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Which inequalities can I use with stochastic integration?

For example, with the standard lebesgue integral we have $$\left|\int_\Omega f(x) dx\right| \le M |\Omega|$$ (where $M$ is the maximum of $|f|$ on $\Omega$ is exists)

Also, $$\left|\int_\Omega f(x) dx\right| \le \int_\Omega |f(x)| dx$$

Plus the ever-useful Holder and Jensen inequalities

Which of these inequalities are still valid if for $$\int_0^\infty H_s dM_s$$ where $M$ is a local martingale and $H \in L^2_{\text{loc}}(M)$? Or maybe there are other type of inequalities in this case?

Can you provide a good reference to this topic, which I imagine is pretty well known? Feel free to take as many additional restriction you want on $H$ and $M$ (also the extreme of integration can very well be taken finite). A special case of interest is $M_t = W_t$ the Brownian Motion

Ant
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1 Answers1

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Neither of them holds, in general, for stochastic integrals.

The trouble already starts if you consider measures which need not to be non-negative, i.e. signed measures. For a signed measure $\mu: (\Omega,\mathcal{A}) \to \mathbb{R}$ we cannot expect that the triangle inequality $$\left| \int f(x) \, \mu(dx) \right| \leq \int |f(x)| \, \mu(dx) \tag{1}$$ holds. To see this, just consider the case that $f$ is an elementary function, i.e. $f$ is of the form

$$f(x) = \sum_{j=1}^n c_j 1_{A_j}(x).$$

Then $(1)$ reads

$$\left| \sum_{j=1}^n c_j \mu(A_j) \right| \leq \sum_{j=1}^n |c_j| \mu(A_j).$$

Note the right-hand side does not even need to be non-negative (but the left-hand side is), so this doesn't make any sense. With the same reasoning, we find that

$$\left| \int f(x) \, \mu(dx) \right| \leq \|f\|_{L^{\infty}} \mu(\Omega) \tag{2}$$

does, in general, not hold true for signed measures. However, one can show that

$$|\mu|(A) := \sup \left\{ \sum_{n \in \mathbb{N}} |\mu(A_n)|; A_n \in \mathcal{A} \, \text{disjoint}, \bigcup_{n \in \mathbb{N}} \subseteq A \right\}$$

defines a non-negative measure, the so-called total variation norm, and that

$$\left| \int f \, d\mu \right| \leq \|f\|_{\infty} |\mu|(\Omega)$$

and

$$\left| \int f \, d\mu \right| \leq \int |f| \, d|\mu|.$$

These two are the natural generalizations of $(1)$ and $(2)$ for signed measures.

Since stochastic integrals are "randomized" signed measures, the situation becomes even more complicated. For example if $(M_t)_{t \geq 0}$ is a Brownian motion, then the stochastic integral

$$\int_0^t H_s \, dM_s$$

is not a pointwise integral and this means that we cannot simply use the above considerations for fixed $\omega$. Additionally, there is the trouble that the Brownian motion has infinite total variation, so, as far as I can see, there is no chance to get such inequalities for stochastic integrals with respect to Brownian motion. Very important inequalities for stochastic integrals (with respect to martingales) are e.g.

  • Doob's inequality
  • the Burkholder-Davis-Gundy inequality

but they don't provide any pointwise estimates.

The only exception I can think of are processes with bounded variation. In this case, we can define the stochastic integrals as a Riemann-Stieltjes integral and obtain similar estimates as for signed measures. This works in particular for processes with non-decreasing sample paths, e.g. subordinators.

saz
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    another great answer! You deserve more upvotes! Thanks :-) – Ant Feb 09 '16 at 20:06
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    @Ant Thanks a lot for your enthusiasm. I don't care too much about the upvotes, but it is good to know that there is someone who appreciates the answer. – saz Feb 09 '16 at 20:08
  • Very insightful. What do you mean when you say that this is not a pointwise integral? What number do we get when some $\omega$ occurs ? – Speaker Mar 04 '19 at 18:07
  • @Moja The stochastic integral is defined as an $L^2$-limit (and not as a pointwise limit), see this question for details. – saz Mar 04 '19 at 18:10
  • Right, some $L^2$ sequence of random variables. How do you think about the realisation of the intrgral for some $\omega$ in this case? Area under a graph and total change etc which are nice Riemann interpretations donst seem to be vaild. – Speaker Mar 04 '19 at 18:21
  • @Moja Since the integral is defined as an $L^2$-limit it doesn't make any sense to fix $\omega \in \Omega$. And yes, we cannot use standard inequalities and technqiues because we are not dealing with a pointwise integral. – saz Mar 04 '19 at 18:23
  • I ment how do we think of a relisation of the limit random varible $\int f dB$. This is indeed a random variable! But doesnt it have any meaningful interpretation? – Speaker Mar 04 '19 at 18:28
  • @Moja Yeah, sure, it is. What exactly do you mean by "how do we think of it"? The stochastic integral is an abstract object, there is not much we can do about that.... You might want to take a look at this question or this question. – saz Mar 04 '19 at 18:34
  • TBH I think one could think of it as a random height and width sum, and thus an area, BUT the L2 limit of that, whatever that means. Or dosnt random infinitley small widths make sense? – Speaker Mar 04 '19 at 19:00
  • Or do we need the $L^2$ limit since random infinitley small widths dosnt make sense? – Speaker Mar 04 '19 at 19:32
  • What about if we have a the random function integrated against the lebegue meaure. In other workds $\mathcal{I}(\omega)=\int_{0}^{T}f(\omega,t)dt$. Do we have estimnates of type (2) then? – Number4 Mar 15 '19 at 16:29
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    @user7534 As long as you are dealing with "classical" integrals with respect to Lebesgue measure, you have the standard inequalities, yes. Just apply the "deterministic" inequalities to $\int_0^T f(\omega,t) , dt$ with $\omega$ fixed. – saz Mar 15 '19 at 16:33
  • what about estimates for stochastic integrals that are not martingales? both BGD and Doob uses the martingale property. Do you know of estimates such as $L^{p}$ or others estimates that does not assume martingale property? – user123124 Jan 20 '20 at 06:38
  • @user1 What kind of stochastic integrals are you thinking of? For many stochastic integrals you can use a compensator to get a martingale which, then, allows you to use BGD/Doob/... – saz Jan 20 '20 at 07:20
  • @saz driven by fractional brownian for instance. Also can you elaborate on that idea? a reference would also be nice. – user123124 Jan 20 '20 at 07:22