Given two independent random variables $X$ and $Y$ that both follow an exponential distribution of parameter $\lambda > 0$, I am trying to find the joint density function of $(X+Y,X)$.
I have managed to prove that the density function of $X+Y$ is the density function of a $\Gamma(2,\lambda)$ distribution but I am struggling to find the joint density function of $(X+Y,X)$. I have tried to find its distribution function in order to calculate the derivative, but I have not succeeded.
Thanks in advance for the help !
Best Answer
Here is one approach. We will assume $a \geq b \geq 0$. First condition over the value of $X$: $$P(X+Y \leq a, X \leq b) = \int_{-\infty}^{b}P(X+Y \leq a|X=x)f_X(x) dx$$ Then use the density function of $X$ and the fact $X$ and $Y$ are independent $$=\int_{0}^{b}P(x+Y \leq a)\lambda e^{-\lambda x}dx=\int_{0}^{b}P(Y \leq a-x)\lambda e^{-\lambda x}dx$$ $$=\int_{0}^{b}(1-e^{-\lambda (a-x)})\lambda e^{-\lambda x}dx=\lambda \int_{0}^{b} e^{-\lambda x} - e^{-\lambda (a-x)-\lambda x}dx$$ $$=\lambda((\frac{e^{-\lambda b}}{-\lambda} - \frac{e^{0}}{-\lambda})-e^{-\lambda a}(b-0)) = 1-e^{-\lambda b}-b\lambda e^{-\lambda a}$$ Now to find the joint density we need $$\frac{\partial^2}{\partial a \partial b} (1-e^{-\lambda b}-b\lambda e^{-\lambda a}) = \lambda^2 e^{-\lambda a}.$$ Now note if $a < b$, $\{X+Y \leq a, X \leq b\} \implies Y<0$, and so $$P(X+Y \leq a, X \leq b)=0$$ And also if $b<0$, $$P(X \leq b)=0$$ Combining this all together yields $$f_{X+Y,X}(a,b) = \begin{cases} \lambda^2 e^{-\lambda a} & a \geq b \geq 0 \\ 0 & \text{otherwise} \end{cases}$$