Show that if $X_1$ and $X_2$ are distributed $\exp(\theta)$ then $X_1-X_2$ is distributed Laplace$(0,\theta)$ I want to show that if $X_1,X_2$ are independent and distributed Exp($\theta$), then the difference $X_1-X_2$ is distributed Laplace($0$,$\theta$).  I want to use the method of moment generating functions (I tried the method of cumulative distribution functions and it just wasn't working out.), so I would like to get the moment generating function of the Laplace distribution.  If $Y\sim\text{Laplace}(0,\theta)$ then
$M_Y(t)=E[e^{tY}]$
$=\int_{-\infty}^{\infty}\frac{1}{2\theta}e^{ty}e^{-|\frac{y}{\theta}|}$
$=\int_{-\infty}^{\infty}\frac{1}{2\theta}e^{\theta t{\sqrt{\frac{y^2}{\theta^2}}}-\sqrt{\frac{y^2}{\theta^2}}}$
$=\int_{-\infty}^{\infty}\frac{1}{2\theta}e^{(\theta t-1)\sqrt{\frac{y^2}{\theta^2}}}$
and I'm stuck.  Any help would be appreciated.

$$M_{{Y}} \left( t \right) =\int _{-\infty }^{\infty }\!\frac{1}{2\theta}\,{{\rm e}^{t
y}}{{\rm e}^{-{\frac { \left| y \right| }{\theta}}}}{dy}
$$
$$M_{{Y}} \left( t \right) =\int _{-\infty }^{0}\!\frac{1}{2\theta}\,{{\rm e}^{ty}}{
{\rm e}^{{\frac {y}{\theta}}}}{dy}+\int _{0}^{\infty }\!\frac{1}{2\theta}\,{{\rm e}^{ty}}{{\rm e}^{-{\frac {y}{\theta}}}}{dy}
 $$
$$M_{{Y}} \left( t \right) =\frac{1}{2}\, \left( t\theta+1 \right) ^{-1}-\frac{1}{2}\,
 \left( -1+t\theta \right) ^{-1}
$$
$$M_{{Y}} \left( t \right) =\left(1- {t}^{2}{\theta}^{2} \right) ^{-1}$$
Do you agree?
