# Can an SDE hit a point where its covariance is not positive definite?

Question: Does there exist a matrix function $$\sigma:\mathbb{R}^n \to \mathbb{R}^{n\times m}$$ such that

1. $$\sigma$$ is locally Lipschitz
2. The "covariance" $$\Sigma =\sigma \sigma^T$$ is not positive definite (PD) at some point $$a\in\mathbb{R}^n$$ i.e. $$\Sigma(a)$$ fails to be PD.
3. The process $$X$$ solving the SDE $$dX_t = \sigma(X_t) dB_t$$, $$B\in BM(\mathbb{R}^m)$$ can non-trivially hit $$a$$? Meaning, if $$X_0=x\neq a$$ and $$T_a=\inf\{t>0:X_t=a\}$$ then $$\mathbb{P}(T_a<\infty|X_t=x)>0$$.

If I formulated this question correctly, this should preclude processes that start at $$X_0=a$$ and stay there forever, or can only asymptotically reach $$a$$.

Attempts: All of my examples I tried fail on part $$3$$. For example, in one-dimension, $$\sigma(x)=x$$ and $$a=0$$ fail because $$dX_t = X_t dB_t$$ is a GBM with zero drift hence $$X_t = X_0 e^{-0.5 t+B_t}$$ can never hit zero unless $$X_0=0$$ and then we're stuck there forever. We can arbitrarily approach zero however, but then $$T_a$$ is not finite. I tried shifting but then we get $$X_t = a+(X_t-a)e^{-0.5 t+B_t}$$ and the same problem occurs.

If there are no examples, then I think we must prove $$3$$ is false--if there is a point $$a$$ where the covariance fails to be PD, then we cannot non-trivially hit it. Or in other words, if the covariance is everywhere PD, then we should be able to non-trivially hit every point of its domain.

Context: I know (locally) Lipschitz coefficients is enough to ensure a strong solution to an SDE. This is Ito's famous result. A theorem by Phillips and Sarason show that if $$\Sigma(x)$$ is PD everywhere and $$C^1$$ then $$\sigma(x) = \Sigma(x)^{1/2}$$ is locally Lipschitz and further that if $$\Sigma$$ is only PSD but $$C^2$$ we also get locally Lipschitz for its square-root matrix. So these are sufficient conditions on the covariance to ensure the diffusion coefficient is locally Lipschitz, and they involve smoothness and PD/PSD.

So, basically I am sort of asking about the converse. Typing it out makes me realize, by the second part of Phillips and Sarason, that we can have a locally Lipschitz diffusion coefficient whose covariance is not everywhere PD but I have not been able to cook up an example obeying property $$3$$. The reason I am thinking about this property is because, sure, in my example above, $$dX_t = X_t dB_t$$ has a variance that fails to be positive at $$0$$, but I can never reach zero non-trivially anyway, so what do I care?

• How much do you care about local Lipschitzness of $\sigma$? I think if you allow non-Lipschitz $\sigma$, you can get strong uniqueness by Yamada-Watanabe and construct an example of such a process. Feb 5 at 21:37
• @JoseAvilez Interesting! I'm certainly open to the non-Lipschitz case, if you would like to go into more detail. Feb 6 at 0:14

I can think of an example that, as advertised in the comments, fails to be locally Lipschitz at $$a=0$$. Consider the following dynamics for a Cox-Ingersoll-Ross (CIR) process:

$$d v_t = \sqrt{\max (v_t, 0)} dB_t = \sqrt{v_t^+} dB_t$$

where $$v_0 > 0$$. Notice that here we have a square-root diffusion, where the diffusion coefficient is locally Lipschitz everywhere but zero. I claim this process hits zero, where the diffusion is not positive definite, in finite time almost surely.

First, note that by the Yamada-Watanabe existence theorem, a strong solution for $$v_t$$ exists. Now, for $$\epsilon, M$$ with $$\epsilon < v_0 < M$$, define the following stopping times: $$\tau_\epsilon = \inf \{t > 0 \, : \, v_t \leq \epsilon\}$$, $$\tau_M=\inf\{ t> 0\, : \, v_t \geq M \}$$, and $$\tau = \tau_\epsilon \land \tau_M$$. We wish to apply the optional stopping theorem with $$\tau$$; to that end, we show it is bounded in expectation:

\begin{align} M^2 \geq E(v_{t\land \tau}^2) &= E \left[ \left( v_0 + \int_0^{t \land \tau} \sqrt{v_s^+} dB_s \right)^2 \right] \\ &= v_0^2 + E\left( \int_0^{t \land \tau} v_s^2 ds \right) \\ &\geq v_0^2 + \epsilon E(t \land \tau) \\ &\to v_0^2 + \epsilon E(\tau) \end{align} as $$t \to \infty$$. In particular, $$E(\tau) < \infty$$ and, a fortiori, $$\tau < \infty$$ almost surely.

Being a bounded local martingale, $$v_{t\land \tau}$$ is in fact a true martingale, thus the optional stopping theorem applies and yields:

$$v_0 + 1 = (\epsilon + 1) P(\tau_\epsilon < \tau_M) + (M+1) P(\tau_M < \tau_\epsilon)$$

Notice that as $$\epsilon \to 0$$, $$\tau_\epsilon \to \tau_0$$, where $$\tau_0$$ is the first hitting time of zero. Sending $$\epsilon \to 0$$, we get: $$v_0 + 1 = P(\tau_0 < \tau_M) + (M+1)P(\tau_M < \tau_0)$$ Sending $$M \to \infty$$, we note $$\tau_M \to \infty$$ and that since the left-hand side is finite, we must have that $$P(\tau_M < \tau_0) \to 0$$. Thus, $$P(\tau_0 < \infty) = 1$$.

Finally, we note that at $$0$$, the volatility is zero, thus not positive definite, as desired.

• +1 very nice. The bounty is yours. Feb 10 at 21:59