I want to find the density of the random vector $(X,X+Y)$ where $X\sim\text{Exp}_\alpha$ and $Y\sim\text{Exp}_\beta$ are exponentially distributed to different rate parameters $\alpha\ne \beta$.
I don't think I can use the product-density, as $X, X+Y$ are certainly not independent, but this is the only way I've ever seen a joint density done.
How can I proceed to calculate this density? Is there a rule I can use?
EDIT: More precisely, how can I get from $P(X\le a, X+Y\le b)$ to a place where I can actually use the known probability densities of $X,Y$? I can't split the probability because of dependence issues.
Best Answer
You have to assume independence between $X$ and $Y$ then you can use the fundamental transformation theorem (Jacobian). For our knowledge, you have also to tell us if $\alpha$ and $\beta$ are "scale" or "rate" parameter because this modifies the density expression
Let's set
$$\begin{cases} v=x+y\\ u=x \end{cases}\rightarrow \begin{cases} x=u\\ y=v-u \end{cases}$$
The Jacobian is evidently $|J|=1$ thus
$$f_{UV}(u,v)=\alpha e^{-\alpha u}\beta e^{-\beta(v-u)}=\alpha \beta e^{-(\alpha-\beta)u}e^{-\beta v}$$
The only difficulty now is to understand which is the $(U,V)$ support. It is quite easy to understand that it must be $0<u<v<\infty$ thus the requested joint density is
$$f_{UV}(u,v)=\alpha \beta e^{-(\alpha-\beta)u}e^{-\beta v}\cdot\mathbb{1}_{[0;\infty)}(u)\cdot\mathbb{1}_{[u;\infty)}(v)$$
Useful observation
Now let's set $\alpha=\beta$ thus we have
$$f_{UV}(u,v)=\alpha^2 e^{-\alpha v}\cdot\mathbb{1}_{[0;\infty)}(u)\cdot\mathbb{1}_{[u;\infty)}(v)$$
Which integrated in $du$ gives us
$$f_V(v)=\int_{0}^v \alpha^2 e^{-\alpha v}du=\alpha^2 v e^{-\alpha v}\sim Gamma(2;\alpha)$$
as already known by Gamma distribution properties.