Timeline for Does the conditional expectation of a measurable process always have a progressive measurable version?
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| Jan 3, 2017 at 18:48 | comment | added | John Dawkins | @Dan : The question was posed in the context of Brownian motion, where optional and predictable coincide. In general, the predictable projection satisfies $E[X_T\mid\mathcal F_{T-}]$ on $\{T<\infty\}$ for predictable $T$, as you note, and one should use instead the optional projection, which is progressive. | |
| Jan 3, 2017 at 18:36 | comment | added | Dan | To be precise, John, the predictable process $Y$ satisfies only $Y_T=E[X_T | \mathcal{F}_T]$ on $\{T < \infty\}$ only for predictable stopping times $T$. An alternative is the optional projection, which is the essentially unique optional process $Z$ satisfying $Z_T=E[X_T | \mathcal{F}_T]$ on $\{T < \infty\}$ for every stopping time $T$, not just predictable ones. Of course, if you're interested only in deterministic times, you might as well work with the predictable projection. | |
| Jan 2, 2017 at 17:32 | comment | added | John Dawkins | See the second volume of Probabilités et Potentiel by Dellacherie and Meyer, or the book of Jacod and Shiryaev. Notice that in the Brownian motion case of the question at hand, the notions of predictable and optional coincide. The existence is by monotone classes; if $X_t(\omega) =D(t)F(\omega)$ with $D$ and $F$ bounded and measurable, then $Y_t(\omega):=D(t)H_t(\omega)$ does the trick, where $H$ is a right-continuous version of the martingle $\Bbb E[F|\mathcal F_t]$. | |
| Jan 2, 2017 at 17:26 | history | edited | John Dawkins | CC BY-SA 3.0 |
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| Jan 2, 2017 at 16:51 | comment | added | Nate Eldredge | How does one prove this, or where can one find a proof? | |
| Jan 2, 2017 at 16:10 | history | answered | John Dawkins | CC BY-SA 3.0 |