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 $b,a,\rho$.



Is $g(t)=\mathbb P[\inf_{0\le s\le t}X_s>0]$ differentiable with respect to $t$?
**Yes**, $g(t)$ is differentiable with respect to $t$ under the assumption that $a(s) > 0 $ for $s \in [0,t]$.




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*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, \begin{equation} 
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}
\end{equation}
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{equation} \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}\end{equation}
which is differentiable with respect to $t$ without any additional assumptions. $\Box$




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This proof also answers this MO question: [Probability that a drifted Gaussian process does not hit zero](https://mathoverflow.net/q/409414/64449)


