Formal proof that X and X squared random variables are dependent. Intuitively I know that any $X$ and $Y = X^2$ random variables are not independent, but I can't come up with a formal proof. In the case I'm most interested in, $X$ is uniformly distributed on $[-1,1]$.
 A: There are cases where $X$ and $X^2$ are independent! Consider $X$ with $P(X=-1)=P(X=+1)=\frac12$, for example.
Let $X$ be a random variable and assume that $X^2$ is not a.s. constant (which is the case especially for $X$ with uniform distribution, in fact for anything that is not of the form $P[X=a]+P[X=-a]=1$ for some $a\in\mathbb R$). Then there exists an interval $U$ with $0<P[X^2\in U]< 1$. Let $V=\{\,x\mid x^2\notin U\,\}$. Then $P[X\in V]=1-P[X^2\in U]>0$ and $P[X\in V, X^2\in U]=0\ne P[X\in V]\cdot P[X^2\in U]$
A: We can prove that $X$ and $X^2$ are independent if and only if $X^2$ is constant.
One possible proof is to observe that, whenever $0 \leq a \leq b$ and $\Bbb{P}(a \leq |X| \leq b) > 0$, we have
$$ a^2 \Bbb{P}(a \leq |X| \leq b) \leq \Bbb{E}(X^2 \mathbf{1}_{\{a \leq |X| \leq b\}}) \leq b^2 \Bbb{P}(a \leq |X| \leq b).$$
Now if $X$ and $X^2$ are independent, the expectation above reduces to
$$ \Bbb{E}(X^2 \mathbf{1}_{\{a \leq |X| \leq b\}}) = \Bbb{E}(X^2) \Bbb{E}( \mathbf{1}_{\{a \leq |X| \leq b\}}) = \Bbb{E}(X^2) \Bbb{P}(a \leq |X| \leq b). $$
Dividing both sides by the probability, this implies
$$ a^2 \leq \Bbb{E}(X^2) \leq b^2.$$
In other words, either $a > \sqrt{\Bbb{E}(X^2)}$ or $b < \sqrt{\Bbb{E}(X^2)}$ we have $\Bbb{P}(a \leq |X| \leq b) = 0$. Therefore we have
$$ |X| = \sqrt{\Bbb{E}(X^2)} \quad \text{a.s.}$$
A: How 'bout $P(0<X^2<1/2) \ne P(0<X^2<1/2\,\mid\,3/4<X)$?
