \quad |t| < 1, \\ &= \frac{1}{1-t^2}. \end{align*}$$ A trivial way to know that your calculation cannot be correct is to observe that for any random variable for which an MGF exists, we must have $$M_X(0) = \operatorname{E}[e^0] = 1,$$ but your function does not satisfy this condition.