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I have a random variable $X$ that is independent of itself. How knowing that $$ \mathbb{E}[X^2] < \infty $$ is suppose to help me finding that $X$ is a constant.

I already found out that if $X$ is independent, $ \mathbb{P}(X \in A) = 0 $ or $ \mathbb{P}(X \in A) = 1 $. And I just pick $x \in \mathbb{R}$ and I have $\mathbb{P}(X \le x)^2 = \mathbb{P}(X \le x) \ge \mathbb{P}(X \le x) = 0 $ or $ \mathbb{P}(X \le x) = 1$

But how is it related to that expected value ? I really tried but I do not come with any ideas.

Many thanks ! If the question is ambiguous or unclear, please edit it :)

Michael Hardy
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Valentin
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1 Answers1

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As I said in the comments, by the independence, we have that $E(X^2)=E(X)E(X)$ and thus $Var(X)=0$. As a result, $X=E(X)$ (constant) almost surely.

Evangelopoulos Foivos
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  • What does "independent of itself" mean in terms of the definition via a joint distribution? – A rural reader Feb 15 '21 at 20:48
  • Well, $X$ is independent of it self iff $F_{X,X}(x,y) = F_X(x) F_Y(y) $ for all $x,y$. This is equivalent to saying that $ F_X(x) = (F_X(x))^2 $ for all $x$ which is equivalent to $F_X(x)=0 $ or $1$ – Evangelopoulos Foivos Feb 15 '21 at 21:20
  • I guess all sorts of odd things can happen with a degenerate distribution. – A rural reader Feb 15 '21 at 21:49
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    @EvangelopoulosF. : There's potentially consequential typo in your comment. It should say this: $$ F_{X,X}(x,y) = F_X(x) F_X(y). $$ No random variable called (capital) $Y$ is involved here. $\qquad$ – Michael Hardy Feb 15 '21 at 21:56
  • @Aruralreader : $\qquad\uparrow\qquad$ See my comment above, and also this: To say that two real-valued random variables $X,Y$ are independent means that for sets $A,B\subseteq\mathbb R,$ the events $\big[ X\in A\big]$ and $\big[ Y\in B]$ are independent. $\qquad$ – Michael Hardy Feb 15 '21 at 22:28