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