We have proven the following in class:

If $X$ is a finite random variable with $X\geq 0$ then $$E(X)=0 \iff P(X=0)=1$$ (By finite I meant that the range has finitely many elements).

Does it hold for any non-negative random variable $X:\Omega\to\mathbb R_{\geq0}$?

(I've tried proving it with the indicator function yet it didn't help.)

  • $\begingroup$ What do you mean? Do you mean that X can now take any real value, rather than restricting X to $\ge 0$ ? Or do you mean substituting some arbitrary value in for E(X), $E(X)=A:A\in\mathbb{R}$ ? In either case, the answer is no. $\endgroup$ – JustAskin Mar 12 '14 at 1:08
  • $\begingroup$ @Justaskin_: I mean not restricting $X\geq 0$ to be finite. $\endgroup$ – Leo Mar 12 '14 at 1:13
  • $\begingroup$ You mean that $X$ can take the value $+\infty$? Assuming that, then your supposition is correct. Suppose that $P[X=0]=1$. Then the expected value is $\int x\,\textrm{d}P$, which is bounded above by $0\cdot 1+\infty\cdot 0=0$. Conversely, suppose that $P[X=0]<1.$ Then there is some $\epsilon>0$ and $\mu>0$ such that $P[X\geq \epsilon]>\mu$. It follows that $E[X]\geq\mu\epsilon>0$. $\endgroup$ – Unwisdom Mar 12 '14 at 1:18
  • $\begingroup$ @Unwisdom: doesn't your integral imply that we have a density function? And by finite I meant that the range has finitely many elements, as opposed to (un)countable, etc. $\endgroup$ – Leo Mar 12 '14 at 1:24
  • $\begingroup$ I don't think that I am assuming the existence of a density. After all, I am integrating with respect to $P$. Anyway, now that I know what you mean by "infinite", I'll work through the details more carefully in an answer. $\endgroup$ – Unwisdom Mar 12 '14 at 1:30

Note that, for every $x\gt0$, $$ X\geqslant x\mathbf 1_{X\geqslant x}. $$ (This is the step where one uses that $X\geqslant0$ almost surely.) It follows that $$ E(X)\geqslant xP(X\geqslant x). $$ If $E(X)=0$ this inequality implies that $P(X\geqslant x)=0$. Finally, $$ [X\gt0]=\bigcup_{n\geqslant1}[X\geqslant1/n], $$ hence, if every event in the RHS has probability zero the event on the LHS has probability zero.

  • $\begingroup$ Is there a nice proof for the other direction? (And many thanks for this neat proof.) $\endgroup$ – Leo Mar 12 '14 at 23:03
  • $\begingroup$ You mean, P(X=0)=1 implies E(X)=0? Well, yes... $\endgroup$ – Did Mar 12 '14 at 23:24
  • $\begingroup$ One question about the step where you use that $X⩾0$ almost surely. If we see $aP(X⩾a)\leqslant E(X)$ as the integrated version of $a\mathbf 1_{X\geqslant a}\leqslant X$ then the issue with almost surely is clear (since the integral won't notice us leaving out the zero measure set). But we didn't have the $\int_\Omega X \ \text{d}\mathbb P$ definition in the course. Is there a way to understand the step without the integral? $\endgroup$ – Leo Mar 13 '14 at 0:24
  • $\begingroup$ How did you define E(X) then? $\endgroup$ – Did Mar 13 '14 at 8:55

Let $\langle \Omega,\mathcal{F},P\rangle$ denote our probability space. The expected value of $X$ is $$E[X]=\int_{\Omega}X\, \textrm{d}P.$$

If we let $U\in\mathcal{F}$ be $X^{-1}\{0\}$, then we can decompose the exectation: $$E[X]=\int_{U}X\, \textrm{d}P+\int_{\Omega\setminus U}X\, \textrm{d}P=\int_{\Omega\setminus U}X\, \textrm{d}P.$$

(The second equality follows from the fact that $X(\omega)=0$ for $\omega\in U$.)

Now, suppose that $P[\Omega\setminus U]>0$. Since $$\forall \omega\in (\Omega\setminus U)\, \big(X(\omega)>0\big)$$ there must be some $\epsilon>0$ and some $V\subseteq(\Omega\setminus U)$ such that $P[V]>0$ and $$\forall \omega\in V \, \big(X(\omega)>\epsilon\big).$$ Thus: $$E[X]=\int_{\Omega\setminus U}X\, \textrm{d}P \geq \int_{V} X\, \textrm{d}P \geq \int_{V}\epsilon\,\textrm{d}P = \epsilon P[V]>0.$$

For the converse direction, suppose that $P[U]=1$. Then $P[\Omega\setminus U]=0$ and so we get: $$E[X]=\int_{\Omega\setminus U}X\, \textrm{d}P = 0.$$


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