We are given the following task:
$X$ and $Y$ are independent random variables and have the same distribution. $U$ and $V$ are independent and $U, V \sim \Gamma(2,1)$. $X + Y$ has the same distribution as $U - V$. Find the distribution of $X$.
I have calculated that the density function of $U - V$ and $X + Y$ is equal to $f(x) = \frac{1 + |x|}{4}e^{-|x|}$. I know that distribution of $X$ is uniquely defined by the distribution of $X + Y$ because $X$ and $Y$ are independent and have the same distribution. So it would suffice to just guess the correct $X$. But I couldn't manage to guess right. Then I used Fourier transform to factorize the convolution in formula for the density function of $X + Y$ and I calculated that the density function of X is $g(x) = \frac{\pi^{1/4}}{2^{3/4}}e^{-|x|}$. But this is almost certainly an overcomplication.
What would be the easiest way to calculate the distribution of $X$ from distribution of $X + Y$?
$U, V$are independent $\Gamma(2,1)$, so we have for the m.g.f $M_U(t)=\frac1{(1-t)^2}$ and $M_{-V}(t)=M_{V}(-t)=\frac1{(1+t)^2}$; then $$ M_{U-V}(t)=M_{U}(t)M_{-V}(t)=\tfrac{1}{(1-t^2)^2}=M_{X+Y}(t)=M_{X}(t)M_{Y}(t)=\left(M_{X}(t)\right)^2,\;|t|<1 $$ because $X, Y$ are i.i.d.
So we have $M_X(t)=\frac{1}{1-t^2}$, and then $f_X(x)=\frac12 \operatorname e^{-|x|}$, i.e. $X\sim \textrm {Laplace}(0,1) $.