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I am working with a continuous process $Y_t$ generating the filtration $(F_t)$ and have (for simplicity) two stopping times $\tau_1$ and $\tau_2$ such that $\tau_2 \leq \tau_1$ and $U:\Bbb R\rightarrow \Bbb R$ is a continuous function which is concave over $\Bbb R^+$ and convex otherwise. My aim is to show that:

$$\sup_{\tau_2\leq\tau_1}\Bbb E\left[ U \left( \sum_{i=1}^2Y_{\tau_i} \right) \mid F_0 \right] = \sup_{\tau_2}\Bbb E \left[ \sup_{\tau_1} \Bbb E \left[ U \left( \sum_{i=1}^2 Y_{\tau_i}\right)\mid F_{\tau_2}\right] \mid F_0\right] $$

I understand that part of it is simply the application of the "tower property" but not really sure how to go around the exchange between the supremum and the Expectation.

Any help would be greatly appreciated. Thanks in advance :)

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  • $\begingroup$ Usually the trick in such situations is to approach the supremum along a sequence, and use something like the (conditional) monotone convergence theorem. If the supremum is "inherently" over an uncountable set, and can't be approached by a sequence, then you should probably suspect the result might not be true, and look for a counterexample. $\endgroup$ Jul 17, 2016 at 16:03

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You need to look in terms of the essential supreumum, rather than the supremum (see for example Follmer and Schied's Stochastic Finance book) otherwise there's no guarantee that the inner supremum is a random variable (it might not be measurable). Given a nice lattice property (which is guaranteed automatically in your context), you know that this essential supremum can be approximated with a sequence.

Then the trick is to use a convergence theorem (eg dominated convergence) to show that everything agrees. For an example of this sort of argument, see Lemma 21.2.5 (for a control problem) in my recent book: (http://books.google.co.uk/books?id=qWb_CgAAQBAJ)

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