Timeline for Conditions for existence of a semi-martingale representing a system of probability measures
Current License: CC BY-SA 4.0
9 events
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| Jan 17, 2021 at 17:46 | comment | added | ABIM | Thanks, John; this is a good starting point for me. | |
| Jan 16, 2021 at 0:00 | comment | added | John Dawkins | It being Friday I'll confine myself to continuous $X$, and seek out reasonable necessary conditions. As a first approximation further suppose $X_t = M_t + V_t$, in which $M$ s a continuous martingale and $V$ is adapted, continuous, and of integrable variation. Then for a bounded $C^1$ function $f$ (with bounded derivative) one has $$ \Bbb E[f(X_t)]=\Bbb E[f(X_0)]+\Bbb E\left[\int_0^t f'(X_s)\,dV_s\right], $$ implying (I think) that $t\mapsto \Bbb E[f(X_t)]$ is of finite variation on $[0,1]$. How far a condition like this is from being sufficient is not clear to me. | |
| Jan 14, 2021 at 16:44 | comment | added | ABIM | Hehe, that's an very clever degenerat example. I guess part of my question (albehit maybe too implicit) is what conditions would such a system of measures require? | |
| Jan 14, 2021 at 13:46 | comment | added | Mateusz Kwaśnicki | Certainly not without extra assumptions: just pick a path $x(t)$ of unbounded variation, and set $\nu_t = \delta_{x(t)}$ to be a point-mass at $x(t)$. Then $X_t = x(t)$ almost surely, and hence $X_t$ is clearly not a semi-martingale. (I assume $\nu_t$ are Borel measures on $\mathbb R$ rather than on $\Omega$.) | |
| Jan 14, 2021 at 13:01 | comment | added | ABIM | @GeraldEdgar Good point, I reframed the question with your suggestion. However, I'm wondering; what is this (partial?) ordering? | |
| Jan 14, 2021 at 13:00 | history | edited | ABIM | CC BY-SA 4.0 |
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| Jan 14, 2021 at 11:57 | comment | added | Gerald Edgar | No. For any $\nu$ there are many $X$ such that $X \sim \nu$. Starting with a martingale $X_i$, we can change one or more $X_i$ to $Y_i$ with $Y_i \sim X_i$ and destroy the martingale property. A more sensible question may be: for which systems $(\nu_i)$ does there exist a martingale $(X_i)$ with $X_i \sim \nu_i\;\forall i$? There is a certain "order" relation $\preceq$ on probability measures, and the condition you want is $\nu_i \preceq \nu_j$ for $i \le j$. That is a different qustion, so I do not say more here. | |
| Jan 14, 2021 at 10:17 | history | edited | YCor | CC BY-SA 4.0 |
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| Jan 14, 2021 at 10:13 | history | asked | ABIM | CC BY-SA 4.0 |