Find the pdf of T = X + Y Let (X,Y) be a random point chosen uniformly on region R = {(x,y) : |x| + |y| <= 1}. I need to find the pdf of T = X + Y.
I know the joint density is just equal to 1/(area) = fxy(x,y) = 1/2 for |x| + |y| <= 1 and 0 otherwise
i also calculated the marginal densities where fx(x,y)=1-|x| and similarly for fy(x,y)=1-|y| for both x and y between -1 and 1
can someone please help me now find pdf of T = X + Y. i dont know how to go about this.
alright so now i understand i should use the convolution integral: so i should use..
$$f_T(t)=\int_{S_X}f_X(x)f_Y(t-x)\,dx=\int_{|x|-1}^{1-|x|}(1-|x|)f_Y(t-x)dx$$
can someone help me from here? i cant do the $$f_Y(t-x)$$ part.. and now that i think about this i dont even know if my integral bounds are correct.
 A: To determine the cdf $F_T(t)=P(T\le t)$, I would proceed geometrically.  Points $(X,Y)$ come from a square with corners at $(1,0), (0,1), (-1,0), (0,-1)$.  To find $F_T(t)$, draw the line $X+Y=t$ and note it is parallel to two of the sides of the square.  $F_T(t)$ is the area inside the square to the left of the line.  Since the line is parallel to the side of the square, the probability should accumulate at a constant rate while the line passes through the square.
So I think:
$$F_T(t)=\left\{ \begin{array}{ccc} 0&\mbox{if}&t<-1\\\frac12 t+\frac12&\mbox{if}&-1\le t\le 1 \\ 1&\mbox{if}&t>1 \end{array} \right.$$
A: So the convolution integral is:
$$\int_{-1}^{1}(1-|x|)(1-|T-x|)dx$$
Now you can simply calculate this integral:
$$\int_{-1}^{1}(1-|x|)(1-|T-x|)dx=\int_{-1}^{1}1dx+\int_{-1}^{1}|x|dx+\int_{-1}^{1}|T-x|dx+\int_{-1}^{1}|Tx-x^2|1dx$$
First two are easy: $\int_{-1}^{1}1dx=2$, $\int_{-1}^{1}|x|dx=1$.
Third (you can eliminate $|\cdot|$ dividing integral into two integral):
$$\int_{-1}^{1}|T-x|dx=\int_{-1}^{T}T-xdx+\int_{T}^{1}x-Tdx$$
The same with fourth integral($x(T-x)>0$ when $x \in [0,T]$ for $T\geq 0$)
$$\int_{-1}^{1}|x(T-x)|dx=\int_{-1}^{0}-x(T-x)dx+\int_{0}^{T}x(T-x)dx+\int_{T}^{1}-x(T-x)dx$$
You can do similar thing when $T<0$.
