Find the density of random variable $X+Y$ for $f(x,y) = 6(x-y)$ if $0 \leq y \leq x \leq 1$

Question

Find the density of random variable $X+Y$ for $f(x,y) = 6(x-y)$ if $0 \leq y \leq x \leq 1$.

I am able to use method of transformation to convert $f(x,y) = 6(x-y)$ if $0 \leq y \leq x \leq 1$, and the result is $g(v,w) = 6(v-2w)$, where $v = x + y$ and $w = y$. However, I am really stuck at the integration to find the marginal distribution of $g(v)$, particularly the limit of integration.

I think I need to divide the interval for $v$ into $[0,1]$ and $[1, 2]$ but I am not sure how to do it.

Answer 1

I am able to use method of transformation to convert $f(x,y)=6(x−y)$ if $0≤y≤x≤1$, and the result is $g(v,w)=6(v−2w)$, where $v=x+y$ and $w=y$.

More correctly, $g(v,w) = 6\,(v-2w)$ if $0\leq w\leq v\leq 1+w\leq 2$, which is $0\leq v\leq 2$ and $\max\{0,v-1\}\leq w\leq\min\{1,v\}$ so $$\begin{align} g(v) &= \int_{\max\{0,v-1\}}^{\min\{1,v\}} 6\,(v-2w)\,\mathbf 1_{0\leq v\leq 2}\,\mathrm d w \\[1ex] &= 6\;\mathbf 1_{0\leq v\leq 1}\int_0^v(v-2w)\,\mathrm dw+6\;\mathbf 1_{1\lt v\leq 2}\int_{v-1}^1(v-2w)\,\mathrm d w\\[1ex] &~~\vdots \end{align}$$