Decouple system of SDEs / handle scaling problem

Question

Consider

$\begin{split} \newcommand{\d}{\mathrm d} \d x &= -yx \d t + x^2 \d B\\ \d y &= -2 y^2 \d t + 2xy \d B. \end{split}$

This is a system of two SDEs driven by the same standard Brownian motion $B$. All differentials are to be interpreted in the Ito sense.

Now we are interested in getting a dynamical description of $y$ based solely on the knowledge about $x$ and the noise $B$, i.e. we're looking for the correct functions $f, g$ s.t.

$ \mathrm d y = f(x) \mathrm d t + g(x) \mathrm d B.$

The approach I tried was via the Euler-Maruyama discretization:

$\begin{align} x_{n+1} - x_{n} &= -h\cdot x_n \cdot y_n + \sqrt{h} \cdot x_n^2\cdot \xi_{n+1}\\ y_{n+1} - y_{n} &= -2h\cdot y_n^2 + 2\sqrt{h} \cdot x_n\cdot y_n\cdot \xi_{n+1} \end{align}$

Now the first equation can be solved for $y_n$:

$y_n = -\frac{x_{n+1}-x_n}{h\cdot x_n}+ \frac{x_n\cdot\xi_{n+1}}{\sqrt{h}}$

When plugging this into the r.h.s of the second line, we obtain

$y_{n+1}-y_n = -2\frac{(\Delta x_n)^2}{h\cdot x_n^2} + 2\cdot\Delta x_n \cdot\frac{\sqrt{h}\cdot \xi_{n+1}}{h}$.

Now at this point there is no problem at all: If we exchange the original iterative scheme for $y_n$ for the new one, at least numerically, we haven't done anything wrong. But we indeed have an expression for $y_{n+1}$ given only $x_n$ and the noise.

The problem arises in the limit $h\to 0$, where Deltas become differentials, $h$ becomes $\mathrm d t$ etc., so (purely informally),

$$dy = -\frac{2}{x^2}\cdot \frac{(dx)^2}{dt} + 2\frac{dx\cdot dB}{d t}.$$

Now I know that this makes no sense at all: We can't give $dx/dt$ a meaning and both terms on the r.h.s. diverge (but cancel each other out), which already happens in the equation for $y_n$.

But is there a way of making this rigorous and obtaining an expression for the dynamics of $y$ consisting only of terms in $x$ and $B$?

Note that the issue already arises in the equation

$y_n = -\frac{x_{n+1}-x_n}{h\cdot x_n}+ \frac{x_n\cdot\xi_{n+1}}{\sqrt{h}}.$

But terms on the r.h.s. are divergent for $h\to 0$ but their difference is well-behaved.

[Sorry for crossposting this from stackexchange but my post there didn't get any response what so ever.]

Answer 1

Alright, I found out about semimartingale decompositions and attempted the following. Can someone check for correctness? From the defining equation of $x$ we obtain $x = -\int xy\d t + \int x^2 \d B$ which yields the unique semimartingale decomposition ($x$ is continuous) into a local martingale $M = \int x^2 \d B$ and a finite variation process $A = -\int xy\d t$. That means, $x-M$ is a process of finite variation and their derivative exists a. e. and we can write $x-M = -\int x y \d t$, hence $$ y = -\frac{1}{x}\cdot \frac{\d(x-M)}{\d t}. $$ Plugging this into the SDE for $y$ yields $$ \d y = \frac{2}{x^2}\cdot \left(\frac{\mathrm d (x-M)}{\mathrm t}\right)^2 - 2\cdot \frac{\d(x-M)}{\d t}\cdot \d B $$ which is a valid SDE for $y$ given in terms of $x$ and $B$.

Answer 2

Here is a slightly different perspective on your decoupling question. To recap, you asked is it possible to decouple the SDEs: $$ \begin{aligned} &dx = - y x dt + x^2 dB \\ &dy = -2 y^2 dt + 2 x y dB \end{aligned} \tag{$\star$} $$ where $B$ is a one-dimensional standard Brownian motion. Specifically, you wish to obtain an SDE for $y$ of the form: $$ dy = f(x) dt + g(x) d B $$ In general, Ito's formula applied to $y(x)$ tells you if this decoupling is possible. Unfortunately, in your case, it appears this decoupling is not possible. Indeed, by Ito's formula: $$ d y = y' dx + \frac{1}{2} x^4 y'' dt = \left( - x y y'+ \frac{1}{2} x^4 y'' \right) dt + x^2 y' dB \tag{$\star \star$} $$ Equating the SDE coefficients for $y$ in ($\star$) with the SDE coefficients in ($\star \star$) we obtain: $$ x^2 y' = 2 x y \;, \quad - x y y' + \frac{1}{2} x^4 y'' = - 2 y^2 $$ These differential equations do not seem to have a solution. To see this, suppose that $x>0$ and eliminate $y'$ from the first equation to obtain that $y''(x)=0$, which implies $y$ is a linear function of $x$. However, a linear function of $x$ cannot satisfy the first differential equation $x^2 y' = 2 x y$.

In contrast, if you replace ($\star$) with the slightly modified SDEs $$ \begin{aligned} &dx = - y x dt + x^2 dB \\ &dy = - y^2 dt + 2 x y dB \end{aligned} $$ and repeat the steps above, you get that $y(x) = x^2$ and $$ d y = -x^4 dt + 2 x^3 d B $$ However, this last SDE is a bit superfluous, since we already have an explicit formula for $y$ given $x$.