Calculating the probability density function of the sum of two independent random variables that have $\frac{1}{2}e^{-|x|}$ as a density function

Question

Suppose you have two random variables $X$ and $Y$, both independent and which follow the Laplace distribution with $x\mapsto > \frac{1}{2}e^{-|x|}$ as density function. Calculate the density function of $X+Y$

We know that if $X,Y$ are both random variables that are independent, then the density of $X+Y$ is the convolution of density functions of $X$ and $Y$.

So I get then if $f(x) = \frac{1}{2}e^{-|x|}$ as the density function of $X,Y$, then $(f\star f)(x) = \int_{\mathbb{R}} f(x-y)f(y) dy$

My approach to calculate $f \star f$ was to first handle the case when $x \geq 0$. We thus have:

$$(f \star f)(x)=\int_{\mathbb{R}}e^{-\frac{1}{2}|x-y|}e^{-\frac{1}{2}|y|}dy $$ $$= \int_{-\infty}^{x}e^{-\frac{1}{2}(x-y)}e^{-\frac{1}{2}|y|}dy + \int_{x}^{+ \infty}e^{\frac{1}{2}(y-x)}e^{-\frac{1}{2} y}dy $$ $$ = \int_{-\infty}^{0}e^{-\frac{1}{2}(x-y)}e^{\frac{1}{2}y}dy+\int_{0}^{x}e^{-\frac{1}{2}(x-y)}e^{-\frac{1}{2}y} dy + \int_{x}^{+ \infty}e^{\frac{1}{2}(y-x)}e^{-\frac{1}{2} y}dy $$

$$=\int_{-\infty}^{0}e^{-\frac{1}{2}x + y}dy + \int_{0}^xe^{-\frac{1}{2}x}dy + \int_{x}^{\infty}e^{-y -\frac{1}{2}x}dy = e^{\frac{1}{2}x} + e^{\frac{-1}{2}x} - e^{-\frac{3}{2}} $$ Is my approach so far correct?

Answer 1

Comment: The Comment by @Math1000 seems correct. The density function in the comment is a good fit to the histogram of a million sums $S = X + Y$ of standard Laplace random variables. I simulated a Laplace random variable as a difference of two standard exponential random variables. (I wonder if the distribution of $S$ is more easily obtained as the difference to two gamma random variables [sums of exponentials].)

x1 = rexp(10^6);  x2 = rexp(10^6);  x = x1-x2
y1 = rexp(10^6);  y2 = rexp(10^6);  y = y1-y2
s = x + y
hist(s, prob=T, br=40, col="skyblue2", ylim=c(0,.25))
curve(.25*(1+abs(x))*exp(-abs(x)), add=T, col="red", n=10001)