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Given a one-parametric random function on a probability space $(\Omega,\mathcal F,\mathbb P)$:

$X:U\times\Omega\to \mathbb R \text{ and } (a,w)\mapsto X(a,w), \text{ with } \sigma(X(a))\subseteq \mathcal F\quad\quad\forall a\in U\subseteq\mathbb R, $

Then the following holds:

$E\left[\sup\limits_{a\in U}X(a)\right]=\sup\Bigr\lbrace E\left[X(A)\right]\Bigr|\sigma(A)\subseteq{\mathcal F},\;A(\omega)\in U\Bigr\rbrace$ and also $E\left[\sup\limits_{a\in U}X(a)\right]=\sup\Bigr\lbrace E\left[X(A)\right]\Bigr|\sigma(A)\subseteq{\bigcup\limits_{a\in U}\sigma(X(a))},\;A(\omega)\in U\Bigr\rbrace$

Proof:

The following holds trivially:

$E[X(A)]\le E[\sup_{a\in U} X(a)]$

it remains to show the other direction. This is done by applying zhoraster's answer:

Clearly, $M(\omega) = \sup_{a\in U} X(a,w)$ is $\mathcal F$-measurable.

Define for $\delta>0$

$\mathfrak A_\delta = \lbrace(a,\omega)\in U\times \Omega\mid X(a,w) >M(\omega)-\delta\rbrace$

This set is in $\mathcal B(\mathbb R)\otimes \mathcal F_t$, and it has a full projection onto $\Omega$. By a measurable selection theorem (which I think one can find in Bogachev Measure Theory) there is an $\mathcal F$-measurable $A_\delta$ such that $(A_\delta(\omega),\omega)\in\mathfrak A_\delta$ almost surely. Hence $E[X(A_\delta)]≥E[M(\omega)]−\delta$. We get the desired statement by letting $\delta\to 0$.

(One can also use Kuratowski--Ryll-Nardzewski theorem to prove the existence of a measurable $A_\delta$.)

After a very good answer of zhoraster, I realized, that my initial question was a mixup of several different things. Thats why I changed it community wiki and clearified the problem.

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Looks good, I don't see a mistake. –  zhoraster Dec 8 '10 at 10:38
    
@Johannes @zhoraster Sorry to be late at the party but I am afraid I do not understand what is going on. Let me stick to the (apparently) definitive version of Johannes' post, above this comment. One starts with $U\subseteq \mathbb{R}$ and with a family $(X_a)$ of real valued random variables $X_a:(\Omega,\mathcal{F})\to(\mathbb{R},\mathcal{B}(\mathbb{R}))$, indexed by $a\in U$, and one introduces $M=\sup_{a\in U}X_a$. I wonder why $M$ should be measurable in the first place, clearly or not. :-) .../... –  Did Feb 15 '11 at 13:27
    
.../... Assume for instance that $\Omega\subseteq U$ and fix $V\subset\Omega$. Define the family $(X_a)$ by $X_a(\omega)=1$ if $a\in V$ and $\omega=a$, and $X_a(\omega)=0$ otherwise. Then $M(\omega)=1$ if $\omega\in V$ and $M(\omega)=0$ otherwise, hence $M=\mathbf{1}_V$ and, since $V$ can be any subset of $\Omega$, it may well happen that $M:(\Omega,\mathcal{F})\to(\mathbb{R},\mathcal{B}(\mathbb{R}))$ is not measurable. If you could explain what I miss, I would be grateful. –  Did Feb 15 '11 at 13:27
    
@Johannes: Not interested in answering the question in my comment? Let me recall it: why should $M$ be measurable? –  Did Jun 19 '11 at 10:40
    
@zhoraster: Not interested in answering the question in my comment? Let me recall it: in the first line of your answer, why should $M$ be measurable? –  Did Jun 19 '11 at 10:42
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2 Answers

up vote 2 down vote accepted

Clearly, $M(\omega) = \sup_{a\in U} g(a,S_t)$ is $\mathcal F_t$-measurable.

