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.
 A: $$M_{{X_{{1}}}} \left( t \right) = \left( 1-t\theta \right) ^{-1}$$
$$M_{{X_{{2}}}} \left( t \right) = \left( 1-t\theta \right) ^{-1}$$
$$Y=X_{{1}}-X_{{2}}$$
$$M_{{Y}} \left( t \right) =M_{{X_{{1}}}} \left( t \right) M_{{X_{{2}}}}
 \left( -t \right)$$
$$M_{{Y}} \left( t \right) = \left( 1-{t}^{2}{\theta}^{2} \right) ^{-1}$$
Now, the moment generating function for Laplace $L \left( \mu,b \right) $ is
$$M_{{Z}} \left( t \right) ={\frac {{{\rm e}^{t\mu}}}{1-{b}^{2}{t}^{2}}}$$
Comparing we obtain $\mu=0$ and $b=\theta$.  Do you agree?
A: $$M_{{Y}} \left( t \right) =\int _{-\infty }^{\infty }\!\frac{1}{2\theta}\,{{\rm e}^{t
y}}{{\rm e}^{-{\frac { \left| y \right| }{\theta}}}}{dy}
$$
$$M_{{Y}} \left( t \right) =\int _{-\infty }^{0}\!\frac{1}{2\theta}\,{{\rm e}^{ty}}{
{\rm e}^{{\frac {y}{\theta}}}}{dy}+\int _{0}^{\infty }\!\frac{1}{2\theta}\,{{\rm e}^{ty}}{{\rm e}^{-{\frac {y}{\theta}}}}{dy}
 $$
$$M_{{Y}} \left( t \right) =\frac{1}{2}\, \left( t\theta+1 \right) ^{-1}-\frac{1}{2}\,
 \left( -1+t\theta \right) ^{-1}
$$
$$M_{{Y}} \left( t \right) =\left(1- {t}^{2}{\theta}^{2} \right) ^{-1}$$
Do you agree?
