4
$\begingroup$

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?

$\endgroup$
2
  • 1
    $\begingroup$ 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. $\endgroup$ Feb 5 at 21:37
  • $\begingroup$ @JoseAvilez Interesting! I'm certainly open to the non-Lipschitz case, if you would like to go into more detail. $\endgroup$ Feb 6 at 0:14

1 Answer 1

2
+100
$\begingroup$

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.

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

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .