How to find pdf of X+Y given X and Y are dependent.

Question

The joint pdf is f(x,y) = $$\frac{2}{5}(2x+3y)$$ for $0\leq x \leq 1,0\leq y \leq 1$ Normally if the random variables are independent, you can apply the convolution definition Z = X + Y which looks like $g(y) = \int_{\infty}^{\infty}f_1(y-z)f_2(z)dz$. So I believe you can remove the dependence of one random variable on another. But I don't really know how to proceed at that point. For reference the answer is $g(y) = {z}^2$ for $0\leq z \leq 1$ and $z(2-z)$ $1\leq z \leq 2$

Answer 1

For dependent $X$ and $Y$ the convolution formula for the density of $Z:=X+Y$ is $$ g(z):=\int_{-\infty}^\infty f(z-x,x)\,dx.\tag1 $$ In this case the integrand in (1) evaluates to $f(z-x,x)=\frac25\left(2(z-x)+3(x)\right)$, which simplifies to $\frac25(2z+x)$, provided $$ 0\le z-x\le1\qquad\rm{and}\qquad 0\le x\le 1\tag2 $$ and equals zero otherwise. So to evaluate (1), which is an integral over $x$, you need to determine, for each fixed value of $z$, the range of $x$ values where the conditions in (2) are satisfied. Draw a picture to see the limits of integration:

From the picture it is clear that you need to argue by cases:

1. If $0\le z\le1$, the limits run from $x=0$ to $x=z$ and therefore $g(z)=\int_{x=0}^z\frac25(2z+x)\,dx$.

2. If $1\le z\le 2$, the limits run from $x=z-1$ to $x=1$ and therefore $g(z)=\int_{x=z-1}^1\frac25(2z+x)\,dx$.

3. For $z$ outside these two ranges the integrand is zero so $g(z)=0$.