What is the moment generating function of the Laplace distribution?

Solution 1:

Use a substitution to break the integral at zero as usual:

\begin{align} E\left[e^{tX}\right]&=\frac{\lambda}{2}\int_{\mathbb R}e^{tx}e^{-\lambda|x-\mu|}\,dx \\\\&=\frac{1}{2}\int_{\mathbb R}\exp\left[t\left(\mu+\frac{y}{\lambda}\right)-|y|\right]\,dy\qquad\qquad,\,\small\text{ substituting }\lambda(x-\mu)=y \\\\&=\frac{e^{\mu t}}{2}\int_{\mathbb R}e^{ty/\lambda-|y|}\,dy \\\\&=\frac{e^{\mu t}}{2}\left[\int_{-\infty}^0 e^{y(1+t/\lambda)}\,dy+\int_0^\infty e^{-y(1-t/\lambda)}\,dy\right] \\\\&=\frac{e^{\mu t}}{2}\left[\underbrace{\int_0^\infty e^{-z(1+t/\lambda)}\,dz}_{\text{ converges for }1+\frac{t}{\lambda}>0}+\underbrace{\int_0^\infty e^{-y(1-t/\lambda)}\,dy}_{\text{ converges for }1-\frac{t}{\lambda}>0}\right] \\\\&=\frac{e^{\mu t}}{2}\left[\frac{1}{1+\frac{t}{\lambda}}+\frac{1}{1-\frac{t}{\lambda}}\right]\qquad\qquad\qquad,\,|t|<\lambda \end{align}

So finally, $$\boxed{E\left[e^{tX}\right]=e^{\mu t}\left(1-\frac{t^2}{\lambda^2}\right)^{-1}}\qquad,\,|t|<\lambda$$