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I want to condition a stochastic process $X$ on the current value of another process $Y$ in a continuous-time setting. So I am looking for a process $Z$ such that for every fixed $t$, $$Z_t = E\big(X_t \big| \sigma(Y_t)\big) \quad \text{a.s.}\tag{1}$$ Uniqueness holds when $Z$ is required to be cadlag. Alternatively, uniqueness follows when $Z$ is required to be optional and $$Z_{\tau} 1_{\{\tau<\infty\}}= E\big(X_{\tau}1_{\{\tau<\infty\}} \big| \sigma(Y_t)\big) \quad \text{a.s.}\tag{2}$$ holds for every stopping time $\tau$. Existence follows by a monotone class argument for $X$ if I can prove that the process $$X(t,\omega)=1_{[a,b)}(t) 1_F(\omega)$$ admits $Z$ as in $(1)$ or $(2)$. Here $[a,b)$ is some interval and $F$ a measurable set. But I am not able to show this. Any hints or ideas?

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You need to be more precise to get a meaningful answer. What exactly is "stochastic process" here? I mean, what kind of joint measurability/continuity/... w.r.t. $\omega$ and $t$ do you assume and want to preserve in the conditional expectation you want to construct? I mean, without any such restrictions, you can just extend to $N_t$ arbitrarily and there is nothing to talk about. –  fedja Dec 15 '11 at 12:54
Thank you. I edited my question to make it more precise. –  pharms Dec 16 '11 at 17:16
Which sigma field is F measurable with respect to in your indicator function? Perhaps more importantly what filtrations for X and Y are in the background? –  BSteinhurst Dec 16 '11 at 17:46
Let me ask a simple test question to understand what you are looking for. Assume that $Y_t=tX_t$ and $X_t$ is some non-degenerate random process that is as nice as you want. What nice properties of $Z$ do you expect to get at and near $t=0$? If I understand things right, the conditional expectation conditioning on $Y_t$ is here $Z_t=X_t$ for $t\ne 0$ and $Z_0=EX_0$. Note that I can put that degeneration to many places simultaneously screwing things up quite a bit. –  fedja Dec 16 '11 at 18:41
You can do this by combining optional/predictable projection in forwards and reverse time, as along as you assume some properties on Y. Namely, that it satisfies a "strong-Markov" property in both the forwards and reverse time directions. Your question does not mention any assumptions on Y though. –  George Lowther Dec 16 '11 at 21:55
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