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.

  • 1
    $\begingroup$ Hint : this SDE has an explicit solution. Please write it down and add it to the question. Then we go from there. $\endgroup$
    – Kurt G.
    Apr 25, 2022 at 12:28
  • $\begingroup$ Thank you! I have added the explicit formula. $\endgroup$
    – Nate River
    Apr 25, 2022 at 13:15
  • $\begingroup$ I think something needed to be fixed a bit. Did it. Please check. $\endgroup$
    – Kurt G.
    Apr 25, 2022 at 13:22

1 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 $t$.

  • $\begingroup$ Ah I am sorry, I think there is an error in the explicit solution - there should be a minus sign somewhere. I’ve made the correction, please check. $\endgroup$
    – Nate River
    Apr 25, 2022 at 14:13
  • $\begingroup$ You are right. I adapted the answer. $\endgroup$
    – Kurt G.
    Apr 25, 2022 at 14:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.