Finding the density $f_Z(z)$ for $Z$ with known simultaneous density function

Question

We have the two continuous stochastic variables $X$ and $Y$ which have the simultaneous density function $$f(x,y)=6(y-x)$$ in the area $0 < x < y < 1$. Outside this area the simultaneous density is 0. Let $$Z = X + Y$$

I want to find the density $f_Z(z)$ for $Z$.

I think I should use this formula $$f_{X+Y}(z)=\int_{-\infty}^{\infty} f(x,z-x) dx$$ but somehow not doing it correctly. I have tried putting the lower limit to zero and the upper to $z$ but this integrates to zero.

What do I miss?

Answer 1

The marginals are given by $$f_X(x)=\int_0^{1} f(x,y) dy=\frac {1-x^{2}} 2-x+x^{2}, 0<x<1$$ and $$f_Y(y)=\int_0^{1} f(x,y) dx=3y^{2}, 0<y<1.$$ Now compute $F_Z(z)$ using the formula $$f_Z(z)=\int_0^{z} f_X(z-y)f_Y(y)dy$$ for $0<z<2$

Answer 2

For a given value of $z\in\mathbb{R},$ we want to find $x\in\mathbb{R}$ such that $f(x, z-x)$ is non-zero. By definition of $f,$ this is the case if $0<x<z-x<1.$ In other words, we must have $x\in(0, z-x)$ and $z-x<1.$ The second condition means that we must have $x>z-1$. The first condition means that $x>0$ and $x<z-x$, which is equivalent to $x<z/2$. Note that these conditions already imply that we must have $z\in(0, 2).$ Hence, we use your formula with the bounds $0$ and $z/2.$

So, $f_z(z) = \int_0^{z/2} f(x, z-x)dx$ if $z\in(0, 2)$ and $0$ otherwise.