• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $\mathbb F=(\mathcal F_t)_{t\ge 0}$ be a filtration on $(\Omega,\mathcal A)$
  • $B=(B_t)_{t\ge 0}$ be a Brownian motion on $(\Omega,\mathcal A,\operatorname P)$ with respect to $\mathbb F$
  • $H=(H_t)_{t\ge 0}$ be $\mathbb F$-adapted and $\mathbb F$-progressively measurable with $$\operatorname E\left[\int_0^\infty H_t^2\right]<\infty$$

Under this conditions, the Itô integral of $H$ with respect to $B$ $$\int_0^\infty H_s\;dB_s$$ is well-defined. Now, let

  • $b,\sigma:[0,\infty)\times\mathbb R\to\mathbb R$ be Borel measurable

and suppose we are considering a "process" $X=(X_t)_{t\ge 0}$ whose local behavior in time can be described by a differential equation $$dX_t=b\left(t,X_t\right)dt\;.\tag{1}$$ However, maybe $(1)$ is not accurate cause the process is disturbed by a random influence.

The Brownian motion at time $t$ is $\mathcal N_{0,\;t}$-distributed and the increments $B_t-B_s$ are $\mathcal N_{0,\;t-s}$-distributed. So, since the normal distribution occurs almost naturally in many practical problems, it makes sense to me, that we somehow want to integrate $X$ by $B$ and add this term to $(1)$.

Lebesgue–Stieltjes integration would make perfectly sense to me. The integration interval would be weighted by a normally distributed factor (the increments of the Brownian motion).

However, we all know that Lebesgue–Stieltjes integration with the Brownian motion as the integrator is impossible.

So, we use the Itô integral and model our problem by $$dX_t=b(t,X_t)dt+\sigma(t,X_t)dB_t\tag{2}\;.$$ While I know, that the Itô integral has beautiful properties like being a continuous $\mathbb F$-martingale, I ask myself in which terms $(2)$ is still appropriate for our problem.

Let $X$ be the solution of $(2)$. How can we motivate, that $X$ is really a solution for our problem? $X$ should be the solution of $(1)$ (which is the perfect model function without any disturbance) plus some normally distributed distortion whose intensity depends on $\sigma$.

Honestly, I don't see that $X$ has this property.


1 Answer 1


There is not one answer to this question. Your question is really one about modeling, rather than being strictly about mathematics, so the best answer depends on what you're trying to model.

Two answers that come to mind for me are as follows. One would be to instead consider applying small iid mean zero normally distributed perturbations every $\Delta t$ and sending $\Delta t \to 0$. A scaling argument shows that this will either diverge or have no effect unless the variance of the perturbations is a multiple of $\Delta t$ (maybe dependent on time, but only as a factor). This is the same scaling argument that takes place when you derive the heat equation from the simple symmetric random walk. If the variance doesn't have a time dependence, then you get a multiple of Brownian motion as $\Delta t \to 0$. If there is a time dependence but no $x$ dependence, then you get an Ito integral (or whatever other stochastic integral, it doesn't matter). Only when there is an $x$ dependence is there any real subtlety.

The other is that the property:

$$\lim_{\Delta t \to 0} \mathbb{E} \left ( \frac{X_{t+\Delta t}-X_t}{\Delta t} \mid X_t = x \right ) = b(t,x)$$

is desirable. This property essentially says "$b$ is the average drift of the process", which at least to me seems essential in order to think of $b$ as drift and $\sigma$ as diffusion. This requirement forces the noise to be a local martingale, and one can prove that no local martingale has bounded variation.

This is also the only sense in which the process is described by the unperturbed solution plus a noisy disturbance. It actually isn't that simple globally: for instance, you might have two stable equilibria and the noise can take you between them. In this case the perturbed process has qualitatively different behavior from the unperturbed process. But it is like that locally, in that the analogue of the derivative of the process is indeed $b$ at each point.

  • $\begingroup$ thank you very much for your efforts. Please consider the ordinary differential equation $$\frac{dN_t}{dt}=\alpha N_t\;\tag{1}$$ $(1)$ may be the model of a population size. Now, suppose $\alpha$ is disturbed by a random influence, i.e. $$\alpha=r+\text{noise}\;.$$ I've read, that "the easiest possible noise term would be $$\sigma\frac{d B_t}{dt}\;\tag{2}$$ But what does $\frac{dB_t}{dt}$ stand for? The Brownian motion is almost surely nowhere differentiable. So, how do I need to interpret $(2)$? $\endgroup$
    – 0xbadf00d
    Commented Jul 11, 2015 at 10:20
  • $\begingroup$ In this context, I also ask myself, in which sense $$\int_0^t\;X_s\;dB_s$$ is locally described by $X_t$ (since that is what one expects from an integral) Please note, that I know that $dX_t=f(t,X_t)\;dt+g(t,X_t)\;dB_t$ is a shorthand for $$X_t=X_0+\int_0^tf(s,X_s)\;ds+\int_0^tg(s,X_s)\;dB_s\;$$. $\endgroup$
    – 0xbadf00d
    Commented Jul 11, 2015 at 10:20
  • $\begingroup$ The first part of my answer gives an idea of what $dB_t$ stands for in terms of what you can approximate it with. In other words, our intuitive understanding of "$\frac{dB_t}{dt}$" is in terms of the limit process in the Ito integral. Provided the noise intensity doesn't depend on $x$, this is just the limit that you would have in the Lebesgue-Stieltjes context, but where the convergence is not pathwise. Instead about the best convergence notion you can get here is mean square convergence, which also implies convergence in probability. $\endgroup$
    – Ian
    Commented Jul 11, 2015 at 14:42
  • $\begingroup$ As for the second thing, that's actually a somewhat confusing thing about the notation: $\int_0^t X_s dB_s$ is related to $X_0$ (since $B_s$ is starting out at zero), but it is also related to values of $X_s$ for both positive and negative $s$ (since $B_s$ crosses the $s$ axis many times in a neighborhood of zero). $\endgroup$
    – Ian
    Commented Jul 11, 2015 at 14:43
  • $\begingroup$ When the noise intensity does depend on $x$, we find that we have an additional modeling concern, because the mean square limit of the "Riemann sums" depends on the choice of the point where we evaluate the noise intensity. In the Ito convention, we evaluate on the left. This has the nice effect of making the process a martingale, which is what the second part of my answer is about. In any other convention for stochastic integrals, the noise creates an additional drift (i.e. an additional contribution to the limit in my answer), which is undesirable for certain purposes. $\endgroup$
    – Ian
    Commented Jul 11, 2015 at 14:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .