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I think the following is true but I cannot prove it.

Let $Z_1, Z_2$ are two random variables defined on the same sample space $\Omega$. Suppose that $Z_1(\omega) < Z_2(\omega)$ for all $\omega\in \Omega_0$ and $Z_1(\omega) = Z_2(\omega)$ for all $\omega\in \Omega\setminus\Omega_0$. We have:

If $P(\Omega_0)=0$, i.e. $Z_1=Z_2$ almost surely, then $E(Z_1)=E(Z_2)$.

If $P(\Omega_0)>0$, then $E(Z_1)<E(Z_2)$.

Could you show if it holds or not?

TrungDung
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  • If you define $X(\omega) = Z_2(\omega) - Z_1(\omega)$ then $X(\omega)\geq 0$ for all $\omega \in \Omega$. You might use the Markov inequality on the random variable $X$. – Michael Mar 12 '20 at 18:30
  • Are you familiar with the result that if $f \ge 0$ then $\int f d \mu = 0$ iff $f$ is zero ae. $[\mu]$? – copper.hat Mar 12 '20 at 18:39
  • @Michael: I use that inequality and prove it. – TrungDung Mar 16 '20 at 20:20
  • @Michael: I am wondering what if we assume $P(X(\omega)\geq 0)=1$ instead of $X\geq 0$? – TrungDung Mar 23 '20 at 14:46
  • Sets of probability 0 do not affect expectations, so the result is the same. Equivalently you could define a new RV $Y=\max[X,0]$ which is always nonnegative, and $E[Y]=E[X]$. – Michael Mar 23 '20 at 16:35

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Write $$ \mathsf{E}Z_1=\int_{\Omega}Z_1(\omega)\mathsf{P}(d\omega)=\int_{\Omega_0}Z_1(\omega)\mathsf{P}(d\omega)+\int_{\Omega\setminus\Omega_0}Z_1(\omega)\mathsf{P}(d\omega).\tag{1}\label{1} $$ Since $$ \int_{\Omega\setminus\Omega_0}Z_1(\omega)\mathsf{P}(d\omega)=\int_{\Omega\setminus\Omega_0}Z_2(\omega)\mathsf{P}(d\omega), $$ the result depends on the probability of $\Omega_0$. If $\mathsf{P}(\Omega_0)=0$, then the first integral on the RHS of $\eqref{1}$ is $0$ (see, e.g., this question). Otherwise, $$ \int_{\Omega_0}Z_1(\omega)\mathsf{P}(d\omega)<\int_{\Omega_0}Z_2(\omega)\mathsf{P}(d\omega). $$