$X \sim R(0,1)$ and $Y \sim R(0,1)$ , where $X$ and $Y$ are independent.

Question

It is given that $X \sim R(0,1)$ and $Y \sim R(0,1)$ , where $X$ and $Y$ are independent. Find the joint distribution of $X+Y$ and $X-Y$ and then find the marginals. Use only transformations.

So, we have $f_{X,Y}(x,y)=1 I_{0<x<1,0<y<1}$. Now we take $U=X+Y$, $V=X-Y$. I am having a doubt about the ranges of $U,V$. I am getting something like $$0<v<u<2-v<2$$ Is this correct? After this I can solve. Please verify

Answer 1

Is this correct?

No.   $V$ may be negative.   Indeed the marginal domain is $-1<V<1$ .  

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Given $U=X+Y$, $ 0<X<1$, and $0<Y<1$, therefore $0<U<2$

Also given $V=X−Y$, therefore $X=(U+V)/2, Y=(U-V)/2$

Now, to find the joint domain you seek $(U,V)\in S$ where $$S=\{(u,v):0<u<2~,0<u+v<2~, 0<u-v<2\}$$

(Which shall be the union of two triangles.)

$${S~{=\{(u,v):0<u<2~, \max(-u,u-2)<v<\min(2-u,u)\}\\ =\{(u,v):0<u<1~, -u<v<u\}\cup\{(u,v):1\leqslant u<2~, u-2<v<2-u\} \\ = \triangle(0,0)(1,-1)(1,1)\cup\triangle(1,1)(1,-1)(2,0)}}$$

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After this I can solve.

Good luck.