Probability that a geometric Brownian motion with additional determinstic drift ever hits zero

Question

Let $W$ be a standard Brownian motion, and let $X_t$ be the solution to the following SDE

$$dX_t = (\mu X_t - Cke^{-kt}) \, dt + \sigma X_t \, dW_t$$

where $\mu, \sigma, C, k > 0$ are constants, with initial condition $X_0 = x_0 > C$ a.s.

Question: For fixed $T > 0$, can we estimate, or compute the probability

$$\mathbb P(\underset{0 \leq t \leq T}{\text{min}} X_t \leq 0)?$$

That is, the probability that $X_t$ ever hits zero before time $T$.

As suggested by Kurt G. in the comments, this SDE has an explicit solution, which may be helpful in estimating the given probability.

The explicit solution is given by

$$X_t = e^{\mu t + \sigma W_t-\sigma^2t/2} \left (x_0 - \int_0^t e^{-\mu s - \sigma W_s+\sigma^2s/2}\,Cke^{-ks} ds \, \right ) $$

Remark: I tried to apply Girsanov’s theorem to remove the drift, but the conditions for the density process $Z_T$ to be a martingale are not satisfied, due to the determinstic term blowing up when $X_t$ is small.

Answer 1

Clearly, the component $Y_t=e^{\mu t +\sigma W_t-\sigma^2 t/2}$ of the explicit solution never hits zero. This boils down the problem to the question if $$ Z_t:=\int_0^t\frac{Cke^{-ks}}{Y_s}\,ds $$ ever reaches $x_0\,$.

Case $C<0$. Then $Z_t\le 0$ and $Z_t$ can reach $x_0>0$ only when $Z_t=0$ for some $t$. This is however impossible because $$ |Z_t|=|C|\int_0^t\frac{ke^{-ks}}{Y_s}\,ds $$ is zero if and only if $Y_s=+\infty$ for all $s\in[0,t]$ but we know that this is not true.

Case $C=0$. In this case $X_t=x_0Y_t$ which never hits zero.

Case $C\ge 0$. In this case it is conceivable that $$ Y_tZ_t=x_0 $$ for some $t$.