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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!

Rick Sanchez
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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:

$$f_Z(z)=-\frac{2z^2(z-3)}{3}\cdot\mathbb{1}_{(0;1)}(z)+\frac{2z^3-6z^2+8}{3}\cdot\mathbb{1}_{[1;2)}(z)$$

enter image description here

tommik
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  • Wow! Thank you every much for your justification. – Rick Sanchez Jul 26 '21 at 08:33
  • @Henry: thanks for your comment. I did a typo when copying the density. Now I have amended my error and showed a drawing of $f(z)$ – tommik Jul 26 '21 at 09:01
  • +1. $f_Z(z)=\dfrac{2z^2(3-z)}{3}\cdot\mathbb{1}{(0;1)}(z)+\dfrac{2(2-z)^2(1+z)}{3}\cdot\mathbb{1}{[1;2)}(z)$ is the same, illustrating the symmetrical nature of the density slightly more – Henry Jul 26 '21 at 09:30