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.
 A: To complement Nawaf's answer (1 2), 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). 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.
