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There might be different definitions of what a generalized Langevin equation is, but let us consider the following expression:

$$ \dot{x}_i = \frac{dx_i}{dt} = f_i(\mathbf{x}) + \sum\limits_{m=1}^{n} g_i^m(\mathbf{x}) \eta_m(t) $$

where $\mathbf{x}=\{x_i|1\le i\le k\}$ is the set of unknowns, the $f_{i}$ and $g_{i}$ are arbitrary functions and the $\eta_m$ are random functions of time, often referred to as "noise terms".

Now generally higher order equations can generally be transformed into first-order equation, so take an arbitrary first-order stochastic differential equation. Can we always write it in the form above?

In the first place, it seems that assuming a finite number of noise terms $n$ might be too restrictive. Take for instance noise that takes the form $\exp(\eta(t) + \mathbf{x})$. If we expand this as a series, each of the terms can be written in the form $g^m(\mathbf{x}) \xi_m(t)$ for some stochastic variable $\xi_m(t) \sim \eta(t)^m$, but the series would in principle be infinite.

Hence, it is not clear that in general a noise term that depends on $\mathbf{x}$ can be separated into a deterministic term that depends on $\mathbf{x}$ and a stochastic term that only depends on $t$.

Nevertheless, one could still imagine that there might be transformations that can put any arbitrary SDE into the above form. Are there any proofs on whether such transformations exist for general classes of stochastic differential equations?

EDIT: for simplicity, let us assume that the stochastic system under consideration is Markovian.

PianoEntropy
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  • I'm not sure I really understand the question. $\exp(\eta)$ is already one function; writing it in powers of $\eta$ does not change that. – Ian May 15 '17 at 12:13
  • What does raised index $m$ on $g_i^m$ mean? – mathreadler May 15 '17 at 12:34
  • @Ian You have a point there. I guess I was more thinking about functions like $\exp(\eta(t) + \mathbf{x})$, where dependence on the variable under consideration is hard to separate. If we expand this as a series, each of the terms can be written in the form $g^m(\mathbf{x}) \xi_m(t)$ for some stochastic variable $\xi_m(t)$, but the series would be infinite. ~Corrected in question – PianoEntropy May 15 '17 at 12:36
  • @mathreadler This is just notation (see Einstein convention), the sum index is over $m$, $g_i^m$ is the deterministic term corresponding to the noise term $\eta_m$. In physicists' notation, the sum can be left out in case of one raised and one lower index ($m$ in this case). – PianoEntropy May 15 '17 at 12:39
  • @Yiteng so the functions are tensors? Is there some place I can read more about this? – mathreadler May 15 '17 at 13:03
  • You could see them as tensors, yes. The point is that there are $n$ noise terms $\eta_m(t)$ with $1 \leq m \leq n$. The evolution of each of the $k$ variables $x_i$ is assumed to depend linearly on these noise terms through the functions $g_i^m(\mathbf{x})$. Hence there are $n \times k$ such functions. There is no different between superscript and subscript indices, but sometimes physicists leave out the summation - see https://en.wikipedia.org/wiki/Einstein_notation. – PianoEntropy May 15 '17 at 13:28
  • Can something like the differential limit of $x_{n+1}=x_n+\sin(\sqrt{\Delta t} Y_n)$, where $Y_n$ are iid N(0,1), be written in your form? I'm not sure but I doubt it. Here "differential limit" means the process obtained by fixing $t$, setting $n$ equal to the nearest integer to $t/\Delta t$, and then sending $\Delta t \to 0^+$. For instance Brownian motion is the differential limit of $x_{n+1}=x_n+\sqrt{\Delta t} Y_n$. – Ian May 15 '17 at 13:46
  • Sorry, I made a sloppy error, I meant $x_{n+1}=x_n+\sin(\sqrt{\Delta t} Y_n x_n)$. – Ian May 15 '17 at 14:53
  • I'm a bit confused about the $\Delta t$ notation, but I guess in my notation this would be something like $\dot{x} = \sin(\eta(t) x)$ for a single variable $x$. This does not seem to be reducible to the form I gave, since the sine gives an infinite series in $\eta(t)$, and if they are iid N(0,1), then each power $\eta(t)^k$ gives a new random variable (product of normal distributions, I'm not sure whether they are all different for different $k$). – PianoEntropy May 15 '17 at 15:31
  • See https://math.stackexchange.com/questions/101062/is-the-product-of-two-gaussian-random-variables-also-a-gaussian – PianoEntropy May 15 '17 at 15:37
  • @Yiteng: Ok thank you. I have heard about the summation convention, I just did not expect the functions to be tensor valued. – mathreadler May 17 '17 at 14:25

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