Physical meaning of Ito integrals

Solution 1:

I'm far from an expert here, but I've just been learning about this field so thought I might share! NB I'm not trying to be rigorous, just get the ideas across.

Let's first think of a 'standard' Riemann integral,

$$\int\limits_{a}^{b} f(x)\,\text{d}x$$

we interpret this as the limit of a sum of the areas of equally spaced boxes, each with width $\Delta x$ and height equal to the value of the function, then let $\Delta x\rightarrow 0$. So, as an approximation:

$$\int\limits_{a}^{b} f(x)\,\text{d}x \approx \sum\limits_{i=0}^{n-1} f(t_i)(x_{i+1}-x_i),$$

where the $x_i$s are a partition of the interval $[a,b]$ and $t_i$ is a value somewhere in the interval $[x_i,x_{i+1}]$.

Now let's take a stochastic Ito integral of a non-random process, $f(t)$:

$$\int\limits_{0}^{t} f(s)\,\text{d}B(s),$$

where $B(s)$ is a random process (normally Brownian motion). This isn't a PDF, it's actually a random variable itself. We can approximate this in the same way as a Riemann integral, i.e. a sum of the areas of boxes:

$$\int\limits_{0}^{T} f(s)\,\text{d}B(s) \approx \sum\limits_{k=0}^{n-1}f(t_i)(B(t_{i+1})-B(t_i)),$$

where the $t_i$s are a partition of the time interval $[0, T]$ into $n$ segments. Obviously, the larger $n$ is, the better the approximation. Looking at that expression, we can see that it's quite similar to the Riemann integral, but now the intervals themselves, whilst regularly spaced in $t$, are of random width. And a sum of random variables is itself a random variable.

It would be neat if there were a diagram showing these ideas (i.e. drawing the boxes) but I'm not sure how exactly to draw it.

If we now take the Ito integral of a random process, $X(t)$, then we have to replace $f(t_i)$ with a random variable $X(t_i)$ (I think there are several conditions you have to meet as well, I'm ignoring these here for the general picture), so the Ito integral becomes a sum of products of random variables, but again that's a random variable.

To answer your final question: changing the process $B$ means changing the process that generates our box widths, and hence changing the statistical properties of the integral.

Hope this helps, I await someone far more knowledgeable to come along and tell me it's all wrong!

Solution 2:

I think the problem here is notational (or abuse of notation). I have never convinced myself that what is called $dW_t$ is a proper k-form (1-form) differential.

Start with a Wiener process, $W_t$. If we demand $W_t=0$ at $t=0$, and at $t=1$ we have $W_t=W_1\sim N(0,1)$ (distributes a a normal RV with mean zero and unit variance), imagine first breaking up that first unti time in half. Then, for each half, $\Delta W_t \sim N(0,\frac{1}{2})$. Now just keep breaking it up, and the infinitessimal change becomes what we call $dW_t$. Now we have our 'proper' Wiener process, where $W_t$ is the sum of infinitely many little $dW_t$'s.

Suppose that we build a new function $f(t,W_t)$ as a function of the two variables. In the deterministic case we can say $df=\frac{\partial f(t,W_t)}{\partial t}dt+\frac{\partial f(t,W_t)}{\partial W_t}dW_t$, since we can ignore higher-order Taylor terms. However, in the stochastic world one of the second partials is, unfortunately, not $O(dt^2)$ but $O(t)$: $\frac{\partial^2 f(t,W_T)}{dW_t^2}$. (See Oskedahl or Shreve.)

Fortunately, in the "box calculus" (Google it) $dW_T^2=dt$, so to speak.

Thus, a Taylor expansion has what appears to be a second-order term that must be preserved.

All of stochastic calculus is simply taking that additional term into account. To see this search for 'quadratic variation' as well as looking at the above references. At least one pdf I found did a great job of explaining why the 'deterministic' part, or determinstic Taylor expansions, don't have this oddity: quite simply, for a deterministic function, rearrange the Taylor expansion and $f(x)-f(0)=f'(x)dx+f''(x)dx^2+...$ and we know we can throw away terms with $dx^2$ if we integrate the Taylor expansion (and, in the multivariate case, $f_{W_t W_t}$ does not get thrown away...).

Hope that helps.

Solution 3:

I'm confused about why Ito integrals depend on two stochastic processes (you integrate process A with respect to process B). How does the meaning of the integral change when process B changes?

Already (deterministic) Stieltjes integrals $\int\limits_0^tu(t)\mathrm dv(t)$ depend on two functions $u$ and $v$.