Does the conditional expectation of a measurable process always have a progressive measurable version? For example, X_t is a measurable process, but not progressive measurable, let Y_t=E[X_t|F_t], does Y_t has a progressive measurable version? where F_t is a filtration generated by a Brownian motion and satisfies the usual hypotheses. Thanks a lot!
2 Answers
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Even better (assuming for simplicity that $X$ is a bounded process) there is a (essentially unique) predictable process $Y$ with $Y_T=E[X_T\mid\mathcal F_T]$ a.s. on $\{T<\infty\}$, for each stopping time $T$. In particular, if $X$ is adapted, then $Y_t=X_t$ a.s. for each $t\ge 0$. The process $Y$ is called the predictable projection of $X$.
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$\begingroup$ How does one prove this, or where can one find a proof? $\endgroup$ Commented Jan 2, 2017 at 16:51
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$\begingroup$ 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 $\Bbb E[F|\mathcal F_t]$. $\endgroup$ Commented Jan 2, 2017 at 17:32
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$\begingroup$ 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. $\endgroup$– DanCommented Jan 3, 2017 at 18:36
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$\begingroup$ @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. $\endgroup$ Commented Jan 3, 2017 at 18:48
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with the result in the paper CONDITIONAL EXPECTATIONS ASSOCIATEDWITH STOCHASTIC PROCESSES, and the property that a measurable adapted process have progressively modification, we can get the conclusion.
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$\begingroup$ Which result in the paper are you referring to? $\endgroup$ Commented Mar 8, 2022 at 2:38
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$\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$– Community BotCommented Mar 8, 2022 at 2:38
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$\begingroup$ Sorry, it's lemma 2.2. @MichaelAlbanese $\endgroup$– AlbertCommented Mar 10, 2022 at 0:54