Random variable independent of itself

Question

Could someone help me find a way out:

If $X$ is a random variable and $X$ is independent of itself. Show that there is a constant $a$ such that $P(X=a)=1$ if and only if $E[X_1]$ exists.

Answer 1

$\def\R{\mathbb R} \def\Z{\mathbb Z} \def\P{\mathbb P}$ As was already noticed in the comments the formulation is a bit misleading. Under the assumptions both sides of the equivalence must be true.

$X$ is independent of itself - in particular $\mathbb P(X\in B) = \mathbb P(X\in B, X \in B) = \mathbb P(X\in B) \cdot \mathbb P(X\in B) = \mathbb P(X\in B)^2$ (where $B$ is a borel subset of $\mathbb R$), which means that $\mathbb P(X\in B)$ is just equal to $0$ or $1$ for any $B$.

EDIT (more hints requested):

Now divide $\R$ into a contable number of disjoint intervals e.g. $I_n=[n,n+1)$ for $n\in \Z$. There exists a unique $n$ such that $\P(X\in I_n)=1$ (why?). Then you can divide $I_n$ into two halves and see which half has probability $1$. You can do it inductively and in the limit get a set of probability $1$ and diameter $0$. It must be a point and that's $a$ we were looking for.

Answer 2

Using the definition of independence one can see that for each $t$, we have $\mathbb P\{X\leqslant t\}\in\{0,1\}$. Define $$t_0:=\inf\{t\in\mathbb R\mid\mathbb P\{X\leqslant t\}=1\}.$$ The infimum exists since $\lim_{t\to +\infty}\mathbb P\{X\leqslant t\}=1$ (hence $\mathbb P\{X\leqslant t\}=1$ for $t$ large enough).

If $t_n\downarrow t_0$, then we can see that $\mathbb P\{X\leqslant t_0\}=1$ and since $\mathbb P\{X\leqslant s\}=0$ for $s\lt t_0$, we have $\mathbb P\{X\lt t_0\}=0$.

We conclude that $\mathbb P\left\{X=t_0\right\}=1$.

Answer 3

If $X$ has a finite second moment, i.e. $E(X^2)<\infty$, then $E(X^2)=E(X)^2$, hence the variance of $X$ is $0$ and $X$ is almost surely equal to $E(X)$.

In the general case, since independence in preserved by measurable transformations, $\arctan(X)$ is independent of itself. Since it is bounded, the previous paragraph applies and yields $\arctan(X)=c$ almost surely for some $c\in(-\pi/2,\pi/2)$, hence $X = \tan(c)$ a.s.

P.S.: the trick in the second paragraph was communicated to Gautier Appert by Olivier Catoni.