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.

  • 1
    $\begingroup$ 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. $\endgroup$ – Mark Jul 21 at 17:32
  • $\begingroup$ @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? $\endgroup$ – Pwaol Jul 21 at 17:35
  • 1
    $\begingroup$ math.stackexchange.com/questions/341358/… $\endgroup$ – Mark Jul 21 at 17:36
  • 1
    $\begingroup$ 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) $\endgroup$ – Mark Jul 21 at 17:38
  • $\begingroup$ Perhaps you might say "$U$ and $V+1$ have the same distribution" and "$U$ and $V$ are uncorrelated" $\endgroup$ – 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.

  • $\begingroup$ 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 $\endgroup$ – Pwaol Jul 22 at 6:51
  • 1
    $\begingroup$ @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. $\endgroup$ – Math Lover Jul 22 at 6:58
  • $\begingroup$ 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. $\endgroup$ – Pwaol Jul 22 at 7:06
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    $\begingroup$ See this question math.stackexchange.com/questions/2362434/… $\endgroup$ – Math Lover Jul 22 at 7:34
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    $\begingroup$ @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. $\endgroup$ – 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.

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


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