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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.

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.

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:

(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.

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.

Hi

My question

Given some function $f:\mathbb R^n \rightarrow\mathbb R$ and a compact set $U\subset\mathbb R$, what properties for the process $S_t$, which is adapted to a filtration $\mathcal F$ are needed, so that the following holds:

$E\left[\sup\limits_{a\in U}E\left[f(a,(S_{t_i})_{i< n,t_i\ge t})\Bigr|\mathcal F_t\right]\right]=\sup\limits_{A,\;\sigma(A)\subseteq{\mathcal F_t},\;A(\omega)\in U}E\left[f(A,(S_{t_i})_{i< n,t_i\ge t})\right]$

The second supremum should be taken over all random variables $A$, that are $\mathcal F_t$-measurable and take values in $U$.

My Solution (so far)

If $S$ is Markov, the equality holds.

Idea of Proof: Using iterated conditioning on the r.h.s leads to

$\sup\limits_{A,\;\sigma(A)\subseteq{\mathcal F_t},\;A(\omega)\in U}E\left[E\left[f(A,(S_{t_i})_{i< n,t_i\ge t})\Bigr|\mathcal F_t\right]\right]$

Using the Markov property of S and the $\mathcal F_t$-measurability of $A$, it can be shown that a function $g(a,s)$ exists with

$g(A,S_t)=E\left[f(A,(S_{t_i})_{i< n,t_i\ge t})\Bigr|\mathcal F_t\right]$

leading to

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

It can easily be seen, that the l.h.s of equaility provides an upper bound:

$E\left[g(A,S_t)\right]\leq E\left[\sup\limits_{a\in U}\;g(a,S_t)\right]=E\left[\sup\limits_{a\in U}E\left[f(a,(S_{t_i})_{i< n,t_i\ge t})\Bigr|\mathcal F_t\right]\right]$

It remains to show the other direction:

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

But I am stuck here.

I would appreciate some help, like telling me, if I am heading in the right direction or other hints!

Thanks

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:

(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.