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I am reading "Diffusion, Markov Processes and Martingales" by Roger/Williams, which state the following version of the monotone class theorem applied to bounded elementary processes $b\mathcal{E}$.

Lemma 6.5 from Roger/Williams

Later (section 27) they define the norm enter image description here

and define the stochastic integral through the ito isometry. In the last step the Ito isometry should be extended from the elementary processes to as much as possible in $L^2(M)$. They just say that from Lemma 6.5 it follows that $\overline{b\mathcal{E}} = L^2(M)$.


In my opinion at least $(i)$ and $(ii)$ do not hold for $d[M]_s = ds$ (Brownian Motion case), since $\int_0^\infty c^2 ds = \infty$ for $c \not= 0$.

Also how come Lemma 6.5 only gives us that the bounded predictable processes are in $\mathcal{H}$ and now we just take the whole $L^2(M)$ (no boundedness assumption anymore).

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First, it is important to notice that (not unusually) Roger and Williams have, at the point you are looking at, restricted their integrator $M$ to be a continuous $L^2$-bounded martingale. This means that at this stage, Brownian motion is not an allowed integrator, since it is not $L^2$-bounded. The punchline will be that by localisation, once we have the theory for $L^2$-bounded martingales, we will be able to get it for a wider class that includes Brownian motion.

This is important since it means that the class of martingales they consider satisfies $\mathbb{E}([M]_\infty) < \infty$. This will imply that constant functions are in $L^2(M)$, for example.

Now for the main body of your question. Let $U$ be the set of elementary processes in $L^2(M)$ (coinciding with the notation of Rogers/Williams). We want to show that $\overline{U} = L^2(M)$.

First we check that the space of bounded previsible process $b\mathcal{E}$ satisfies $b \mathcal{E} \subseteq \overline{U}$.

For this, we use their lemma 6.5 (the monotone class theorem). First, constant functions are in $\overline{U}$ since $c 1_{[0,n]} \to c$ in $L^2(M)$ as $n \to \infty$.

For the second condition, for arbitrary $\varepsilon$ we can pick an $N$ large enough that $|H_n(s,\omega) - H(s,\omega)| \leq \varepsilon$ for all $n \geq N$, $s \in (0,\infty)$ and $\omega \in \Omega$. Hence $$\|H_n - H\|_{L^2(M)}^2 \leq \mathbb{E} \bigg [ \int_0^\infty \varepsilon d [M]_s \bigg] = \varepsilon \mathbb{E}([M]_\infty).$$ That is, the uniform convergence implies convergence in $L^2(M)$ so $H \in \overline{U}$ also since $\overline{U}$ is closed.

Finally, it is an easy exercise to use the DCT to see that $\overline{U}$ satisfies condition $3$ in Lemma 6.5 also.

Therefore, $b \mathcal{E} \subseteq \overline{U}$. You are right that general elements of $L^2(M)$ need not be bounded so we aren't quite finished. The point is that a general element of $L^2(M)$ can be approximated by functions in $b \mathcal{E}$.

The crudest way one might try to do this is to take $\phi \in L^2(M)$ and just brutally cut it off at height $N$. That is, let $$\phi_N = \begin{cases} N \qquad |\phi| \geq N \\ \phi \qquad \text{otherwise} \end{cases}$$ Then $\phi_N$ is certainly bounded and it follows by the DCT that $\phi_N \to \phi$ in $L^2(M)$. Hence $L^2(M) \subseteq \overline{b \mathcal{E}} \subseteq \overline{U}$ as desired.

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  • $\begingroup$ Thank you very much. Actually they don't assume continuity yet, but I oversaw the important assumption of $L^2$ boundedness. They also alternate all the time between saying that we have $b\mathcal{E}$ and $L^2(M)$ as integrands which confuses me. Later they use $lb\mathcal{E}$ (loc, bounded elementary), instead of locally $L^2(M)$ as integrands when localizing which confuses me more. Do I understand it right, that for a $M^2_{loc}$ the integrands are "local $L^2(M)$" and for a cts local martingale (and thus $M^2_{loc}$) the integrands even include H s.t. $P[\int H_s d<M>_s < \infty] = 1$? $\endgroup$
    – Lochend
    Jul 3, 2019 at 16:04
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    $\begingroup$ Usually one builds up the integral in stages and so your spaces will get bigger as you go for a fixed integrator, and then you impose extra conditions to allow you to weaken the assumption on your integrator. When you construct for $L^2$-bounded martingales, the right space is $L^2(M)$. Then you construct for local martingales and for this you only need $\mathbb{P}( \forall t, \int_0^t H_s^2 d[M]_s < \infty) = 1$. Then for semimartingales, you also need the integrand to be integrable against the finite variation part so you usually want a locally bounded, predictable integrand. $\endgroup$ Jul 3, 2019 at 16:13

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