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Variables $X$ and $Y$ are indepedent and have the same exponential distribution, $\lambda=1$. Let $U=X+Y, V=X-Y$. Find density function of vector $(U,V)$. Are $U$ and $V$ independent?

My attempt:

Because $X$ and $Y$ are independt, thus

$f_{X,Y}(x,y)=f_X(x)\cdot f_Y(y)=\begin{cases} 0 &\text{ for } x,y \in (-\infty,0) \\ e^{-x-y} &\text{ for } x,y \in [0,\infty) \end{cases}$

Jacobian determinant:

$J=\left|\begin{smallmatrix} 1&1\\ 1&-1 \end{smallmatrix}\right|=-2$

$\begin{cases} U=X+Y \\ V=X-Y \end{cases} \rightarrow \begin{cases} X=\frac{U+V}{2} \\ Y=\frac{U-V}{2} \end{cases}$

Hence

$f_{U,V}(u,v)=\frac{\exp\left(-\frac{u+v}{2}\right)\cdot\exp\left(-\frac{u-v}{2}\right)}{|-2|}=\frac{\exp\left(-u\right)}{2}$

And now is the part where I have a problem.

$u\in [0,\infty)$

But what about the $v$?

  • 1
    Note that both $U+V = 2X$ and $U-V = 2Y$ are non-negative. This forces $V \in [-U,U]$. The complete density is thus $f_{UV}(u,v) = \frac{\exp(-u)}{2} \mathbb{1}_{\{v \in [-u,u]\}}$ – 2017-01-28
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    Oh, this one was quite simple after all. Thanks! – 2017-01-28
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    Uncorrelated: Easy to show $Cov(U,V) = 0.$ _Not_ independent: Easy to see $P(U < 5) > 0,\,P(V > 5) > 0,$ but $P(U<5, V>5)=0.$ – 2017-01-29

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