Itō integral
Partial function
context | ⟨⟨Ω,F,(Ft)t≥0,P⟩⟩ … filtered probability space |
context | Xt … Ft-adapted |
context | Ht … left-continuous, locally bounded, Ft-adapted |
let | hn=b−an |
let | ΔXn,i=Xa+ihn−Xa+(i−1)hn |
definition | todo: type |
definition | ∫baHt−dXt:=plimn→∞∑ni=1Ha+(i−1)hnΔXn,i |
todo:
∙ filtered probability space (Ft being a filter meas with passing time, more and more of the algebra is available)
∙ adapted process (Xt is measurable always w.r.t. the current Ft)
∙ limit in probability: (plimn→∞Xn=X)≡∀(ε>0). lim
where \Pr\left(X\geq x\right) \equiv F_X(x)=\int_{-\infty}^xP_X(y)\,{\mathrm d}y
and for a metric d in a Separable space inducing a new random variable d(X_n,X)
(As X_n,X isn't really in the domain of d, that's abuse of notation here)
\bullet the sum can be done over other partitions of the interval [a,b]
The following definition corresponds to a sequence of equidistant grids. The common more general definition allows for more genial sequences of finer and finer grids.
Discussion
Note that for arbitrary X_t, the expressions \Delta X_{n,i} = X_{a+ih_n}-X_{a+(i-1)h_n} are just a particular kind of stochastic processes. Importantly, it's one where the limit of n to infto goes to zero (as a stochastic process). This is akin to \Delta x_i going to zero in the Riemann integral, where x(t) is some (non-stochastic) continuous function. Indeed for X_t=x(t) a (deterministic) function, \Delta X_i=x(a+ih)-x(a+(i-1)h) are just the lengths of parts of a grid-partitioning.
Comparison with other stochastic integrals
Trivially, H_{a+(i-1)h} is
\dfrac{1}{2}(H_{a+(i-1)h}+H_{a+(i-{\bf{1}})h}).
If we shift the the index of the second term by one,
\dfrac{1}{2}(H_{a+(i-1)h}+H_{a+(i-{\bf{0}})h}),
we get another stochastic integral (Stratonowitsch-Integral).
In the case that we deal with smooth functions instead of stochastic processes, the discretization is irrelevant. Here, if H and X are correlated, this does make a difference.
Motivation
Math
For p>0, we have
\int_\epsilon^1(p\,t^{p-1})\,{\mathrm d}t=\int_\epsilon^1{\mathrm d}t^p = 1-\epsilon^p
and
\lim_{\epsilon\to 0}\int_\epsilon^1(p\,t^{p-1})\,{\mathrm d}t = 1
doesn't even depend on the value of p.
What's not so nice here is that the naive integral \int_0^1(p\,t^{p-1})\,{\mathrm d}t doesn't exist for p\in(0,1). This is because \lim_{t\to 0} of e.g. t^{\tfrac{1}{2}-1}=\tfrac{1}{\sqrt{t}} doesn't exist. We can also formulate that issue by saying that for p\in(0,1), the function t^p in \int_\epsilon^1{\mathrm d}t^p isn't differentiable at t=0.
In conclusion,
“\int_0^t{\mathrm d}\sqrt{s}=\sqrt{t}”
doesn't make sense, only
\lim_{\epsilon\to 0}\int_\epsilon^t{\mathrm d}\sqrt{s}=\sqrt{t}
Physics
The accumulation of the values of a function F along a smooth path x(t) is \int_{t_0}^{t_1} F(x(s))\, {\mathrm d}x(s), which is \int_{t_0}^{t_1} F(x(s))\, x'(s)\, {\mathrm d}s, where
x'(t)=\lim_{\Delta t\to 0}\frac{x(t+\Delta t)-x(t)}{\Delta t}.
E.g. for F being force, the integral along a path is the work.
The Itō integral is a limit of sums not over functions but stochastic processes, and there {\mathrm d}{\mathcal W}_t doesn't rely on differential calculus. For {\mathcal W}_t a Wiener process \sqrt{\left\langle\left(\int_0^t {\mathrm d}{\mathcal W}_s\right)^2\right\rangle}=\sqrt{t}.
In physics, the integral is a means of computing the accumulation of a function along a path in cases where x' isn't defined in a sensible way. As the density associated with a random variable obtained by a stochastic integral follows a diffusion equation, those situations arise everywhere in the microscopic world, and many models in quantum mechanics look just like that as well.
QM
Consider the drift diffusion equation
\frac{\partial}{\partial t}\psi=\mu\frac{\partial}{\partial x}\psi+\kappa^2\frac{\partial^2}{\partial x^2}\psi.
Dimensional analysis tells us that \mu is a characteristic length per time (drift velocity) while \kappa is a characteristic length per square root of time. This small factoid has curious consequences.
In statistical physics, \kappa^2=2D is the diffusion coefficient. What follows also applies to non relativistic quantum mechanics, except the diffusion coefficient is imaginary, \kappa^2=\frac{i\hbar}{2m}.
Example for how the situation arises
Given the value x(t) of a curve/stochastic process at time t, for any time interval \Delta t > 0, we can test for x(t+\Delta t) and the increment \Delta x\equiv x(t+\Delta t)-x(t) is probabilistic and dependents on \Delta t (and possibly on t or even on x(t)).
For example, in the case of a Brownian motion each new \Delta x takes values according to the distribution
P(\Delta x)=\dfrac{1}{\kappa\sqrt{\Delta t}\sqrt{2\pi}}\exp \left( -\dfrac{1}{2}\dfrac{(\Delta x)^2}{\kappa^2\,\Delta t} \right).
