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Let $\Omega = [0,1]$ equipped with the Borel $\sigma$-algebra and Lebesgue measure. Let $X:\Omega\rightarrow\mathbb{R}$ be defined by $X(\omega) = \omega$. Clearly $X$ is uniform $(0,1)$ distributed. Does there exist a random variable $Y:\Omega\rightarrow\mathbb{R}$ which is uniform $(0,1)$ distributed such that $X$ and $Y$ are independent? And if so, can we find a formula for $Y$?

I'm convinced that $Y$ should exist, but I don't know how I would construct it.

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  • $\begingroup$ However, you can extract as many mutually independent uniform random variables $g_1(X), g_2(X), ... , g_k(X)$ as you want, all pure functions of $X$ (and all dependent on $X$). $\endgroup$ – Michael Oct 4 '16 at 17:59
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No, there is no such $Y$.

Suppose there were. Let $A = \{\omega : Y(\omega) \le \frac{1}{2}\}$ be the event that $Y$ is at most $\frac{1}{2}$. Then for any $0 \le a \le b \le 1$ we have $$\begin{align*}m(A \cap [a,b]) &= P(Y \le \frac{1}{2}, a \le X \le b) \\ &= P(Y \le \frac{1}{2}) P(a \le X \le b) && \text{by independence} \\ &= \frac{1}{2} (b-a).\end{align*}$$ In particular, for every $x \in (0,1)$ and every sufficiently small $\epsilon > 0$, taking $a = x-\epsilon$ and $b = x+\epsilon$ we have $\frac{1}{2 \epsilon} m(A \cap [x-\epsilon, x+\epsilon]) = \frac{1}{2}$. So for every $x \in (0,1)$ we have $$\lim_{\epsilon \to 0} \frac{1}{2 \epsilon} m(A \cap [x-\epsilon, x+\epsilon]) = \frac{1}{2}.$$ This contradicts the Lebesgue density theorem, which asserts that this limit equals 0 or 1 for almost every $x$.

In fact, by the same logic, any random variable $Y$ on $\Omega$ which is independent of $X$ must have a degenerate distribution, i.e. be almost surely constant.

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    $\begingroup$ A simpler proof is: Suppose $Y$ is a uniform random variable on the same probability space. Then $Y=g(\omega) = g(X)$ for some measurable function $g$. Then the events $\{Y\leq 1/2\}$ and $\{g(X)>1/2\}$ are not independent. $\endgroup$ – Michael Oct 4 '16 at 17:43
  • $\begingroup$ A similar argument holds to show that for any random variable $X$ and any (measurable) function $g$, the random variable $Y=g(X)$ cannot be independent of $X$ as long as there is no constant $c \in \mathbb{R}$ such that $P[Y=c]=1$. $\endgroup$ – Michael Oct 4 '16 at 17:43

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