Assume that X and Y are independent random variables where X has a pdf given by $f_X(x)=2xI_{(0,1)}(x)$ and Y has a pdf given by $f_Y(y)=2(1-y)I_{(0,1)}(y))$. Find the distribution of X+Y.
I tried to solve the distribution using the convolution formula, by letting Z=X+Y. My problem is that I don't know how to get the integral bounds of the convolution. Also, is there a way to know the exact distribution of X+Y (Normal Dist., Uniform, etc...). Thanks in advance!
The integrand is the following
$$f(x,z-x)=4x(1+x-z)$$
and observe that
$$y=z-x \in(0;1)$$
That is
$$0<z-x<1$$
This region is clearly a parallelogram thus when $z\in(0;1)$ we have
$$f_Z(z)=\int_0^z 4x(1+x-z)dx$$
while when $z\in(1;2)$ we have
$$f_Z(z)=\int_{z-1}^1 4x(1+x-z)dx$$
Concluding:
$$f_Z(z)=-\frac{2z^2(z-3)}{3}\cdot\mathbb{1}_{(0;1)}(z)+\frac{2z^3-6z^2+8}{3}\cdot\mathbb{1}_{[1;2)}(z)$$