While browsing my local library, I found in one textbook the following quite interesting lemma:
If: random variable $X$ is independent to every other random variable defined on the same probabilistic space as X, then: $\exists$ $c:=(const.) \in \mathbb{R}$: $P(X=c)=1$
I made attempt of proving this lemma by contradiction. Intuitively, I thought of creating one specific random variable from all possible random variables defined on the same probabilistic space as random variable $X$. I thought of constructing the contradiction by making use of random variable $Y :=X^{3} $, however I got nowhere.
I checked the content of posts on this website and there are no similiar examples that may, in my humble opinion, help me go further in proving the lemma mentioned above. Besides, I checked google browser and did not find any useful information.
I am very curious how to construct the proof for such lemma. I would be thankful for any constructive help!
Suppose $X$ is independent of itself (or if "every other" excludes $X$ itself, you can make a similar argument with $1+X$, say). Then for any $r\in\mathbb{R}$, $$P(X\leq r)=P(X\leq r\text{ and }X\leq r)=P(X\leq r)P(X\leq r)$$ since $X$ and $X$ are independent, and so $P(X\leq r)$ can only be $0$ or $1$. There must also exist $r$ in which both cases hold (since the limit as $r\to\infty$ must be $1$ and the limit as $r\to-\infty$ must be $0$). Now let $c=\inf\{r\in\mathbb{R}:P(X\leq r)=1\}$. Since $P(X\leq r)$ is continuous from the right, $P(X\leq c)=1$, but $P(X<c)=0$ since $P(X<r)=0$ for all $r< c$. Thus $P(X=c)=1$.