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

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.

• Hint : this SDE has an explicit solution. Please write it down and add it to the question. Then we go from there. Apr 25 at 12:28
• Thank you! I have added the explicit formula. Apr 25 at 13:15
• I think something needed to be fixed a bit. Did it. Please check. Apr 25 at 13:22

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 $$t$$.