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I'm confused with the terminology 'almost surely'(a.s.) in probability theory. If $X$ is a random variable, does the following statement equivalent?

  1. $X$ is almost surely nonegative.

  2. $\mathbb{P}(X>0)>0$

  3. $\mathbb{P}(X \geq 0)=1$ a.s.

If we are given the first and second statement, how to prove $\mathbb{E}(X)>0$?

Cloud
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1 Answers1

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An event is said to happen almost surely if the probability of that event is $1$.

$X$ is almost surely non-negative means that $P(X\ge0)=1$.

The third statement makes no sense.

The second statement $P(X>0)>0$ does not imply $\Bbb E(X)>0$. For example, take $X$ to be the random variable where $P(X=-1)=\frac23, P(X=1)=\frac13$.

When both first and second statement holds, check Expectation of nonnegative RV and A nonnegative random variable has zero expectation if and only if it is zero almost surely

QED
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  • Thanks for your help. If we are given the first and second statement, how to prove $EX>0$? Sorry for that mistake, I have edited it. – Cloud Oct 16 '20 at 17:22
  • @Cloud Note that if $\mathbb{P}(X>0)>0$ then there is some $n\in\mathbb{N}$ such that $\mathbb{P}(X>\frac{1}{n})>0$. Now use Markov's inequality. – Mark Oct 16 '20 at 17:35
  • Note that Markov's inequality holds for non-negative random variables only. – QED Oct 16 '20 at 17:38
  • @QED Yes, but we assume $X$ is almost surely nonnegative. – Mark Oct 16 '20 at 17:38
  • @Mark The question was so basic, it is more elementary than the proof of Markov;s inequality, right? And my above comment was for the OP, stressing on the fact that why non-negativity of the r.v. is necessary. – QED Oct 16 '20 at 17:40