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I found it in Schilling, Partzsch's textbook "Brownian Motion":

only consider in $[0,T]$, they define the Dolean's measure $\mathbb P\times\mu$, and the corresponding $L^2$ norm on $L^2(\Omega\times [0,T],\mathcal F\times\mathcal B_T,\mathbb P\times \mu)$, where $\mu$ is the Lebesgue measure.

then define the elementary predictable set $\mathcal E_T$ and the corresponding stochastic integral.

then extend this definition to the closure of $\mathcal E_T$ under the $L^2$ norm above according to Ito isometry.

finally the author proved closure of $\mathcal E_T$ in $L^2(\Omega\times [0,T],\mathcal F\times\mathcal B_T,\mathbb P\times \mu)$ is $L^2(\Omega\times [0,T],\text{progressive sigma-algebra},\mathbb P\times \mu)$


my question is during the proof of the last statement, I don't find anything to do with the progressive property, the proof is valid if I replace it with "predictable sigma-algebra".

Do these two space coincide? I found that since

$\mathcal E_T\subset L^2(\Omega\times [0,T],\text{predictable},\mathbb P\times \mu)\subset L^2(\Omega\times [0,T],\text{progressive},\mathbb P\times \mu)$

take closure under $L^2$ norm, note that the second is a Hilbert space, we must have

$ L^2(\Omega\times [0,T],\text{predictable},\mathbb P\times \mu)=L^2(\Omega\times [0,T],\text{progressive},\mathbb P\times \mu)\tag{1}$

it seems strange, is there something wrong?

Does this also holds for any continuous square integrable martingale $M$?


GUESS: take

$A:=L^2(\Omega\times [0,T],\mathcal F\times\mathcal B_T,\mathbb P\times \mu_M)$

$B:= L^2(\Omega\times [0,T],\text{progressive},\mathbb P\times \mu_M)$

$C:=L^2(\Omega\times [0,T],\text{predictable},\mathbb P\times \mu_M)$

where $\mathbb P\times\mu_M$ is the corresponding Dolean's measure.

define $f_t=\langle M,M\rangle_t$

if $M$ is continuous square integrable martingale, and $f_t$ is absolutely continuous wrt the Lebesgue measure, then $A=B=C$ and the stochastic integral is defined for $C=B=A$ (motivated by the Shreve's textbook)

if $M$ is continuous square integrable martingale, then $C=B\subset A$ and the stochastic integral is defined for $C=B$

if $M$ is square integrable martingale, then $C\subset B\subset A$, and the stochastic integral is only defined for $C$


I found (1) i.e. (B=C) strange, because for any $f\in B$, (1) indicate that we can find $g\in C$ s.t. $$\mathbb E\left[\int_0^T|f_s-g_s|^2\,ds\right]=0$$

then $f=g$ a.s., that every progressive process is predictable? I think it is impossible.

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For Brownian motion it is true that $B=C$. This means that if $f\in B$ there exists $g\in C$ such that $f(\omega,t)=g(\omega,t)$ for $\Bbb R\times\mu$-a.e. $(\omega,t)\in\Omega\times[0,T]$. This does not mean that $t\mapsto f(\cdot,t)$ is a predictable process. The matter is discussed in some detail in Chapter 3 of Introduction to Stochastic Integration by Chung and Williams.

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  • $\begingroup$ thanks, the book you recommended is very useful! $\endgroup$ – Lookout May 9 '16 at 0:49

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