TITLE: Physical meaning of Ito integrals
QUESTION [17 upvotes]: I'm having trouble getting my head around the meaning of the stochastic Ito integral.  Specifically: the intuitive meaning of "Stochastic Integral" to me is a function that takes a time $t$ and returns a PDF for the integral of a stochastic process on $[0,t]$.  Is this indeed what Ito integrals model?  If so, 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?
Thanks for your help!

REPLY [7 votes]: 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.