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Hi, I'm looking for conditions on $G(t,x)$ such that $$ \sup\limits_{t\in [0,1]}E[G(t,X)]=E[\sup\limits_{t\in [0,1]}G(t,X)] $$ where $X$ is a random variable (it's easy to see that $\sup\limits_{t\in [0,1]}E[G(t,X)]\leq E[\sup\limits_{t\in [0,1]}G(t,X)]$).

Any suggestion or reference is greatly appreciated!

edited Sep 23 at 14:17

asked Sep 23 at 6:47

martin
21●4

	Are $I$ and $X$ fixed? – Davide Giraudo Sep 23 at 9:25
	I talked with a mathematician once who expressed frustration with some modern theoretical physics writing, exactly because of this point. They would use this without justification, or even without noticing. If the random variable $X$ is constant a.s. it would have been OK (in that setting, at least); and in thermodynamics it often turns out that they are constant; but some physicists would forge ahead, maximizing the r.v. by maximizing instead the expectation. – Gerald Edgar Sep 23 at 12:55
	Yes, $I$ and $X$ are fixed. I've also added an assumption that $I=[0,1]$. – martin Sep 23 at 16:00
	Is there anything more you're prepared to specify about $X$? – Yemon Choi Sep 23 at 18:17
	@Yemon Choi: I would like the result to hold for any distribution of $X$, but if you want to impose some conditions on $X$, that's fine. I'm still clueless on how to approach the problem. – martin Sep 24 at 2:21

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Dan · Accepted Answer · 2012-09-25 13:15:43Z UTC

This will require very strong conditions on $G$. The most general result I know of is an "almost upward-filtering" condition, fairly well known in stochastic optimal control theory: Assume $G(t,\cdot)$ is measurable for each $t \in [0,1]$, and $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] < \infty$. Then $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\text{ess}\sup_{t \in [0,1]}G(t,X)]$ if and only if for all $\epsilon > 0$ and $r,s \in [0,1]$ there exists $t \in [0,1]$ such that $\mathbb{E}[(G(t,X) - G(s,X) \vee G(r,X))^-] \le \epsilon$. I believe this result is originally due to J.A. Yan, in the hard-to-find paper "On the commutability of essential infimum and conditional expectation operators". Note that $\sup_{t \in [0,1]}G(t,X)$ need not be measurable, which is why the essential supremum is used in the aforementioned theorem.

A more transparent condition can be derived from the above if we add a continuity assumption: Assume $G(\cdot,x)$ is continuous for each $x$, $G(t,\cdot)$ is measurable for each $t$, and $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] < \infty$. Then $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\sup_{t \in [0,1]}G(t,X)]$ if and only if there exists $T \in [0,1]$ such that $G(T,X) = \sup_{t \in [0,1]}G(t,X)$ almost surely.

PROOF: The continuity assumption guarantees that $\sup_{t \in [0,1]}G(t,X)$ is indeed measurable (e.g. by Theorem 18.19 of Aliprantis & Border), and thus $\text{ess}\sup_{t \in [0,1]}G(t,X) = \sup_{t \in [0,1]}G(t,X)$. The aforementioned theorem and a simple argument using compactness of $[0,1]$ and Fatou's lemma shows that (under our assumptions) $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] = \mathbb{E}[\sup_{t \in [0,1]}G(t,X)]$ if and only if for all $r,s \in [0,1]$ there exists $t \in [0,1]$ such that $G(t,X) \ge G(r,X) \vee G(s,X)$ a.s.. Since the "if" part is trivial, we now prove the "only if". Consider the set $S := \{G(t,X) : t \in [0,1]\}$ with the partial order given by almost sure inequality. Compactness of $[0,1]$ and continuity of $G(\cdot,x)$ for all $x$ yield the existence of an upper bound in $S$ for any chain of $S$, and thus by Zorn's lemma $S$ contains a maximal element. That is, there exists $T \in [0,1]$ such that there is no $s \in [0,1]$ for which $G(s,X) \ge G(T,X)$ a.s. and $P(G(s,X) > G(T,X)) > 0$. For any $t \in [0,1]$ there exists $r \in [0,1]$ such that $G(r,X) \ge G(T,X) \vee G(t,X) \ge G(T,X)$ a.s., which implies $G(r,X) = G(T,X)$ a.s. and thus $G(T,X) \ge G(t,X)$ a.s.. Hence $G(T,X) \ge G(t,X)$ a.s. for any $t \in [0,1]$.

Stanislav · Answer 2 · 2012-09-23 12:04:11Z UTC

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The equality holds if $G(t,x) = f(t) + g(x)$.
If $G(t,x) = f(t) g(x)$, then $$E(\sup(G(t,X))=\sup f(t) E(g(X)1_{g(X)\geq0}) + \inf f(t) E(g(X)1_{g(X)<0})$$ and \begin{equation} \sup E(G(t,X)) = \begin{cases} E(g(X)) \sup f(t) & \text{ if } E(g(X)) \geq 0 \newline E(g(X)) \inf f(t) & \text{ if } E(g(X)) < 0 \end{cases} \end{equation} So the equality holds if $g(X) \geq 0$ a.s. or $g(X) \leq 0$ a.s.

edited Sep 23 at 12:04

answered Sep 23 at 11:56

Stanislav
304●2●6

Thanks, but I would like necessary and sufficient conditions on $G(t,x)$ (or at least a condition that allows for a more general class of functions $G(t,x)$ if possible). – martin Sep 23 at 14:30

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If $X$ keeps its sign, one can also consider the Taylor series of $G(t,x)$: $G(t,x) = \sum_{k,n\ geq 0} a_{kn} t^k x^n$ and analyze the conditions under which $\sup_t E(\sum a_{kn} t^k x^n) = \sup_t \sum a_{kn} t^k E(x^n) = \sum a_{kn} \sup_t(t^k) E(x^n) = E(\sum a_{kn} \sup_t(t^k) x^n) = E( \sup_t( \sum a_{kn} t^k x^n))$. – Stanislav Sep 23 at 19:47

Is $\sup_t\sum a_{kn}t^kE(x^n)=\sum a_{kn}\sup_t(t^k)E(x^n)$ a typo? How can you pass $\sup$ into the summation? – martin Sep 23 at 20:22

This step requires $\sum a_{kn} t^k E(X^n)$ to be an increasing function of $t$. It may be a very strong condition, but I don't see any other way to exchange $\sup(\cdot)$ and $E(\cdot)$ here. – Stanislav Sep 23 at 20:48

That indeed is a very strong condition :) – martin Sep 24 at 0:50

When do maximum and expectation commute?

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