I asked this problem on MSE some while ago, but it has stubbornly resisted any attempts at solving it. Maybe there is someone here who can either close the gap in one of the existing answers or has an independent answer.

Let the following stochastic system be given

$$dX_t=-X_t \, dt+dW_t,$$ $$dY_t=X_tY_t(1-Y_t)(dt+dV_t),$$

where $W_t$ and $V_t$ are independent Wiener processes, $X_0\sim\mathcal{N}(0,1/2)$ (the stationary measure), and $Y_0\in[0,1]$.

I would like to show that $\lim_{t\to\infty}Y_t\in\{0,1\}$ almost surely, or at least that the invariant probability measure for $Y$ is of the form $\alpha\delta_0+(1-\alpha)\delta_1$ for some $\alpha\in[0,1]$.


Given a realization of the Ornstein-Uhlenbeck process $X_t$, the SDE $$ d Y_t = Y_t (1- Y_t) X_t (dt + d V_t) \tag{1} $$ is scalar, nonautonomous, and nonlinear. Note that (1) has two fixed points at $0$ and $1$, which are asymptotically stable in the following sense.

Theorem. For almost all $Y_0 \in (0,1)$, we have $\lim_{t \to \infty} Y_t \in \{0, 1\}$ almost surely.

Intuitive Explanation

This theorem is plausible for the following reason: when $Y_0$ is close to zero, $Y_t$ behaves like the process $\tilde Y_t$ which satisfies the linear SDE: $$ d \tilde Y_t = X_t \tilde Y_t dt + X_t \tilde Y_t d V_t \;. $$ This linear SDE has the pathwise solution: $$ \tilde Y_t = Y_0 e^{ \int_0^t X_s dV_s + \int_0^t ( X_s - \frac{1}{2} X_s^2) ds } \;. $$ By the strong law of large numbers, $$ \lim_{t \to \infty} \frac{1}{t} \left( \int_0^t X_s dV_s + \int_0^t ( X_s - \frac{1}{2} X_s^2) ds\right) = - \frac{1}{4} \quad \text{a.s.} \tag{$\star$} $$ It follows that $\tilde Y_t \to 0$ as $t \to \infty$ almost surely. A similar argument holds if $Y_0$ is close to one.


A key tool in this proof is the function $f(y) = \log(y) - \log( 1-y)$ which bijectively maps the unit interval $(0,1)$ to $\mathbb{R}$ by mapping $0$ to $-\infty$, $1$ to $+\infty$, and $1/2$ to $0$.

By Itô's Lemma, a.s., for all $t \ge 0$, $$ d f(Y_t) = f'(Y_t) Y_t (1-Y_t) X_t (dt + d V_t) + \frac{1}{2} f''(Y_t) Y_t^2 (1-Y_t)^2 X_t^2 dt \;. $$ Since $$ f'(y) y (1-y) = 1 \quad \text{and} \quad \frac{1}{2} f''(y) y^2 (1-y)^2 = -\frac{1}{2} + y $$ we obtain the following SDE for $Z_t = f(Y_t)$ $$ d Z_t = \frac{1}{2} \tanh\left( \frac{Z_t}{2} \right) X_t^2 dt + X_t ( dt + dV_t ) \;. \tag{2} $$

Note from (2), and the fact that $-1 < \tanh(x) < 1$ for all $x \in \mathbb{R}$, $$ \int_0^T \left( - \frac{1}{2} X_t^2 dt + X_t dt + X_t d V_t\right) < Z_T - Z_0 < \int_0^T \left( \frac{1}{2} X_t^2 dt + X_t dt + X_t d V_t\right) $$ and hence, a.s., $$ -\frac{1}{4} < \lim_{T \to \infty} \frac{Z_T}{T} < \frac{1}{4} . \tag{3} $$ Note that (3) limits how fast $Z_T$ can diverge.

