Joint Distribution of Exponentials

Question

Let $X_1,X_2$ be iid from exp$(\lambda)$. I'm trying to find the joint density of $X_1,X_2$ and $X_1+X_2$. The density of the sum is $$f_{X_1+X_2}(x)=\lambda^2xe^{-\lambda x}.$$ Any ideas on how to get the full joint density from it?

Answer 1

If $x,y>0$ and $x+y <z$ then $$P(X_1 \leq x,X_2 \leq y, X_1+X_2 \leq z)$$ $$=\int_0^{x}\int_0^{ y \wedge (z-x_1)} \lambda^{2}e^{-\lambda (x_1+x_2)} dx_2 dx_1.$$ You can split this as follows: $$\int_0^{z-y}\int_0^{y} \lambda^{2}e^{-\lambda (x_1+x_2)} dx_2 dx_1+\int_{z-y}^{x}\int_0^{ z-x_1} \lambda^{2}e^{-\lambda (x_1+x_2)} dx_2 dx_1.$$ Now differentiate partially w.r.t $x$, $y$ and $z$ to find the joint density of $X_,X_2$ and $X_1+X_2$.

Answer 2

Obviously, $(X_1,X_2,X_1+X_2)$ does not have a true probability density function as the 3D vector is distributed over a 2D surface.

The joint probability measure function will be

$$\begin{align}f_{\small X_1,X_2,X_1+X_2}(x,y,z) &= {f_{\small X_1}(x)\,f_{\small X_2}(y)\,{\mathsf P(X_1{+}X_2{=}z\mid X_1{=}x,X_2{=}y)}}\end{align}$$