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.



Is the solution to this SDE always positive?
To complement Nawaf's answer ([1](https://mathoverflow.net/questions/427153/is-the-solution-to-this-sde-always-positive#comment1098413_427153) [2](https://mathoverflow.net/questions/427153/is-the-solution-to-this-sde-always-positive#comment1098415_427153)), I thought I'd present a short argument how with some additional regularity assumptions, the answer is in fact yes.


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](http://www.statslab.cam.ac.uk//%7Ejpm205/teaching/lent2016/lecture_notes.pdf)). 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.


