# Is the solution to this SDE always positive?

Let $$W$$ be a standard one dimensional Brownian motion, and consider the SDE

$$dX_t = \sigma(X_t) \, dW_t, \, \, \, X_0 = 1 \, \text {a.s.}$$

Assume $$\sigma$$ is regular enough that the above SDE admits a globally defined solution.

Suppose $$|\sigma(x)| \to 0$$ as $$x \to 0$$.

Question: Is it true that almost surely, $$X_t > 0$$ for all $$t$$?

It seems like the Dambis-Dubins-Schwarz theorem may help, but I’m not sure how to turn it into a proof.

• Isn't the CIR model a counterexample? Recall that the CIR solution is a.s. positive only if the drift is sufficiently large; see, e.g., en.wikipedia.org/wiki/Cox–Ingersoll–Ross_model Jul 23, 2022 at 11:33
• Hm but that has a $-r_t \, dt$ drift term. Does removing that change anything? Jul 23, 2022 at 11:35
• To be sure, the drift coefficient is of the form $a (b-r_t)$, the noise coefficient is $\sigma \sqrt{r_t}$ and the positivity condition (just an instance of Feller's boundary classification) is $2 a b \ge \sigma^2$. So if $a=0$, this condition cannot be met. Jul 23, 2022 at 11:39
• Ah, this is a hole in my knowledge… what is a good reference to read about Feller’s boundary classification work? Jul 23, 2022 at 11:43
• Name of @NawafBou-Rabee's reference: ben Naouara and Trabelsi - A short review on boundary behavior of linear diffusion processes. Jul 30, 2022 at 2:52

Suppose that $$\sigma\in{C^{2}(\mathbb{R})}$$ with $$\sigma(0)=0$$. Then $$\sigma$$ is locally Lipschitz, which is sufficient to give us existence and pathwise uniqueness for the solution of the SDE: $$dX_{t}=\sigma(X_{t})dB_{t}, \hspace{10pt}X_{0}=x_{0}$$ for any choice of $$x_{0}\in{\mathbb{R}}$$ (see theorem 6.9 of Miller - Stochastic calculus). Since $$X_t=0$$ is a perfectly valid solution to the above SDE when $$X_0=0$$, it is actually THE solution. By the strong Markov property, if $$X_t$$ ever hits $$0$$, it is stuck there. Thus, to show that $$X_{t}>0$$ for all $$t\geq{0}$$ when $$X_{0}=1$$, it suffices to show that $$X_{t}$$ cannot hit $$0$$ in finite time.
Consider the process $$U_t=\log(\sigma(X_t))$$. Notice that $$X_t$$ hits a zero of $$\sigma$$ in finite time iff $$U_t$$ diverges to $$-\infty$$ in finite time. By Itô's formula, $$dU_t = \sigma'(X_t)dB_t + \frac{1}{2}\big(\sigma''(X_t)\sigma(X_t) - (\sigma'(X_t))^2\big)dt.$$ This is a diffusion with bounded coefficients when $$X_t$$ lies in a neighborhood of $$0$$ (as per our regularity assumptions on $$\sigma$$). Thus, while $$X_t$$ lies in a neighborhood of $$0$$, $$U_t$$ cannot run away to $$-\infty$$ in finite time and so $$X_{t}$$ cannot hit $$0$$ in finite time.
• Thanks for the answer! Very cool trick to consider $\log \sigma(X_t)$. Jul 30, 2022 at 7:09