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?
How does one interpret the meaning of a stochastic derivative?
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0For example one could speak of "a random smooth temperature distribution in the plane" (and one would then have to specify a particular probability distribution for that choice) and then ask what is the probability that the temperature gradient at some particular point is greater than 5 kelvin per meter. This question is about a random variable (the temperature gradient) that arises as the derivative of the random temperature distribution. – 2012-01-15
2 Answers
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.
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0The question was about "stochastic derivatives". What you described in a few paragraphs is Ito/Stratonovich *integration*, and ended up with the unhelpful "there is no stochastic derivative". – 2018-07-09
There is some ambiguity associated with the term "stochastic derivative". It can mean "Nelson derivative" and "Malliavin derivative", for example.
The definition I use is the following:
A real-valued stochastic process $\{ X(t) \}$ is said to be stochastically differentiable (differentiable with probability one, differentiable in the $L_p$ sense) at a point $t_0$ if there exists a random variable $\eta$ such that, for $t \to t_0$,
$ \frac{ X(t_0) - X(t) }{t_0 - t} \to \eta$
in probability (with probability one, or in $L_p$ respectively). The random variable $\eta$ is called the stochastic derivative of the process at the point $t_0$, and is denoted $X'(t_0)$.
In other words, even though $\{X(t)\}$ may not be differentiable, there is an $\eta$ from which we can make the approximation
$ X(t) = X(t_0) + (t_0 - t) \eta + r(t) $
where $r(t)$ is a remainder small enough in probability or in a $L_p$ sense.
For differentiability in probabiliy, we have
$r(t) = op(|t_0 - t|)$
which means that for any $\epsilon > 0$,
$\lim_{t \to t_0} P\left( \left|\frac{r(t)}{t_0 - t}\right| \ge \epsilon\right) = 0$