Is $g(t)=\mathbb P[\inf_{0\le s\le t}X_s>0]$ differentiable with respect to $t$?

Consider the SDE

$$dX_t =b(t)dt + a(t)dW_t,\quad \forall t>0,$$

with $$X_0>0$$ has a density function $$\rho:\mathbb R_+\to\mathbb R_+$$. Consider the probability $$g(t):=\mathbb P[\inf_{0\le s\le t}X_s>0]$$. Can we prove $$g'$$ exists and

$$-\frac{C}{\sqrt{t}}\le g'(t)\le 0,\quad \forall t>0,$$

where $$C$$ depends only on $$b,a,\rho$$. If it helps, we may assume any "rational" condition satisfied by $$b,a,\rho$$.

• Do you mean a(t) and b(t) to be functions of t only ?
– mike
Jun 10, 2022 at 5:07
• @mike Yes. They are deterministic functions on $t$
– user478492
Jun 10, 2022 at 7:32

Yes, $$g(t)$$ is differentiable with respect to $$t$$ under the assumption that $$a(s) > 0$$ for $$s \in [0,t]$$.

Proof. Introduce the time-change $$\tau(s) := \int_0^s a(u)^{2} du$$ and set $$c(s) := b(\tau^{-1}(s)) a(\tau^{-1}(s))^{-2}$$ for $$s \in [0,t]$$. In terms of this time change, the process $$X_t$$ can be written equivalently (in law) as $$X_t \overset{d}{=} X_0 + \int_0^{\tau(t)} c(r) dr + W_{\tau(t)} \;.$$ By Girsanov's Theorem, \begin{align*} & g(t) = E\left[ 1(\inf_{0 \le s \le \tau(t)} \left( \int_0^s c(r) dr + W_s \right) \ge -X_0 ) \right] \\ &= E\left[ 1(\inf_{0 \le s \le \tau(t)} W_s \ge -X_0 ) e^{\left(\int_0^{\tau(t)} c(s) dW_s - \frac{1}{2} \int_0^{\tau(t)} c(s)^2 ds \right)} \right] \\ &= E\left[ 1(\sup_{0 \le s \le \tau(t)} (-W_s) \le X_0 ) e^{\left(\int_0^{\tau(t)} c(s) dW_s - \frac{1}{2} \int_0^{\tau(t)} c(s)^2 ds \right)} \right] \\ &= 1 - E\left[ 1(\sup_{0 \le s \le \tau(t)} W_s \ge X_0) e^{\left(-\int_0^{\tau(t)} c(s) dW_s - \frac{1}{2} \int_0^{\tau(t)} c(s)^2 ds \right)} \right] \end{align*} where in the last step we used that $$-\inf_{0 \le s \le \tau(t)} W_s = \sup_{0 \le s \le \tau(t)} (-W_s)$$ together with the reflection symmetry of Brownian motion, i.e., $$-W_s$$ is also a standard BM.

Let $$T = \inf\{ t \ge 0 \mid W_t = X_0 \}$$. By the strong Markov property of BM, $$$$g(t) = 1 - E\left[ 1(T \le \tau(t)) e^{\left(-\int_0^{T} c(s) dW_s - \frac{1}{2} \int_0^{T} c(s)^2 ds \right)} \right] \tag{1} \label{g_T}$$$$ Moreover, since the process $$\{ W_s \}_{s \in [0,T]}$$ is a Brownian bridge from $$(0,0)$$ to $$(T,X_0)$$, we have $$W_s = X_0 \frac{s}{T} + (T-s) \int_0^s \frac{1}{T-r} dB_r \;,$$ where $$B$$ is a standard BM. Hence, \begin{align*} \int_0^{T} c(s) dW_s = \frac{X_0}{T} \int_0^T c(s) ds + \int_0^T \left( c(s) - \frac{1}{T-s} \int_s^T c(r) dr \right) d B_s \end{align*} Inserting this into \eqref{g_T} yields, \begin{align*} &g(t) = \\ & \quad 1 - E\left[ 1(T \le \tau(t)) e^{- \int\limits_0^{T} \left(-\frac{X_0}{T} c(s)-\frac{1}{2} c(s)^2 + \frac{1}{2} (c_s - \frac{1}{T-s} \int\limits_s^T c_r dr)^2 \right) ds} \right] \end{align*} Since the distribution of the hitting time $$T$$ is inverse gamma with parameters $$1/2$$ and $$(1/2) X_0^2$$, \begin{aligned} &g(t) = \\ & \quad 1 - \int\limits_0^{\tau(t)} \frac{X_0 e^{- \frac{X_0^2}{2 z}- \int\limits_0^{z} \left(-\frac{X_0}{z} c(s)-\frac{1}{2} c(s)^2 + \frac{1}{2} (c_s - \frac{1}{z-s} \int\limits_s^z c_r dr)^2 \right) ds}}{\sqrt{2 \pi} z^{3/2}} dz \end{aligned} \tag{\star} \label{final_g} which is differentiable with respect to $$t$$ without any additional assumptions. $$\Box$$

This proof also answers this MO question: Probability that a drifted Gaussian process does not hit zero

• Deriviatives of the upper and lower bounds aren't upper and lower bounds on the derivative. I think it is not as messy as all that in this case because after a time change this process is a brownian motion plus a deterministic function.
– mike
Jun 12, 2022 at 5:37
• @NawafBou-Rabee Thank you very kindly for your reply and sorry for my super late feedback. I agree with mike that the bounds for $g$ do not yield the differentiability of $g$.
– user478492
Jun 15, 2022 at 13:39
• @NawafBou-Rabee Indeed, for the constant drift case, where we denote $g\equiv g_c$ to highlight the dependence, with $c\in \mathbb R$ being the constant drift. We see the $c\mapsto g_c$ is continuous (quite regular). From this observation I do not understand "the desired bound does not hold". I do appreciate an analysis for the general drift. Thank you very kindly for your consideration.
– user478492
Jun 15, 2022 at 14:06
• Thank you very kindly for the answer. This is very meaningful to me
– user478492
Aug 24, 2022 at 18:27