(I set \mu=0 and note that usually one uses a variable \sigma=\kappa\sqrt{\Delta t})
The Gauss curve distribution for \Delta x says that even for very small \Delta t, there is a non-vanishing change that x(t+\Delta t) is far away from x(t). For bigger \Delta t, the distribution flattens out and the chance for bigger net deviation grows.
Note that this weight also arises in the quantization of L(q,{\dot q})\propto {\dot q}^2: \frac{(\Delta x)^2}{\Delta t}=\left(\frac{\Delta x}{\Delta t}\right)^2\Delta t\approx \int_0^{\Delta t} \left(\frac{{\mathrm d}x}{{\mathrm d}t}\right)^2{\mathrm d}t
Now, for the above P, we have:
\langle \Delta x\rangle=0
\langle \left|\Delta x\right| \rangle=\sqrt{\tfrac{2}{\pi}}\,\kappa\,\sqrt{\Delta t}
\langle (\Delta x)^2\rangle=\kappa^2\,\Delta t
This says that the movement has no preferred direction, but for a finite waiting time \Delta t and if x(t) is some mean path, we expect x(t+\Delta t)=x(t)+\kappa\sqrt{\Delta t}, see picture. The intuition is that for a very small waiting time, you could possibly already have a big deviation and the longer the wait the farther you get away from the center - however this movement is sub-linear because with more time, more and more cancellation occur as well. The non-differentiability of the curve manifests itself here: While we know the overall deviation goes as \sqrt{\Delta t}, we can't make a good estimate for the instantaneous growth, because at \Delta t=0 the slope of the square root function isn’t finite! There is no x'(t)!
Itō process
An Itō process is a stochastic process X_t which is the sums of a Lebesgue and an Itō integral:
X_t = X_0 + \int_0^t \mu_s(X_s, s)\,\mathrm ds + \int_0^t \sigma_s(X_s, s) \,\mathrm dW_s
One writes
{\mathrm d}X_t = \mu_t(X_t, t) \, {\mathrm d}t + \sigma_t(X_t, t) \, {\mathrm d}W_t
If X_t isn't known, this is called a stochastic differential equation in X_t. Being an Itō process is the stochastic analog of being differentiable.
If \mu_t and B_t are time independent, we speak of Itō diffusion.
A geometric Brownian motion is characterized via \mu_t(X_s, s)=X_s\,\mu and \sigma_t(X_s, s)=X_s\,\sigma, i.e. both are “just” \propto X_s.
Itō lemma
{\mathrm d}f(t,X_t) = \left(\dfrac{\partial f}{\partial t} + \dfrac{\sigma_t^2}{2}\dfrac{\partial^2f}{\partial x^2}\right){\mathrm d}t + \dfrac{\partial f}{\partial x}\,{\mathrm d}X_t
As this really is an integral relation, it corresponds to a version of the fundamental theorem of calculus. If we know how to integrate against X_t, we can compute f(t,X_t) as such an integral (plus an ordinary integral).
See the references for a higher dimensional version of the lemma.
The following comment is foreplay to the next subsection: Note that for f(x,t)=\frac{1}{2}x^2 and \frac{\sigma_t^2}{2}=\kappa^2 we get
{\mathrm d}\left(\frac{m}{2}X_t^2\right) = m\,\kappa^2{\mathrm d}t + X_t\,m\,{\mathrm d}X_t
Manifestation of the commutation relation
Let p_{\Delta t}(t)=m\frac{x(t+{\Delta t})-x(t)}{{\Delta t}}. If the limit \lim_{\Delta t\to 0}p_{\Delta t}(t) exists, then for auxiliary \delta, we have \lim_{\Delta t\to}x(t+\delta^2{\Delta t})=x(t).
Hence, for ever smaller time grid size {\Delta t}, e.g. an expression like x(t+\delta_1^2{\Delta t})\,x(t+\delta_2^2{\Delta t})\,x(t+\delta_3^2{\Delta t}) converges to x(t)^3.
However, for x(t+\Delta t)\approx x(t)+\kappa{\sqrt{\Delta t}}. We find x(t+\delta^2{\Delta t})\,p_{\Delta t}(t)=\delta^4\,m\,\kappa^2+x(t)\,p_{\Delta t}(t).
The result says that two naively equivalent approximation schemes (e.g. \delta=0 vs. \delta=1) systematically differ by an additive diffusion term (e.g. m\kappa^2 here). I.e. the Itō integral is defined with most left of the grid cells as in the explicit Euler-method numerical approximation scheme. The implicit Euler-method simply corresponds to a different notion of integral here and would give a different result.
In quantum mechanics, the difference of the products
[x,p]_{\Delta t}:=x(t+\delta^2{\Delta t})\,p_{\Delta t}(t)-x(t)\,p_{\Delta t}(t)
equals m\,\kappa^2=m\frac{i\hbar}{2m}=\frac{i\hbar}{2}.
Fractional quantum mechanics
So we had
\frac{\partial}{\partial t} \psi = \kappa^2 \frac{\partial^2}{\partial x^2} \psi
(note the imbalance of dimensions, t vs. x^2) and in turn
P(\Delta x) \propto \exp\left(c\frac{(\Delta x)^2}{\Delta t}\right)
as next-step distribution, and then \langle |x|\rangle\propto t^{1/2} gives the non-smooth curve.
You may want to look at other next-step distributions, (effectively giving the theories with \langle |x|\rangle\propto t^{1/\alpha}) and this is all there is to „fractional quantum mechanics“. The imbalance t vs. x^\alpha for non-integer alphas forced fractional derivatives on you.
Reference
Wikipedia: Itō calculus, Itō lemma, Riemann–Stieltjes integral, Stratonovich integral, Path integral formulation, Fractional quantum mechanics
Stochastic Calculus Cheatsheet: http://furius.ca/cqfpub/doc/stocalc/stocalc.pdf