How does one interpret the meaning of a stochastic derivative?

My understanding of derivatives is in the difference quotient limit sense... How does one interpret the meaning of a stochastic derivative? How can one possibly differentiate a random variable? What is its physical meaning, if it has any?

-

migrated from scicomp.stackexchange.comJan 15 '12 at 4:10

This question came from our site for scientists using computers to solve scientific problems.

What are you differentiating your random variable with respect to? – Henning Makholm Jan 15 '12 at 4:17
Let's say that it's a derivative with respect to time... – Paul Jan 15 '12 at 18:29
Then it could sound like what you're looking at is actually a continuous-time stochastic process, which is formally a random variable that takes values in a function space, interpreted as a function of time. One can certainly speak about differentiating the particular function that is the value of the process in one experiment, which will give you another function-valued random variable. (I'm not saying that is what you're looking at -- you're providing precious little context for your question, so it's impossible to be sure of anything). – Henning Makholm Jan 15 '12 at 18:37
And if it were a derivative with respect to space? How different would its meaning be? Can we still think of it as an instantaneous rate of change? If it is a truly random variable, it's instantaneous rate of change can take on an infinite number of possible values (i.e. it's not unique)... Thus, how can one use this information meaningfully? – Paul Jan 15 '12 at 18:50
I think you'll need to disclose more context if you want any truly satisfying explanation. The best I can say without more is that if you have given a random function $F$ (formally a map from the sample space into a space of functions), then they derivative of that function is another random function $G$, such that for each particular point in the sample space the $G$ at that point in the sample space is the derivative of the $F$ at that point in the sample space. This is not a priori meaningless. – Henning Makholm Jan 15 '12 at 19:46

As pointed out in the comments, there is some context missing in your question, so I'll just guess to fill it in: Let's talk about one-dimensional Brownian motion, which is a stochastic process: It is a family of random variables indexed by a continuous parameter, which is usually called "time" and is written as $B_t$.
Another point of view is that Brownian motion is a probability measure on a suitable set of functions. Since it can be shown that Brownian motion has continuous sample paths with probability one, we can think of it as probability measure on the set $C[0, T]$, the set of of continuous functions $$f: [0, T] \to \mathbb{R}$$ In addition, one can prove that Brownian motion has with probability one sample paths that are not differentiable and not even of bounded variation. This means it is not possible to define a Riemann-Stieltjes integral with respect to the sample paths. This is why one needs to develop a new concept of an integral with respect to Brownian motion, for example the Ito or the Stratonovich integral. It is possible to give precise meaning to the expressions like this one: $$X_T = \int_0^T f(t, x) d B_t$$ and prove (with appropriate assumptions for $f$) that there is a unique stochastic process $X_T$ satisfying this relation. These integral equations are usually abbreviated, with an abuse of notation, as $$d X_t = f \; d B_t$$ but one has to keep in mind that the symbol $d B_t$ is actually undefined. Only the integral with respect to Brownian motion is defined in the Ito- or the Stratonovich calculus. This means that there is no "stochastic derivative", and that the notion of "velocity" is undefined for Brownian motion. There just is no room for the interpretation of a "velocity" in physical terms in the theory.