A SDE that "forgets" its input

Question

Let $W$ be the Wiener process.

Consider the coupled SDE $$ dX_t = - X_t dt + \sqrt{2} dW_t$$ $$ d\bar W_t = -\sqrt{2} X_t dt + dW_t$$

with initial conditions $\bar W_0 = 0$ and $X_0 \sim N(0, 1)$. The symbol $\bar W$ is arbitrary and I could have labeled the second process $dY_t = -\sqrt{2} X_t + dW_t$, but it will be apparent later why this symbol was chosen.

Under these conditions, $X$ is a stationary Ornstein Uhlenbeck process. What's interesting is that $\bar W$ seems to be a standard Wiener process, driven by both $W$ and $X$. Even more, $X_t$ seems to be independent of $\bar W_s$ for $s \le t$ so in some way, $\bar W$ remembers nothing about $X$, even though $X$ enters into its SDE.

Is there a way to (easily) prove this? The above facts can be justified if one can show uniqueness of the solution to the associated Fokker-Planck equation, but I haven't been able to prove that. Instead, I have some computer simulations.

Sample Paths of W
These are just sample paths of the Wiener process. Nothing too surprising here.

Sample Paths of X
These show that, indeed, $X$ is the stationary Ornstein Uhlenbeck process.

Sample Paths of $\bar W$
Here is a surprise. These paths look like they follow a Wiener process.

Here are plots of the sample path variances of $W$ and $\bar W$, where the horizontal axis is time, and the vertical axis is the variance of all sample paths. This is is further evidence that $\bar W$ is also a Wiener process.

However, $W$ is correlated with $X$, since it is the driving process behind $X$. Weirdly, $\bar W$ is uncorrelated with $X$.

Answer 1

It's easy to deduce the closed-form expression of the OU process $X_t$: $$X_t = X_0 e^{-t} + \sqrt{2}\int_0^t e^{s-t}dW_s \tag{1}$$ and we observe that $$d\bar{W}_t = dW_t -\sqrt{2}X_t dt =dW_t +\sqrt{2} \left(dX_t - \sqrt{2} dW_t \right) = \sqrt{2}dX_t-dW_t$$ then, we deduce the closed-form expression of $\bar{W}_t$ $$\implies \bar{W}_t =\sqrt{2}(X_t - X_0)-W_t = \sqrt{2}X_0(e^{-t}-1)+\int_0^t\left(2e^{s-t} - 1 \right)dW_s \tag{2}$$

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The stochastic integral of $(2)$ follow a normal distribution $\mathcal{N}\left( 0, \int_0^t(2e^{s-t}-1)^2 ds\right)$ and is independent to $X_0$, then, $\bar{W}_t$ follows also a normal distribution $$\bar{W}_t \sim \mathcal{N}\left( 0,2(e^{-t}-1)^2+ \int_0^t(2e^{s-t}-1)^2 ds\right) = \mathcal{N}\left( 0,t\right)$$ The process $(\bar{W}_t)_t$ is then a standard Wiener process.

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For the correlation between $X_t$ and $\bar{W}_s$ for $s \le t$, as $X_t$ and $\bar{W}_s$ follows both normal distributions, the dependence of $(X_t,\bar{W}_s)$ can be check by calculating the covariance $\mathbb{E}(X_t\bar{W}_s)$: $$\begin{align} \mathbb{E}(X_t\bar{W}_s) &= \mathbb{E} \left(\left(X_0 e^{-t} + \sqrt{2}\int_0^t e^{u-t}dW_u \right)\cdot\left(\sqrt{2}X_0(e^{-s}-1)+\int_0^s\left(2e^{u-s} - 1 \right)dW_u \right) \right)\\ &=\sqrt{2}e^{-t}(e^{-s}-1)\mathbb{E} \left(X_0^2 \right)+ \mathbb{E} \left(\left( \sqrt{2}\int_0^t e^{u-t}dW_u \right)\cdot\left(\int_0^s\left(2e^{u-s} - 1 \right)dW_u \right) \right)\\ &=\sqrt{2}e^{-t}(e^{-s}-1) + \sqrt{2} \mathbb{E} \left(\left( \int_0^\color{red}{s} e^{u-t}dW_u \right)\cdot\left(\int_0^s\left(2e^{u-s} - 1 \right)dW_u \right) \right)\\ &=\sqrt{2}e^{-t}(e^{-s}-1) + \sqrt{2} e^{-s - t} (-1 + e^s)\\ &=0 \end{align}$$

The covariance is null, we deduce that $(X_t,\bar{W}_s)$ is uncorrelated.

Answer 2

Some comments on NN2's answer.

A. To verify the closed form $(1)$ $ X_t = X_0 e^{-t} + \int_0^t \sqrt{2} e^{s-t}\, dW_s$ from NN2's answer:

Set $Y_t = e^{t} X_t$. Setting $y(x, t) = x e^t$, we have by the Ito Lemma $dY_t = \left[\frac{\partial y}{\partial t}|_{(X_t, t)} - X_t \frac{\partial y}{\partial x}|_{(X_t, t)} + \frac{\partial^2 y}{\partial x^2}|_{(X_t, t)}\right]dt + \sqrt{2} \frac{\partial y}{\partial t}|_{(X_t, t)} dW_t$. Hence $dY_t = \sqrt{2} e^t\, dW_t$. So $Y_t - Y_0 = e^t X_t - X_0 = \int_0^t \sqrt{2} e^s\, dW_s$.

B. To verify Wiener-ness of $\bar W$ we also need to talk about the increments $\bar W_{t + \Delta t} - \bar W_t$. But these are distributed as $N(0, \Delta t)$ appealing to the $t=0$ case and that $X$ is stationary.