# 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

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

Mathematics Asked by override on November 12, 2021

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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