# How to check if functions of i.i.d random variables are dependent or independent?

i'm new to this forum and the science of statistic.This is my question:

Let's say that we have two i.i.d random variables X and Y, which both follow a Rayleigh distribution. Then, we define two new random variables U and V as follows: $U = \frac{X^{2}}{Y^{2}+a}, V = \frac{Y^{2}}{X^{2}+a},$ where $a$ is a constant. Are $U$ and $V$ are independent or dependent?

Intuitively speaking, i believe that the two are dependent, as i can write $X^{2} = U(Y^{2}+a)$ and substitute this into $V$ to get $V=\frac{Y^{2}}{U(Y^{2}+a)+a}$. But another man told me that they are independent since they are created independently from the same distribution. So i'm pretty confuse right now.

• For the special case $a = 0$, $U = \frac{X^2}{Y^2}$ and $V = \frac{Y^2}{X^2}$ are reciprocals of each other, and so are very definitely dependent random variables. More generally, your "another man" might have been thinking of the notion that $g(X)$ and $h(Y)$ are independent whenever $X$ and $Y$ are independent, but that is not the case here. It might be that $U$ and $V$ can be proven to be independent when $a\neq 0$, but I very much doubt it. Commented Apr 12, 2016 at 15:23
• If $X$ and $Y$ are iid Rayleigh,then $X^2$ and $Y^2$ are iid exponential random variables. This might help in figuring the joint density of $U$ and $V$. Commented Apr 14, 2016 at 2:39

## 1 Answer

Independence of random variables $U$ and $V$ implies the distribution of $U$ is the same regardless of what value $V$ might have.

In some cases, checking independence requires working out the joint distribution of $(U,V)$. But if you suspect there might be lack of independence, it suffices to find enough values of $V$ for which the conditional distribution of $U$ differs.

("Enough" means there has to be nonzero probability of achieving values of $V$ where the conditional distribution of $U$ varies.)

In this case, algebra tells us that

$$Y^2 = V(X^2 + a),$$

whence

$$\frac{1}{U} = \frac{Y^2+a}{X^2} = \frac{V(X^2+a)+a}{X^2} = V + a \frac{V+1}{X^2}.\tag{1}$$

With the Rayleigh distribution, $X^2$ has positive probability density for all $X^2 \gt 0.$ As $X^2$ ranges through all positive numbers, the right hand side of $(1)$ ranges over the interval $(V, \infty)$ when $a(V+1)\gt 0$, over the interval $(-\infty,V)$ when $a(V+1) \lt 0$, and otherwise is fixed at $V$. This immediately implies that the range of values of $U$ that have some chance of happening depends on $V$, and that we cannot get rid of this problem by eliminating a set of $V$ having just zero probability.

Because the ranges of possible $U$ differs with $V$, the conditional probability distribution of $U$ obviously varies with $V$, too. Therefore $U$ and $V$ are not independent.

The "other man" can be confuted by considering a simplified version of his assertion where there is just one variable, say $X$. We may "independently" construct many random variables from $X$, such as $U=2X$ and $V=4X$, but I hope it's obvious the resulting variables are not themselves independent. In this example, for instance, $V=2U$ exhibits the dependence explicitly. The same argument applies to multivariate random variables and for the same reasons.

Finally, there are some special cases where sets of variables constructed from the same "core" of independent variables are independent. The best-known (and arguably most important) example consists of an orthogonal transformation of independent and identically distributed Normal variables: the resulting variables are still independent and identically distributed.