Self-independent random variable

Question

Let $X$ be a self-independent random variable. Show that $X$ is almost sure constant.

My proof (by condradiction): Assume that there are two disjoint Borel set $A$,$B$ such that: $\Pr(X \in A)>0$ and $\Pr(X \in B)>0$

Since $X$ is self-independent, there must be $\Pr(X \in A)=1$ and $\Pr(X \in B)=1$. It's a contradiction, because then $\Pr(X \in A \cup B)>1$.

Is everything ok?

If $A$ and $B$ are disjoint then they cant both have probability 1. — , Jun 18 '15 at 22:06
Because $\Pr(X \in A)=\Pr(X \in A, X \in A)=\Pr(X \in A)(PX \in A)$, there must be $\Pr(X \in A)=0$ or $\Pr(X \in A)=1$ — mrnobody, Jun 18 '15 at 22:08
@dreamer94 Your statement in the comments is sufficient, as pre-kidney shows. — Teepeemm, Jun 19 '15 at 00:12
Why did you accept (25 minutes after it was posted) an answer which does not answer the question? — Did, Jun 19 '15 at 08:10
https://math.stackexchange.com/questions/464264/random-variable-independent-of-itself — StubbornAtom, Nov 29 '19 at 15:54

pre-kidney · Accepted Answer · 2019-07-31T02:09:10.597

A direct proof:

For any measurable set $A$, applying independence of the events $\{X\in A\}$ and $\{X\in A\}$ gives $$ \mathbb P(X\in A)=\mathbb P(X\in A)\mathbb P(X\in A)\implies \mathbb P(X\in A)\in \{0,1\}. $$ Hence $\mathbb P$ is a dirac measure. Therefore $X$ is a.s. constant.

Edit, 4 years later.

From a certain perspective, the more interesting part of the answer is the passage from "$\mathbb P(X\in A)\in \{0,1\}$" to "there exists $x\in\mathbb R$ such that $\mathbb P(X=x)=1$" (the latter is the more common definition of a dirac measure), which my original answer did not address. In fact, it is a special case of a more abstract result, with $\mathbb R$ replaced by any separable metric space.

Lemma. Let $\mu$ be a Borel probability measure on a separable metric space $M$ such that $\mu(A)\in \{0,1\}$ for all Borel subsets $A\subseteq M$. Then there exists $x\in M$ such that $\mu(\{x\})=1$.

Proof. In a separable metric space, every open cover has a countable subcover. Consider the cover of $M$ by open balls of radius $\epsilon$ around every point, and extract a countable subcover. There must exist a ball $B_\epsilon$ in the countable subcover with $\mu(B_\epsilon)=1$, otherwise all balls would be null sets and thus $M$ would be null as well, a contradiction. Observe that the family of sets with measure $1$ is closed under finite (and therefore countable) intersections, from which it follows that $$\mu(B)=1,\qquad B=\bigcap_{n\in\mathbb N}B_{1/n}.$$ Finally, diam$(B)=0$ and thus $B=\{x\}$, yielding the claim.

Nicolas · Answer 2 · 2015-06-19T08:58:55.013

5

$$\mathbb{E}\left[\left(X-\mathbb{E}\left[X\right]\right)^2\right] =\mathbb{E}\left[X^2\right]-\mathbb{E}\left[X\right]^2 =\mathbb{E}\left[X\right]^2-\mathbb{E}\left[X\right]^2=0$$ thus $X-\mathbb{E}\left[X\right]=0$ a.e. .

IMPORTANT EDIT : This requires $X$ to be square integrable.

edited Jun 19 '15 at 08:58

answered Jun 18 '15 at 22:13

Nicolas

3,326

What a beautiful proof :D – mrnobody Jun 18 '15 at 22:15
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How do you show your upper line? $:$ (Note that the problem does not assume $\mathbb{E}[X\hspace{.03 in}]$ exists.) $\hspace{.93 in}$ – Jun 19 '15 at 00:05
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Indeed, even less that X is square integrable. Sorry but this is not the proof required (and votes on MSE are as baffling as ever). – Did Jun 19 '15 at 08:09
Sorry, I forgot to make this point clear : indeed, this quick proof is available provided that all quantities exist, i.e. provided that $X$ is square integrable. I add this crucial point in my post, thanks to have pointed it out. – Nicolas Jun 19 '15 at 08:57

Self-independent random variable

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