# A nonnegative random variable has zero expectation if and only if it is zero almost surely

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 $$Y$$.

• What is the meaning of $P(Y=0) = 1$ for non-discrete $Y$? Commented Aug 15, 2014 at 2:04
• For a non-discrete random variable isn't $P (Y=c)=0$ where c can be zero or anyother value $Y$ can take on? Commented Aug 15, 2014 at 2:08
• @nishrito I don't really know. You're asking the same question I am. I'm wondering if the question itself is incorrect. Commented Aug 15, 2014 at 2:12
• If $Y$ is non-discrete but positive, we can say $P(Y = 0) = 1$ iff $P(Y < \epsilon) = 1$ for all $\epsilon > 0$. Commented Aug 15, 2014 at 2:21
• @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) Commented Jul 28, 2023 at 16:53

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 $$E(Y) = \int Y dP = 0$$

• It means $Y = 0$ except on a negligible set of probability measure $0$.
– RRL
Commented Aug 15, 2014 at 13:58
• 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.
– RRL
Commented Aug 15, 2014 at 14:17
• I do not understand why the converse is "obvious", i.e. why does $Y=0$ a.s. implies that $E(Y)=0$. Commented Dec 25, 2020 at 18:35
• @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 $\int_\Omega \varphi \, dP = \sum_{j=1}^n a_jP(A_j)$.
– RRL
Commented Dec 25, 2020 at 22:07
• 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 $\int_\Omega \varphi \, dP = \int_{\Omega \setminus E} \varphi \, dP + \int_E \varphi \, dP = 0$.
– RRL
Commented Dec 25, 2020 at 22:14

Sufficiency is obvious. For necessity, assume for contradiction that $$E[Y] = 0$$ and $$P(Y > 0) = c > 0$$. From total expectation, we have:

$$0 = E[Y] \ge P(Y>0)E[Y\mid Y >0] = cE[Y \mid Y>0] > 0.$$

• nice. i knew there had to be some way to do this without continuity of measure/probability.
– BCLC
Commented Apr 15, 2021 at 13:17