Let $U=X+Y$, $V=X-Y$, while $X,Y\sim U[0,1]$ and independent. Prove or disprove..

Let $$U=X+Y$$, $$V=X-Y$$, while $$X,Y\sim U[0,1]$$ and independent. Prove or disprove:

$$(U,V)$$ has a uniform distribution on some area in the plane.

$$U$$ and $$V+1$$ are distributed the same (sorry if the translation is bad, would be happy to know how it's usually written).

$$U,V$$ are independent.

$$U,V$$ are (uncoordinated - not sure of the translation), but what it means is $$Cov(U,V)=0$$

My work:

• For first statement:
Intuitively this is true, but I wanted to find the CDF:
$$F_{U,V}(u,v)=P(X+Y \le u, X-Y \le v)=P(Y \le u-X)P(X \le v+Y)=(u-X)(v+Y)$$ whenever $$u,v\le 1$$.
I'm confused if what I did is correct and would love to hear feedback.

• For second statement:
$$P(V+1 \le v)=P(V \le v-1)=0$$
$$P(U \le u) = P(X+Y \le u)=P(X \le u-Y)=$$.. I'm a little stuck here, what does it mean that $$X$$ is less than $$u-Y$$ since $$Y$$ could be anything, this is giving me some problems.

• For third statement:
I need to either prove that $$F_UF_V=F_{U,V}$$ or disprove it.
My intuition says that they're dependent, since they both depend on $$X,Y$$.
$$F_U(u)F_V(v)=P(X+Y \le u , X-Y \le v)$$, again I'm struggling with calculating these, How do I reach $$X,Y$$ or stuff that I know how to deal with, without complicating myself?

• The last one was not hard, all I've done is $$Cov(X+Y,X-Y)=Var(X)-Var(Y)=0$$.

Any help and feedback is really appreciated, thanks in advance.

• 2. True. $1+V = X+(1-Y). 1-Y$ has the same distribution has $Y$. 3. False. Suppose $U = 2$. Then $X=1, Y=1 \implies V = 0$, thus $V$ is known, determined by $U$, thus not independent. – Mark Jul 21 at 17:32
• @Mark Could you please emphasize on that a little more, I can't see why that's true, or maybe could you tell me the subject that I should look into to find that out? – Pwaol Jul 21 at 17:35
• math.stackexchange.com/questions/341358/… – Mark Jul 21 at 17:36
• also U has symmetric triangular distribution (symmetric version of en.wikipedia.org/wiki/Triangular_distribution) (en.wikipedia.org/wiki/… the 2nd to last related distribution) – Mark Jul 21 at 17:38
• Perhaps you might say "$U$ and $V+1$ have the same distribution" and "$U$ and $V$ are uncorrelated" – Henry Jul 21 at 17:46

To prove or disprove the first statement, I will directly find the joint density of $$(U, V)$$ using Jacobian transformation.

If $$\ U = X + Y, V = X - Y, \ \ X = \cfrac{U+V}{2}, Y = \cfrac{U-V}{2}$$

$$|J| = \cfrac{1}{2}$$

As $$f_{XY} (x, y) = 1$$,

$$f_{UV}(u, v) = f_{XY} \left(\cfrac{u+v}{2}, \cfrac{u-v}{2}\right) |J| = \cfrac{1}{2}$$

Now as $$\ 0 \leq x \leq 1, 0 \leq y \leq 1$$, $$(U,V)$$ is uniformly distributed over region $$R$$ defined below,

i) $$-u \leq v \leq u \$$ for $$\ 0 \leq u \leq 1 \$$ and,
ii) $$u - 2 \leq v \leq 2 - u \$$ for $$\ 1 \leq u \leq 2$$

The above also shows that $$U, V$$ are not independent.

• Thanks! could you please explain what makes the last sentence true? I guess it has something to do with being able to split the PDF to two different function of $u,v$, but I'm not quite seeing how did you know it – Pwaol Jul 22 at 6:51
• @Pwaol the last statement meaning that $U, V$ are not independent? That is because $- u \leq v \leq u$... if they were independent, the support of joint density function will be a rectangle with sides parallel to $u, v$ axes. – Math Lover Jul 22 at 6:58
• Could you please tell me the subject I need to revise to see that? Feels like a weak spot to me and I'm a little lost. – Pwaol Jul 22 at 7:06
• See this question math.stackexchange.com/questions/2362434/… – Math Lover Jul 22 at 7:34
• @Pwaol $U$ and $V$ will be distributed over the square with vertices $(0,0),(1,1),(2,0),(1,-1)$ in the u-v plane with $f_{u,v}(u,v)=\frac{1}{2}$ . Now integrate over $v$ to get the distribution function for $U$ . Which is :- for $u\in(0,1]\,\,\int_{-u}^{u}\frac{1}{2}d\,v$ and for $u\in (1,2]\,\,\int_{u-2}^{2-u}\frac{1}{2}d\,v$. Which when summed up gives :- $f_{u}(u) = u\,\,,0<u\leq 1$ and $f_{u}(u)=2-u\,\,,1<u\leq 2$ . Similarly you can find the distribution function of v by integrating over $u$ . Multiplication of them won't yield $f_{u,v}(u,v)=\frac{1}{2}$. Hence they are not independent. – Arghyadeep Chatterjee Jul 22 at 7:41
1. Intuitively this is true, but I wanted to find the CDF:

I'd suggest working with the pdf rather than the CDF.

Use the Jacobian change-of-variables transformation: $$f_{U,V}(u,v)=\begin{Vmatrix}\tfrac{\partial (u+v)/2}{\partial u}&\tfrac{\partial (u+v)/2}{\partial v}\\\tfrac{\partial (u-v)/2}{\partial u}&\tfrac{\partial (u-v)/2}{\partial v}\end{Vmatrix} f_{X,Y}(\tfrac {u+v}2,\tfrac {u-v}2)$$

This is much easier to evaluate and work with.

$$\phantom{f_{U,V}(u,v)=\tfrac 12(\mathbf 1_{0\leqslant u\lt 1}\mathbf 1_{-u\leqslant v\leqslant u}+\mathbf 1_{1\leqslant u\leqslant 2}\mathbf 1_{u-2\leqslant v\leqslant 2-u})}$$