Let $Y$ be a non-negative random variable. Prove that $E(Y) = 0$ if and only if $P(Y=0)=1$.
My understanding is that while you can prove it for discrete $Y$, the result is true for all $Y$.
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1$\begingroup$ What is the meaning of $P(Y=0) = 1$ for non-discrete $Y$? $\endgroup$– NishritoCommented Aug 15, 2014 at 2:04
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$\begingroup$ For a non-discrete random variable isn't $ P (Y=c)=0$ where c can be zero or anyother value $ Y $ can take on? $\endgroup$– dylan7Commented Aug 15, 2014 at 2:08
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$\begingroup$ @nishrito I don't really know. You're asking the same question I am. I'm wondering if the question itself is incorrect. $\endgroup$– user137481Commented Aug 15, 2014 at 2:12
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$\begingroup$ If $Y$ is non-discrete but positive, we can say $P(Y = 0) = 1$ iff $P(Y < \epsilon) = 1$ for all $\epsilon > 0$. $\endgroup$– JimmyK4542Commented Aug 15, 2014 at 2:21
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$\begingroup$ @JimmyK4542 I think you intended $P(|Y|<\epsilon)=1$ but I would still describe this as a discrete random variable (its support is a finite or countably infinite number of points) $\endgroup$– HenryCommented Jul 28, 2023 at 16:53
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2 Answers
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If $Y$ is a non-negative random variable defined on a probability space $\Omega$ and
$$E(Y) = \int_{\Omega} Y dP=0$$
then $Y = 0$ almost surely and $P(\{\omega \in \Omega: Y(\omega)=0\})=1$.
Proof: For any $m \in \mathbf{N}$, let
$$E_m = \{\omega \in \Omega:Y(\omega) > 1/m\}$$
then, since $Y$ is non-negative, we have $Y = Y1_{\Omega} \geq Y 1_{E_m}$ and then
$$0 = \int_{\Omega} Y dP \geq \int_{E_m} Y dP\geq \frac1{m}P(E_m) \geq 0,$$
and $P(E_m) = 0$.
So
$$0 \leq P(\{\omega \in \Omega:Y(\omega) \neq 0\})= P\left(\bigcup E_m\right) = \lim_{m \rightarrow \infty}P(E_m)=0$$
Hence
$$P(\{\omega \in \Omega:Y(\omega) \neq 0\})=0 \implies P(\{\omega \in \Omega:Y(\omega) = 0\})=1 $$ QED
Conversely, if $Y=0$ a.s. then $E(Y) = \int Y dP = 0$
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2$\begingroup$ It means $Y = 0$ except on a negligible set of probability measure $0$. $\endgroup$– RRLCommented Aug 15, 2014 at 13:58
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1$\begingroup$ The random variable $Y$ is a measurable function on the probability space, $Y:\Omega \rightarrow [0,\infty)$. Yes -- at a single point $P(Y(\omega_0)=c)=0$. But $(Y=0)$ is an event: $(Y=0):=\{\omega:Y(\omega)=0\}.$ not a singleton. $\endgroup$– RRLCommented Aug 15, 2014 at 14:17
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1$\begingroup$ I do not understand why the converse is "obvious", i.e. why does $Y=0$ a.s. implies that $E(Y)=0$. $\endgroup$ Commented Dec 25, 2020 at 18:35
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2$\begingroup$ @MichaelBaudin: As in the general theory of integration of functions defined on a measure space, the expected value (integral) of a nonnegative random variable $Y$ is defined as $E(Y) = \int_\Omega Y \, dP = \sup_{\varphi \leqslant Y} \int_\Omega \varphi \, dP$ where the supremum is taken over integrals of all simple random variables dominated by $Y$. A simple random variable has a finite number of distinct values on disjoint events, i.e., $\varphi(\omega) = \sum_{j=1}^n a_j \mathbf{1}_{A_j}(\omega)$ and $\int_\Omega \varphi \, dP = \sum_{j=1}^n a_jP(A_j)$. $\endgroup$– RRLCommented Dec 25, 2020 at 22:07
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2$\begingroup$ Now if $Y=0$ a.s., then if $0 \leqslant \varphi \leqslant Y$ you should be able to show easily that $\int_\Omega \varphi \, dP = 0$ for any such $\varphi$ and the supremum of integrals is also zero. Let $E = \{\omega \, | \, Y(\omega) \neq 0\}$ and since $P(E) = 0$, show $\int_\Omega \varphi \, dP = \int_{\Omega \setminus E} \varphi \, dP + \int_E \varphi \, dP = 0$. $\endgroup$– RRLCommented Dec 25, 2020 at 22:14
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Sufficiency is obvious. For necessity, assume for contradiction that $E[Y] = 0$ and $P(Y > 0) = c > 0$. From total expectation, we have:
$$ 0 = E[Y] \ge P(Y>0)E[Y\mid Y >0] = cE[Y \mid Y>0] > 0. $$
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$\begingroup$ nice. i knew there had to be some way to do this without continuity of measure/probability. $\endgroup$– BCLCCommented Apr 15, 2021 at 13:17