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Proving a self independent random variable can get only one value

I’m doing a course in probability and was asked the prove the following. I’d appreciate some feedback on it, thank you.

Let $X$ be a self independent r.v. . Prove that exists $cin mathbb{R}$ s.t. $mathbb{P}(X=c)=1$

My proof:

Let A denote the support of $X$.

Lets assume that exists $x,yin A$ such that $xneq y$.

Let $B_1= {x}$ and $B_2={y}$.

So we get:
$mathbb{P}(Xin B_1 cap B_2) = mathbb{P}(Xin B_1, ;Xin B_2) = mathbb{P}(Xin B_1) cdot mathbb{P}(Xin B_2)=q_1 cdot q_2 >0$

(this is true because $x,y in A$ and have a non zero probability and $X$ is self independent).

But $B_1 cap B_2 = emptyset Rightarrow mathbb{P}(X in B_1 cap B_2)=0$ – Contradiction!

And since the support of a r.v. can not be empty we get that $A={x}$ and by definition

$mathbb{P}(X=x)=1$

Edit: Thank you for your comments, I’ll fix it

Mathematics Asked by override on November 12, 2021

1 Answers

One Answer

By "self independent" I presume you mean $X$ and $X$ are independent.

Suggestion: Rather than looking at $P(X = x)$ which might always be $0$, consider the cumulative distribution function $F(x) = P(X le x)$. Show that this is always $0$ or $1$.

Answered by Robert Israel on November 12, 2021

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