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I try to reduce my problem to the following question, which is stated rather sloppy (without possibly necessary additional conditions).

Let $Y_t$ be a real stochastic process for $t \in [0, T]$ and $\mathscr{F}_t$ some filtration. Does there exist a random variable $X$ such that $$Y_t = E [ X | \mathscr{F}_t ] \quad \text{ a.s.} $$ for all $t \in [0, T]$?

If this is true, can $X$ be explicitly given in terms of $Y_t$ in some way?

yada
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  • An interesting question because as far as I know you can always create a stochastic process by taking a conditional expectation with respect to a filtration, but proving the reverse (may be trivial) I have not seen. Therefore +1. – Chinny84 Jun 26 '15 at 07:26
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    This can be so only when $(Y_t)$ is a martingale -- then a pretty complete answer comes from the martingale convergence theorem. – Did Jun 26 '15 at 07:28
  • Thank you. I see why $Y_t$ is then necessarily a $\mathscr{F}_t$-martingale. But I don't see why it is necessary to apply the martingale convergence theorem, since $T < \infty$. I think, setting $X := Y_T$ is then fine enough. – yada Jun 26 '15 at 11:55

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