0

In the Book of Schilling (Brownain motion), there is the following theorem

enter image description here

I'm quite surprised by this theorem. It looks to mean that all $L^2(\Omega ,\mathcal F_T,\mathbb P)$ is a Martingale (or local martingale). Is this really true ?

Walace
  • 710
  • 4
  • 12
  • Note that $Y$ isn't a process, it is a fixed random variable and so it doesn't make sense to ask if it is a (Local) Martingale. – Rhys Steele May 12 '20 at 09:55
  • @RhysSteele: But this hold for all $t\in [0,T]$, i.e. $$Y_t=y+\int_0^t X_sdB_s$$ for all $t\in [0,T]$ – Walace May 12 '20 at 10:20
  • $X$ depends on your choice of $T$ though. You have no reason to believe in general that the same process $X$ works for every $Y_t$. – Rhys Steele May 12 '20 at 11:26

1 Answers1

0

Given such a $Y$, consider the martingale $M_t:=E[Y|\mathscr F_t]$, $0\le t\le T$. Because of the assumption on the filtration, $(M_t)_{0\le t\le T}$ can be taken to be right continuous. And because of the martingale representation theorem for Brownian motion, you then have $M_t=M_0+\int_0^t X_s\,dB_s$, $\forall t\in[0,T]$, a.s., for a square-integrable process $(X_s)$ as you have indicated. Because $Y$ is $\mathscr F_T$ measurable, $$ y+\int_0^T X_s\,dB_s = M_T=E[Y|\mathscr F_T]=Y, $$ where $y:=M_0=E[Y]$.

John Dawkins
  • 25,733
  • This is a sketch of the proof for the Theorem mentioned in the question but that isn't an answer to the question actually posed there? – Rhys Steele May 12 '20 at 16:41
  • You're right. I forgot to preface it with "No, but..." – John Dawkins May 13 '20 at 00:15
  • I'm just suggesting that a more valuable answer to this question would be "no, because... but..." because it's clear OP has a source of confusion, access to a textbook that probably contains this proof and this proof does nothing to resolve that confusion. – Rhys Steele May 13 '20 at 09:23
  • I was trying to let the OP make the connection for himself. – John Dawkins May 13 '20 at 15:38