Define for $\delta>0$ $$ \mathfrak A_\delta = \{(a,\omega)\in U\times \Omega\mid g(a,\omega)>M(\omega)-\delta\} $$ This set is in $\mathcal B(\mathbb R)\otimes \mathcal F_t$, and it has a full projection onto $\Omega$. By a measurable selection theorem (which I think one can find in Bogachev Measure Theory) there is an $\mathcal F_t$-measurable $A_\delta$ such that $(A_\delta(\omega),\omega)\in \mathfrak A_\delta$ almost surely. Hence $E[g(A_\delta,S_t)]\ge E[M(\omega)]-\delta$. We get the desired statement by letting $\delta\to 0$.

(One can also use Kuratowski--Ryll-Nardzewski theorem to prove the existence of a measurable $A_\delta$.)

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This looks like it would also work for non-markovian processes $S$ and more general $f$, with arbitrary dependence on the path of $S$. Do you agree that your argument can also be used to show: $E\left[\sup\limits_{a\in U}E\left[f(a,\lbrace S\rbrace)\Bigr|\mathcal F_t\right]\right]=\sup\limits_{A,\;\sigma(A)\subseteq{\mathcal F_t},\;A(\omega)\in U}E\left[f(A,\lbrace S\rbrace)\right]$ –  Johannes Dec 4 '10 at 9:58
    
I would say, that the Markov property of $S$ would further imply that it sufficies to take the supremum over $\left\lbrace A|\sigma(A)\subseteq\sigma(S_t)\right\rbrace$ –  Johannes Dec 4 '10 at 10:01
    
Yes, you are right. Don't see the reason why it shouldn't work for a non-Markov process (though I didn't check thoroughly). –  zhoraster Dec 4 '10 at 14:45
    
I changed the original question and included your answer. Do you agree with the assertion? –  Johannes Dec 4 '10 at 17:21
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I think it is quite easy to show

$E\left[\sup\limits_{a\in U}\;g(a,S_t)\right]\leq \sup\limits_{A,\;\sigma(A)\subseteq{\mathcal F_t},\;A(\omega)\in U} E\left[g(A,S_t)\right]$

Proof:

$S$ is $\mathcal F$-adapted, which implies: $\sigma(S_t)\subseteq\mathcal F_t$

From that it follows,

$\sup\limits_{A,\;\sigma(A)\subseteq{\mathcal F_t},\;A(\omega)\in U} E\left[g(A,S_t)\right]\geq \sup\limits_{A,\;\sigma(A)\subseteq\sigma(S_t),\;A(\omega)\in U} E\left[g(A,S_t)\right]$

the supremum is taken over all $\sigma(S_t)$-measurable $A$.

We know that, for any $S_t$ there exists a increasing sequence $a^{S_t}_n \in U$ with

$\lim\limits_{n\rightarrow\infty}g(a^{S_t}_n,S_t)=\sup\limits_{a}\;g(a,S_t)$

For every $n$, $a^{S_t}_n$ is clearly a random variable measurable on $\sigma(S_t)$. From that it follows:

$\sup\limits_{A,\;\sigma(A)\subseteq\sigma(S_t),\;A(\omega)\in U} E\left[g(A,S_t)\right]\geq E\left[g(a^{S_t}_n,S_t)\right] \quad\forall n$

i.e.

$ \sup\limits_{A,\;\sigma(A)\subseteq{\mathcal F_t},\;A(\omega)\in U} E\left[g(A,S_t)\right]\geq E\left[\sup\limits_{a\in U}\;g(a,S_t)\right]$

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"... clearly a random variable measurable ..." For me, it's not clear at all. Moreover, one needs $g(a_n,S_t)$, not $a_n$, to be increasing. –  zhoraster Dec 3 '10 at 5:50
    
Thanks for pointing that out, but do you know how it would work? –  Johannes Dec 3 '10 at 11:23
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