The difference of two IID random variables (exponential) So I'm self-studying probability distributions a bit, and have run into several trouble figuring out the sum and difference and product (etc) of distributions.
Let $X_1$and $X_2$ be independent and exponentially distributed with parameter = 1, so the pdf is $e^{-x}$ in both cases for $x \ge 0$. 
What is the pdf for $Z = X_1 - X_2$? I know how to do sums, but how to do differences? 
My attempt was to first calculate the pdf for $W = -X_2$ and then use the method for the sum. So we get that $$F_W(w) = P(-X_2 \le w) = 1 - P(X_2 \le -w) = 1 - (1 - e^{- (-w)}) = e^{w} $$ Which we differentiate, and that also gives the same result. Is this correct so far? What are the limits on $w$ here? And would I then calculate the pdf for $Z = X_1 + X_2$ by $$\int_{-\infty}^\infty e^{-x}e^{z-x} dx$$
What would then my limits be for this integral and for $z$? 

As usual, writing down explicitely the limits of PDFs in the formulas for PDFs would simply make the problem dissolve by itself... Here, $$f_X(x)=\mathrm e^{-x}\mathbf 1_{x\gt0},\qquad f_W(w)=\mathrm e^{w}\mathbf 1_{w\lt0},$$ hence, for every real number $z$, $$f_Z(z)=\int_\mathbb Rf_X(x)f_W(z-x)\mathrm dx=\int_\mathbb R\mathrm e^{-x}\mathbf 1_{x\gt0}\mathrm e^{z-x}\mathbf 1_{z-x\lt0}\mathrm dx=\mathrm e^z\int_\mathbb R\mathrm e^{-2x}\mathbf 1_{x\gt0,x\gt z}\mathrm dx,$$ that is, $$f_Z(z)=\mathrm e^z\int_{\max(z,0)}^\infty\mathrm e^{-2x}\mathrm dx=\mathrm e^z\cdot\tfrac12\mathrm e^{-2\max(z,0)}=\tfrac12\mathrm e^{-|z|}.$$
