Assume you have a stochastic base $(\Omega, \mathcal{F},P,\mathbb{F})$. A submartingale is usually defined as

an adapted process

for each $t$ $E(|X_t|)<\infty$

and $E(X_t|\mathcal{F}_s)\ge X_s$ a.s.

However, in a book I am reading I have come over another definition, which is almost the same, but it does not require integrability, but reguires that $E(X_t^+)< \infty$. (The author also adds càdlàg paths as a requirement, I am not sure if that is relevant for my question.)

This creates a problem for me. Because I am not sure then how to construct the conditional expectation $E(X_t|\mathcal{F}_s)$ using the Radon-Nikodym theorem. The most natural step is constructing $E(X_t^+|\mathcal{F}_s)$ and $E(X_t^-|\mathcal{F}_s)$ and defining $E(X_t|\mathcal{F}_s)=E(X_t^+|\mathcal{F}_s)-E(X_t^-|\mathcal{F}_s)$.

Because of the integrability, there is no problem using Radon-Nikodym to define $E(X_t^+|\mathcal{F}_s)$, the problem is defining $E(X_t^-|\mathcal{F}_s)$. The first step is defining a measure on $(\Omega,\mathcal{F_s},Q)$ such that $Q(A) = E(X^-_t\mathcal{X}_A), A \in \mathcal{F}_s$. In order to use the Radon-Nikodym theorem we need $\sigma$-finiteness. But I can not see that we have that in this case? We could define $E_n = \{X^-_t<n\}$, since $X_t^-$ are $\mathcal{F}_t$ measurable, these sets are in $\mathcal{F}_t$, so we do have that $(\Omega, \mathcal{F}_t,Q)$ is $\sigma$-finite. But the sets may not be $\mathcal{F}_s$-measurable, so how do we get $\sigma$-finiteness on the space $(\Omega,\mathcal{F}_s,Q)$?

Update: I get the same problem in this book: Continuous Martingales and Brownian motion. However I can't find how they construct the conditional expectation in this book either. There must be a simple answer for this since it is used in many books.

  • $\begingroup$ In which book did you find this definition? I'd like to take a look because this does seem a bit unusual. $\endgroup$ – Math1000 Sep 9 '16 at 18:26
  • $\begingroup$ @Math1000 amazon.com/Stochastic-Integration-Theory-Graduate-Mathematics/… page 29. $\endgroup$ – user119615 Sep 9 '16 at 18:43
  • $\begingroup$ The author addresses this when discussing the predictable prediction in chapter 3.1 by introducing another operator which is then used to define the generalized and extended conditional expectation. I'm not 100% sure if this solves your problem, but at least it's a start. $\endgroup$ – Olorun Sep 12 '16 at 8:11
  • $\begingroup$ @Olorun I looked at this thanks, but I see that he introduces this in Definition 3.25, but in Theorem 3.24 before this, he says something about ordinary conditional expectation that should be enough to solve what I have, and he says that it is a direct consequence of the Radon-Nikodym theorem. There he says that if $\eta$ is a non-neg r.v., and $\mathcal{F}$ is a sigma algebra, then $E(\eta|\mathcal{F})$ exists, but does this follow directly form the Radon-Nikodym theorem?[continued] $\endgroup$ – user119615 Sep 12 '16 at 9:22
  • $\begingroup$ If we define the measure $\mu=E(\eta\cdot I_F)$ on $(\Omega,\mathcal{F})$. we need that $\mu$ is $\sigma$-finite. But if $\eta$ is a r.v. on $(\Omega,\mathcal{A},P)$, where $\mathcal{F}\subset \mathcal{A}$, do we still have that $\mu$ is $\sigma$-finite on $(\Omega,\mathcal{F})$ so that we can use the Radon-Nikodym theorem? $\endgroup$ – user119615 Sep 12 '16 at 9:23

For any non-negative random variable $\xi$, if $E(\xi)<\infty$, the conditional expectation $E(\xi\mid \mathscr{G})$ is well defined, where $\mathscr{G}$ is a sub-$\sigma$-algebra. On the other hand, if $E(\xi)=\infty$, we consider, for $n\ge 1$, the random variable \begin{align*} \xi_n = \xi\, \mathbb{I}_{\xi < n}. \end{align*} Note that, $\{\xi_n\}_{n=1}^{\infty}$ is non-decreasing and \begin{align*} \lim_{n\rightarrow \infty}\xi_n = \xi, \end{align*} $P$-a.s. For each $n$, we define the set function $Q_n$ on the sub-$\sigma$-algebra $\mathscr{G}$ defined by \begin{align*} Q_n(A) = \int_A \xi_n dP, \end{align*} for $A \in \mathscr{G}$. Then $Q_n$ is a finite measure, and the Radon-Nikodym derivative $dQ_n/dP$ exists. That is, the conditional expectation $E(\xi_n \mid \mathscr{G})$ is well defined. Moreover, it can be shown that, for $n \ge 1$, \begin{align*} E(\xi_n \mid \mathscr{G}) \le E(\xi_{n+1} \mid \mathscr{G}), \end{align*} $P$-a.s. See Page 195 of this book. We then define \begin{align*} E(\xi\mid \mathscr{G}) = \lim_{n\rightarrow \infty}E(\xi_n \mid \mathscr{G}). \end{align*}

For any random variable $\eta$, the conditional expectation $E(\eta\mid \mathscr{G})$ is considered to be defined if \begin{align*} \min\left(E(\eta^+\mid \mathscr{G}), \, E(\eta^-\mid \mathscr{G}) \right) < \infty, \end{align*} $P$-a.s.

  • 1
    $\begingroup$ The problem is I do not really see how we have $\sigma$-finiteness on the sub-sigma algebra. If our probability space is $((0,1),\mathcal{B}((0,1)), \lambda|_{(0,1)})$. And assume our r.v. is $X(\omega)=1/\omega$. Let the sub-sigma algebra $\mathscr{G}$ be $\mathscr{G}=\{\emptyset, (0,1), (0,0.5],(0.5,1)\}$, then $((0,1),\mathscr{G},Q)$ is not a -sigma-finite measure space, where Q is defined as you have. $\endgroup$ – user119615 Sep 20 '16 at 21:36
  • $\begingroup$ See revision above. $\endgroup$ – Gordon Sep 21 '16 at 12:51
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    $\begingroup$ It is still the same problem, the sets $A_n$ are measurable in the original sigma-algebra of the space. But we do not know if they are measurable in the sub-sigma algebra, and hence the measure may not be sigma-finite there? Because you are not sure if $A_n \in \mathscr{G}$? $\endgroup$ – user119615 Sep 21 '16 at 13:09
  • $\begingroup$ Indeed, your observation is correct. $\endgroup$ – Gordon Sep 21 '16 at 16:16
  • $\begingroup$ See further revision above. $\endgroup$ – Gordon Sep 21 '16 at 18:09

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