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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.

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    $\begingroup$ 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 $\endgroup$ Jul 23, 2022 at 11:33
  • $\begingroup$ Hm but that has a $-r_t \, dt$ drift term. Does removing that change anything? $\endgroup$
    – Nate River
    Jul 23, 2022 at 11:35
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    $\begingroup$ 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. $\endgroup$ Jul 23, 2022 at 11:39
  • $\begingroup$ Ah, this is a hole in my knowledge… what is a good reference to read about Feller’s boundary classification work? $\endgroup$
    – Nate River
    Jul 23, 2022 at 11:43
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    $\begingroup$ Name of @NawafBou-Rabee's reference: ben Naouara and Trabelsi - A short review on boundary behavior of linear diffusion processes. $\endgroup$
    – LSpice
    Jul 30, 2022 at 2:52

1 Answer 1

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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.

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  • $\begingroup$ Thanks for the answer! Very cool trick to consider $\log \sigma(X_t)$. $\endgroup$
    – Nate River
    Jul 30, 2022 at 7:09

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