Suppose $X$ and $Y$ have the joint density function $f(x,y) = x + y$ for $0 < x < 1$, $ 0 < y < 1$. If $Z = X + Y$, then $Z$ has the density function (______)?
I try it as follow but that's not the answer the book gives:
$$Pr(Z<z)=H(z)=\iint_{\Omega}{x+y}\,dy\,dx\\ \Omega:{x\in(0,1),\,y\in(0,1),\,x+y\le z}$$ $$H(z) = \int_0^z[xy+1/2y^2]_0^{z-x}\,dx \\ = \int_0^z[x(z-x)+1/2(z-x)^2]\,dx \\ = 1/2\int_0^z{z^2-x^2}\,dx =1/2[z^2x-1/3x^3]_{x=0}^z \\ =1/2[z^3-1/3z^3] \\ =1/3z^3$$
so $$h(z)=H'(z)=z^2$$ but why is there another branch?
ps: The book's answer is:
Just write down the probability as an integral.
$P(Z \leq z) = P(X+Y \leq z) = \iint_{x+y \leq z} f_{X,Y}(x,y) dx dy =\iint_{x+y \leq z \text{ and } 0 < x < 1 \text{ and } 0 < y <1 } (x+y) dx dy$.
Now, simply parameterize the region $\{x+y \leq z \text{ and } 0 < x < 1 \text{ and } 0 < y <1 \}$ and do the integral.