Are $U + V$ and $UV$ independent when $U,V$ are independent and standard uniform?

Question

This is related a previous question I posted on the product of two independent variables here. As an alternative method, one could note that if $X,Y \sim U(-1,1)$ and $U,V \sim U(-1,1)$ then $Z = XY = 4UV - 2(U+V) + 1 = S - T + 1$. It is easy to show that $$ f_S(s) = -\frac{1}{4} \ln \frac{s}{4}, \quad f_T(t) = \frac{1}{4} \begin{cases} t & \text{if } 0 < t < 2 \\ 4-t & \text{if } 2 < t < 4 \end{cases}. $$ Then, it remains to consider the difference of these variables and then the shift (+1). At this point, I realize that (while maybe not necessary), it is unclear to me whether $S$ and $T$ are independent. Intuitively, I would think potentially not since $T > 2$ implies that $S > 0$ and so there should be some dependence.

So this question really has two parts: 1) Is there some way to show dependence? 2) Regardless, is there a way to proceed here to arrive at the distribution of $Z$?

Answer 1

Here's a very simple solution. It involves no integration and only the easiest algebra.

Let $X=2U-1$ and $Y=2V-1.$ These are iid uniform random variables on $[-1,1]$ and therefore are symmetric about $0:$ that is, $-X$ and $-Y$ have the same distribution, too. Thus

$$\begin{aligned} \operatorname{Cov}(XY, X+Y) &= \operatorname{Cov}((-X)(-Y),\ (-X)+(-Y)) \\ &= \operatorname{Cov}(XY,\ -(X+Y)) \\ &= -\operatorname{Cov}(XY,\ X+Y). \end{aligned}$$

Since $X$ and $Y$ are bounded, their covariance exists and is finite. The only number equal to its own negative is $0:$ that must be the covariance.

Now exploit the basic rules of covariance (bilinearity) to relate this result to what you want:

$$\begin{aligned} 0 &= \operatorname{Cov}(XY,\ X+Y) = \operatorname{Cov}((2U-1)(2V-1),\ (2U-1)+(2V-1)) \\ &= \operatorname{Cov}(4UV-2(U+V),\ 2U+2V)\\ &= 8\operatorname{Cov}(UV,\ U+V) - 4\operatorname{Cov}(U+V,\ U+V) \end{aligned}$$

Because $U+V$ is not constant, it has positive covariance, whence the left hand side of the last line cannot be zero. This means $UV$ and $U+V$ have positive covariance and therefore cannot be independent, QED.

BTW, if you wish to do the integrals you can compute that the common variance of $U$ and $V$ must be $1/12$ and conclude (from the independence of $(U,V)$) that $$\operatorname{Cov}(UV, U+V)=\frac{4}{8}\operatorname{Cov}(U+V,U+V) = \frac{4}{8}\left(\frac{1}{12}+\frac{1}{12}\right)=1/12.$$

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To find the distribution of $Z=XY,$ observe that it, too, must be symmetric about $0$ and the distribution of its positive part must be that of $UV.$ That distribution function is

$$\Pr(UV \le t) = \iint^{(1,1)}_{uv \le t} \mathrm{d}u\mathrm{d}v = t(1-\log(t))$$

for $0 \lt t \le 1.$ Consequently

$$\Pr(-t \le Z \le t) = t(1-\log(t)),$$

which is a useful formula for the distribution of $Z.$ In particular, its density at both $\pm t$ must be half the derivative of this expression, giving

$$f_Z(t) = -\frac{1}{2}\log|t|,\ -1 \le t \le 1; t\ne 0.$$

(The density is not defined at $0.$)

The figure is a histogram of a million draws of $XY$, over which is plotted in red a graph of $f_Z.$ They match.