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Let $X$ be a random variable such that $X\in[0,1]$. I was wondering if $\mathbb{E}[X]$ must be $\geq0$. Since $X$ is a positive random variable, we can apply the Markov-inequality: for each positive random variable $X$ and for each $\delta>0$ it has $$P(X\geq \delta)\leq \frac{\mathbb{E}[X]}{\delta}$$ but $P(X\geq \delta)\geq 0$ and so $$\mathbb{E}[X]\geq 0.$$ Is it correct?If yes, is there another way to prove this statement, without using the Markov-inequality?

Thanks.

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    $\begingroup$ My, my, my, if this is not turning things around, I do not know what is. Hint: $X∈[0,1]$ (almost surely) hence $X\geqslant0$ almost surely hence $E(X)\geqslant E(0)=$ $____$. $\endgroup$ – Did Jul 29 '15 at 8:01
  • $\begingroup$ Which textbooks and/or sources are you using? $\endgroup$ – Did Jul 29 '15 at 8:01
  • $\begingroup$ Nobody, I'm a self-taught, I joined this forum to learn some math :-) (Excuse me if my questions are trivial or incorrect!) $\endgroup$ – Lely Jul 29 '15 at 8:07
  • $\begingroup$ The question was not "who?" but "what?" Anyway, your approach seems to meet some limits. $\endgroup$ – Did Jul 29 '15 at 8:09
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If $X$ is a nonnegative random variable then by definition:

$\mathbb{E}X:=\sup\left\{ \mathbb{E}Y\mid Y\text{ is rv with finite range and suffices: }0\leq Y\left(\omega\right)\leq X\left(\omega\right)\text{ for each }\omega\in\Omega\right\} $

If $Y$ is a random variable as described above then it can be written as: $$Y=\sum_{k=1}^{n}r_{k}1_{A_{k}}$$ where the $r_{k}$ are nonnegative and the $A_{k}$ are measurable and form a partition of $\Omega$.

That gives $\mathbb{E}Y=\sum_{k=1}^{n}r_{k}P\left(A_{k}\right)\geq0$.

So $\mathbb EX$ is the supremum of a subset of $[0,\infty)$ hence $\mathbb EX\geq0$.

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$$X\leq Y\implies \mathbb E[X]\leq \mathbb E[Y],$$

and thus $$X\geq 0\implies \mathbb E[X]\geq \mathbb E[0]=0.$$

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