# The integral is the area under the curve. Is there a similar notion for stochastic integrals?

As discussed in the answers to this question, the integral is defined to be the (net signed) area under the curve. The definition in terms of Riemann sums is precisely designed to accomplish this.

Now the stochastic integral in Ito calculus is more formally defined and the result of the integration is another stochastic process. Is there a similar geometric Interpretation of a stochastic integral?

Are there special cases which are simpler to understand? For example what about Brownian Motion? Are there restrictions which allow pathwise integration?

Edit: I found a related question here which asks how to compute $\int W_sdW_s$.

• Not a geometric interpretation but the stochastic integral is your wealth as a result of following the integrand as a betting/trading strategy in a fair game characterized by the martingale integrator. Feb 20, 2017 at 8:27
• Here is a simpler question for you: if the underlying measure is not Lebesgue's, how would you interpret the value of integral geometrically? Feb 20, 2017 at 8:32
• @zhoraster I would not interpret the value of the integral geometrically at all. Perhaps there are standard cases wich are easy to understand? Feb 20, 2017 at 9:48
• You can still think of it as an area under a curve, but now this area is going to be a random variable, the first picture in the article en.wikipedia.org/wiki/Itô_calculus actually shows a Brownian motion and the integrated process. So for a given trajectory you still have the geometric intuition of the area under that trajectory, however I am not sure how far this gets you. Also when properties of the Ito integral such as it's mean and variance are considered it is often done by first considering something very much like a Riemann sum approximation Feb 20, 2017 at 10:03

For the geometric interpretation there is a nice result in the book by Karatzas and Shreve (chapter 3 2.29): Having a partition $\Pi$ with $0=t_0<t_1<\dots<t_n=T$ and defining $$S_\varepsilon(\Pi) = \sum_{i=0}^{n-1} \big[(1-\varepsilon)W(t_i)+\varepsilon W(t_{i+1})\big]\big(W(t_{i+1})-W(t_i)\big)$$ then $$\lim_{\|\Pi\|\to 0} S_\varepsilon(\Pi) = \frac 12 W^2(t)+(\varepsilon-1/2)t.$$ For $\varepsilon=0$ the right-handed side is the Itô integral $\int_0^T W(s)dW(s)$ and for $\varepsilon=1/2$ this is the Fisk-Stratonovich integral $\int_0^T W(s)\circ dW(s)$. So this is a nice interpretation for the stochastic integral as a geometric area. The only drawback is that the limit is in $L^2$-sense and we are dealing with stochastic processes (not just real functions) depending on a stochastic state $\omega$.
The general stochastic integral is first defined for simple (step-)functions exactly using the notion of area as in the Riemann case. For arbitrary functions approximations and limits are used. Thus again the notion of area is in the integral however somehow hided by the $L^2$-convergence and possible exclusions of null-sets where the integral can be defined completely arbitrary.
Let $X$ be a semimartingale (this covers in particular the case of a Brownian motion) and let $H$ be a predictable and locally bounded process. In this case, the integral process of $H$ with respect to $X$ is well-defined. For each $n$, let $t^n_0,\ldots,t^n_{K_n}$ be a partition of $[0,t]$ and assume that the mesh (i.e. size) of the partitions tends to zero, so that the partitions become arbitrarily fine. It then holds that $$\sum_{k=1}^{K_n} H_{t^n_{k-1}}(X_{t^n_k} - X_{t^n_{k-1}}) \to \int_0^t H_{s-}dX_s$$ where the convergence is in probability. See for example the book "Stochastic integration and differential equations" by P. Protter for more on this. The right-hand side is very similar to a Riemann sum, and represents an approximation to "the net signed area under the curve" where $H$ is the curve and $X$ is the weight assigned per time. A more precise analogy is to the Lebesgue integral of a function with respect to a measure. In any case, this yields a "geometric interpretation" of the integral very similar to the classical (Riemann / Lebesgue) case.
• It's not $X$ that is assigned as weight per time but $dX$. And this is what makes it impossible to have a geometric picture, in my opinion Nov 28, 2017 at 18:18