$X$ and $X^2$ independent, show that there is $c \in \mathbb{R}$ such that $P(|X|=c)=1$ $X$ and $X^2$ independent real valued, random variables, show that there is $c \in \mathbb{R}$ such that $P(|X|=c)=1$
I would like to receive a hint on how to approach this problem.
 A: Intuitive reasoning: $X$ and $X^2$ being independent means that knowing $X$ tells you nothing about $X^2$. This cannot be true if $|X|$ can attain more than one value.
A: Hint: For any set $A\in\mathbb{R}$ we have
$$
P(|X|\in A) = P(|X|\in A ; X^2\in A^2) = P(|X|\in A)P(X^2\in A^2) = P(|X|\in A)^2.
$$
Therefore $P(|X|\in A)$ is either zero or 1.
A: Note that for any $x\in{\bf R}$,
$$
P(|X|\leq x)=P(|X|\leq x,X^2\leq x^2)=P(|X|\leq x)\cdot P(X^2\leq x^2)=P(|X|\leq x)^2.
$$
Using the properties of probability distribution function one can show that $|X|$ is constant (almost surely).
A: Note that for $a>0$ $$P(X^2\leq a^2\mid X>b)=P(|X|\leq a\mid X>b)=P(|X|\leq a)\space\text{if $b\leq -a$}\\=P(b\leq X\leq a)\space\text{if $-a< b\leq a$}\\=0\space\text{if $b>a$}$$
But since $X^2$ and $X$ are independent $$P(X^2\leq a^2\mid X>b)=P(|X|\leq a)$$
Then if $-a<b\leq a$, $P(-a\leq X<b)=0$. If we choose $b=a$ then we get $P(X=a)>0$
Similarly if $a<0$ we get $P(X=-a)>0$.
Hence $P(|X|=a)>0$ o.w. $P(|X|=x)=0$ for all $x\in \mathbb{R}-\{a\}$.
Then $P(|X|=a)=1\space\space\space\blacksquare$
A: There are two independently useful facts which are being used here:


*

*if $X$ and $Y$ are independent then $f(X)$ and $g(Y)$ are independent for any continuous $f,g$;

*if $X$ is independent of itself then it is almost surely a constant.


The proofs are both simple. For the first, we have
$$P(f(X)\in A,g(Y)\in B)=P(X\in f^{-1}(A),Y\in g^{-1}(B))=P(X\in f^{-1}(A))P(Y\in g^{-1}(B))=P(f(X)\in A)P(g(Y)\in B).$$
For the second,
$$P(X\le x)=P(X\le x,X\le x)=P(X\le x)^2,$$
so $P(X\le x)\in\{0,1\}$. This implies $P(X=c)=1$ where $c=\min\{x:P(X\le x)=1\}$.
Using the independence of $|X|$ and $X^2$ is actually a bit redundant, since the proof of this is the same as proving $X^2$ is independent of itself, which implies $X^2=c$ and thus $|X|=\sqrt{c}$ with probability one.
