# Probability density for continous function

I am stuck on a question with two parts :

For $$f(x,y) = \begin{cases} a^2 e^{-ay} & 0 \le x \le y, \\ 0 & \text{otherwise} \end{cases}$$

a) Compute the distribution function and density of $Z = X+Y$

b) Find the joint distribution function and the densty of $(Y, X+Y)$

I have been able to calculate part a) by integrating the function by taking limits for $x$ from $0$ to $z/2$ and for $y$ $x$ to $z-x$ and I get the correct solution.

For part b) I am applying the theorem that $$f(y_1,y_2) = f(x_1(y_1,y_2),x_2(y_1,y_2))|J(y_1,y_2)|$$ where $J$ is the Jacobian.

But I am unsure about the limits for the integral. How do I proceed?

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(Y, Z) = (Y, X+Y) is a linear transformation of (X,Y), so checking where (X,Y) is allowed to find the region in question. But confused – user669083 Dec 3 '12 at 18:21
In part b) there is no integration to be performed at all, so I don't know what it means when you say "I am unsure about the limits for the integral". But you do need to determine the region in which your formula applies. – Dilip Sarwate Dec 3 '12 at 18:39
For calculating the density I will have to integrate over the area, but how to calculate the area. Please help – user669083 Dec 3 '12 at 19:23

Here, $f_{X,Y}(x,y)=g(y)\cdot[0\leqslant x\leqslant y]$ for some function $g$ whose exact expression will be irrelevant, and $(z,t)=(y,x+y)$ hence the Jacobian is $1$ and $(x,y)=(t-z,z)$.
This yields $f_{X,Y}(x,y)=g(z)\cdot[0\leqslant t-z\leqslant z]$ hence $$f_{Y,X+Y}(z,t)=g(z)\cdot[z\leqslant t\leqslant 2z].$$ Note that if $z\leqslant t\leqslant 2z$ then $z\leqslant 2z$ hence $z\geqslant0$ and $f_{Y,X+Y}$ is also $$f_{Y,X+Y}(z,t)=g(z)\cdot[z\geqslant0,t\geqslant0,z\leqslant t\leqslant 2z].$$