Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!

share|improve this question
    
Are $I$ and $X$ fixed? –  Davide Giraudo Sep 23 '12 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 '12 at 12:55
    
Yes, $I$ and $X$ are fixed. I've also added an assumption that $I=[0,1]$. –  martin Sep 23 '12 at 16:00
    
Is there anything more you're prepared to specify about $X$? –  Yemon Choi Sep 23 '12 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 '12 at 2:21

2 Answers 2

up vote 6 down vote accepted

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 $t \in [0,1]$.

share|improve this answer
    
Thanks Dan! In the statement of the necessary and sufficient condition, did you mean there exists $T\in [0,1]$ such that $G(T,X)=\sup_{t\in [0,1]}G(t,X)$ a.s. ? I think there's another way of coming up with the condition, which also follows from pgassiat's suggestion above: if $\sup_{t\in [0,1]}\mathbb{E}[G(t,X)]$ is obtained at $t=T$ then we have $\mathbb{E}[\sup_{t\in [0,1]}G(t,X)−G[T,X]]=0$. This implies that $G(T,X)=\sup_{t\in [0,1]}G(t,X)$ a.s. I'm not sure if this condition is very useful for me in action because I'm not sure how to check it for a given function $G(t,x)$. –  martin Sep 25 '12 at 2:14
1  
Yes, you're right. I edited the answer to reflect this, and also to change "the dominated convergence theorem" to "Fatou's lemma", to be a bit more precise. But your argument is much better! Last night I missed this simple proof: If $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)] =\mathbb{E}[\sup_{t \in [0,1]}G(t,X)] < \infty$, then Fatou's lemma and continuity of $G$ in $t$ imply that $t \mapsto \mathbb{E}[G(t,X)]$ is upper-semicontinuous. Thus $\sup_{t \in [0,1]}\mathbb{E}[G(t,X)]$ is attained, and the rest is as you said. I knew Zorn's lemma felt like overkill for this problem... –  Dan Sep 25 '12 at 13:46
  1. The equality holds if $G(t,x) = f(t) + g(x)$.

  2. 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.

share|improve this answer
    
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 '12 at 14:30
1  
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 '12 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 '12 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 '12 at 20:48
    
That indeed is a very strong condition :) –  martin Sep 24 '12 at 0:50

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.