(X,X) independent and moments I have a question regarding a random variable X being independent of itself.
I have to proof, knowing that X has no moment, that the following real number exists:
$$ t_{0} = \inf_{} \{t\in \mathbb{R} | F_X(t) = 1 \}$$
Please, if my question is ambiguous or unclear, do not hesitate to edit it !
Thank you so much !
 A: Intuitively, if any event regarding X is independent with itself, when it realises, the probability of it realising doesn't change, so either it never realises, either it was already sure that it was gonna realise. Let us put this formally.
Consider the events $X\leq t$ for $t\in \mathbb{R}$. Since X is indepedent with itself, we know that $P(X\leq t) = P(X\leq t \cap X\leq t)= P(X\leq t)P(X\leq t)$ so $P(X\leq t) = 1$ or $P(X\leq t) = 0$.
So $\forall t \in \mathbb{R}, F_X(t) = 0$ or $F_X(t) = 1$.
Now, can it always be 0? No it cannot, because otherwise X would always be greater than any real number, which is impossible. Formally:
$P(X \leq n) = P(\cap_{k \leq n}X \leq k) \to P(X \in \mathbb{R}) = 1$
From this, we know that $\exists t_0 \in \mathbb{R}, F_X(t_0) = 1$. So the considered set is not empty.
Also, with the same argument, we know that $\exists t_1 \in \mathbb{R}, F_X(t_1) = 0$, and since $F_X$ is increasing, $\forall t \leq t_1, F_X(t) = 0$. So the set is bounded from the bottom.
Therefore, it has an inf. This is what we wanted to show.
