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I am having trouble with this question and feel like my supports are not entirely correct:

Given $f_u(u)= 1_{[1,0]}(u)$ and $f_v(v)= 2v*1_{[1,0]}$ with U and V being independent and $U,V \in \mathbb{R}$

with the Transformations $ X = U, Y = U-V$

What is the density of $f_{X,Y}(x,y)$ AND what is the median of Y $f_Y(y$)

I have the joint density of $f_{U,V}(u,v)= 2v*1_{{[1,0]^2}}(u,v)$

then with the transformations of $h_1(u,v)=u=x$ and $h_2(u,v)=u-v=y$ with inverses of $h_1^{-1}(x,y)=x=u$ and $h_2^{-1}(x,y)=x-y=v$ & a Jacobian of 1

with supports $\{(x,y)\in\mathbb{R}:x\in[0,1],y\in[x-1,x]\}$

which gives me a Joint PDF for X,Y of $f_{X,Y}(x,y)= 2(x-y)*1_{{[1,0]^2}}(x,y)$

To get the marginal PDF of Y, I have integrated $\int^1_0 f_{X,Y}(x,y)$ to get $f_Y(y)=1-2y$ which seems incorrect, then further, attempting to find the Distribution Function of Y, I don't know if I should be integrating $\int^x_{x-1}f_Y(y)$

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