Let $z^{(0)} < z^{(1)}$ be two different initial conditions for (2), and let $Z^{(0)}_t$ and $Z^{(1)}_t$ be the corresponding paths emanating from these initial conditions which satisfy: \begin{align*} Z^{(0)}_T &= z^{(0)} + \int_0^T \left\{ \frac{1}{2} \tanh\left(\frac{Z_t^{(0)}}{2}\right) X_t^2 dt + X_t ( dt + dV_t ) \right\} \;, \\ Z^{(1)}_T &= z^{(1)} + \int_0^T \left\{ \frac{1}{2} \tanh\left(\frac{Z_t^{(1)}}{2}\right) X_t^2 dt + X_t ( dt + dV_t ) \right\} \;. \\ \end{align*} We stress that these paths are driven by the same realization of Brownian motion $V_t$ and Ornstein-Uhlenbeck process $X_t$. Hence, a.s., $$ Z^{(1)}_T - Z^{(0)}_T = z^{(1)} - z^{(0)} + \frac{1}{2} \int_0^T \left\{\tanh\left(\frac{Z_t^{(1)}}{2}\right) - \tanh\left(\frac{Z_t^{(0)}}{2}\right) \right\} X_t^2 dt \;. $$ Since $\tanh$ is increasing, the difference $Z^{(1)}_T - Z^{(0)}_T$ itself is increasing and \begin{align*} \lim_{T \to \infty} \frac{Z^{(1)}_T - Z^{(0)}_T}{T} &= \lim_{T \to \infty} \frac{1}{T} \int_0^T \frac{1}{2} \left\{ \tanh\left(\frac{Z_t^{(1)}}{2}\right) - \tanh\left(\frac{Z_t^{(0)}}{2}\right) \right\} X_t^2 dt \\ &\ge \frac{1}{2} \left\{ \tanh\left(\frac{z^{(1)}}{2}\right) - \tanh\left(\frac{z^{(0)}}{2}\right) \right\} \lim_{T \to \infty} \frac{1}{T} \int_0^T X_t^2 dt \\ &\ge \frac{1}{4} \left\{ \tanh\left(\frac{z^{(1)}}{2}\right) - \tanh\left(\frac{z^{(0)}}{2}\right) \right\} >0 \tag{4} \end{align*} In other words, the difference $Z^{(1)}_T - Z^{(0)}_T$ a.s. diverges as $T \to \infty$.

Now suppose that $$ \lim_{T \to \infty} \frac{Z^{(0)}_T}{T} = 0 \;. $$ This can happen if, e.g., the realization $Z^{(0)}_T$ asymptotes to a finite value or diverges at a sublinear rate. However, this can happen at most once, since for any $z^{(\star)} \ne z^{(0)}$ the previous result in (4) implies that $$ \begin{cases} \lim_{T \to \infty} \frac{Z^{(\star)}_T}{T} < 0 ~~\text{if $z^{(\star)} < z^{(0)}$ }\\ \lim_{T \to \infty} \frac{Z^{(\star)}_T}{T} > 0 ~~\text{if $z^{(\star)} > z^{(0)}$ } \end{cases} $$ where $Z^{(\star)}_T$ is the realization with initial condition $z^{(\star)}$. In other words, realizations corresponding to initial conditions: (i) to the left of $z^{(0)}$ diverge to $-\infty$; and (ii) to the right of $z^{(0)}$ diverge to $+\infty$. Hence, there can be at most one initial condition $z^{(0)}$ such that $\lim_{T \to \infty} \frac{Z^{(0)}_T}{T} = 0$ -- otherwise one gets the contradiction that some realizations diverge to $\pm \infty$ simultaneously.

In the original variables, this implies that: for all, but at most one initial condition, we have $\lim_{t \to \infty} Y_t \in \{0,1 \}$ almost surely -- as required.

  • $\begingroup$ Thanks! Could you spell out the strong approximation property for me? Does it come with any bounds on $Y_t-\tilde Y_t$? $\endgroup$
    – S.Surace
    Jan 17 '18 at 7:53
  • $\begingroup$ Thanks! This looks great! I will have a deeper look soon. From a brief look, it seems that one can even obtain exponential convergence rate of $Y_t$ from your argument by using the fact that $Z_t$ diverges to either extreme like $t$. $\endgroup$
    – S.Surace
    Jan 19 '18 at 23:06
  • $\begingroup$ No worries. Thanks also for giving a more appropriate title. I actually asked a generalization of this question earlier, if you are interested: mathoverflow.net/questions/287418/…. I would like to think that one could define a stochastic Lyapunov function as if the whole system had a stable fixpoint. But maybe one can instead suitably generalize the process $\tilde Z_t$. $\endgroup$
    – S.Surace
    Jan 19 '18 at 23:42
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    $\begingroup$ @S.Surace I revised the proof according to your feedback. $\endgroup$ Jan 27 '18 at 15:53
  • 1
    $\begingroup$ This is very elegant and also preserves the symmetry of the problem! I learned a lot from this! Thank you! $\endgroup$
    – S.Surace
    Jan 27 '18 at 18:11

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