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Mar 8, 2019 at 21:10 history edited YCor
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Nov 10, 2017 at 21:00 comment added Tartrate Thank you for your comments! Indeed, my statement was inexact. I corrected the question and added the assumption that X is a martingale.
Nov 10, 2017 at 20:58 history edited Tartrate CC BY-SA 3.0
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Nov 10, 2017 at 16:26 comment added John Dawkins 1. Under mild conditions on $E$, the Kolmogorov extension theorem does grant the existence of a unique $\Bbb Q$ on $\mathcal F$ such that $d\Bbb Q|_{\mathcal F_n}/d\Bbb P|_{\mathcal F_n}=M_n$, $\Bbb P$-a.s., for each $n$. The probability measure $\Bbb Q$ need not be absolutely continuous with respect to $\Bbb P$ on $\mathcal F$, and indeed it may be that $\Bbb Q\perp\Bbb P$. 2. ``All martingale transforms of $X$ are martingales": You seem to be implicitly assuming that $X$ is a martingale.
Nov 9, 2017 at 17:16 history asked Tartrate CC BY-SA 3.0