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Let $(\Omega, \mathfrak A, P)$ be a probability space and $\mathscr P$ the set of probability measures. Let $X$ be a nonnegative random variable with $\sup_{P \in \mathscr P}E_P[X]<\infty$.

Question: Is it true that $$\operatorname{ess sup}_{P\in \mathscr P}E_P[X]=\sup_{P \in \mathscr P}E_P[X] \ ?$$

I would guess this is true since $E_P[X]$ is always a finite number depending on $P$ but I can't explain it rigorously.

Any hints or ideas?

scholar
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    Unless mistaken you don't "ess sup" over a collection of probabilities nor over expectation operator. The ess sup is the lowest maximum of a random variable over non negligible sets. So your left hand side is not properly using the ess sup and the question is not set in a way that can receive an answer, as a matter of fact I can't see a way to turn this right : https://en.wikipedia.org/wiki/Essential_infimum_and_essential_supremum – TheBridge Nov 21 '22 at 21:55
  • @TheBridge Thank you for your answer. The context of this question is the following: Let $U_t=\text{ess sup}{P \in \mathscr P}E_P[X|\mathcal F_t]$, where $\mathcal F$ with $\mathcal F_0={\emptyset, \Omega}$ is a filtration. The text that I am reading claims $U_0=\sup{P\in \mathscr P}E_P[X]$ without further explanation. So I thought that the mentioned identity in my post must be the explanation. – scholar Nov 21 '22 at 22:08
  • @scholar $E_P[X|\mathcal{F}_t]$ is a conditional expectation relative to a $\sigma-$algebra. If you consider $t=0$, then $E_P[X|\mathcal{F}_t]$ becomes $E_P[X|\mathcal{F}_0]$, which by definition is simply the usual expectation $E_P[X]$. Maybe you're not familiar with the definition of conditional expectation wrt a $\sigma-$algebra. If that is the case, then start by looking that up. https://math.stackexchange.com/questions/2870000/what-is-the-intuition-behind-conditional-expectation-in-a-measure-theoretic-trea/4459431#4459431 – Oscar Nov 22 '22 at 08:07
  • @Oscar Thank you for your response. I do understand how $E_P[X|\mathcal F_t]=E_P[X]$ when $t=0$, cause $\mathcal F_0$ has no information. But if you take $U_t=\text{ess sup}E_P[X|\mathcal F_t]$ and simply plug in $t=0$ then we get $U_0=\text{ess sup}E_P[X|\mathcal F_0]=\text{ess sup}E_P[X]$ , how do we get from "ess sup" to "sup"? – scholar Nov 22 '22 at 12:05

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