Let $W$ be a Brownian motion and $u:\mathbb{R}_+ \to \mathbb{R}_+$ an upper barrier and $l:\mathbb{R}_+ \to \mathbb{R}_-$ a lower barrier.

Let $$\tau_u(\mu) = \inf\{ t \colon \mu t + W_t \geq u(t)\}$$ the first hitting time of the upper barrier and $$\tau_l(\mu) = \inf\{ t \colon \mu t + W_t \leq l(t)\}$$ the first hitting time of the lower barrier.

Define $p(\mu) = \mathbb{P}[\tau_u(\mu) \leq \tau_l(\mu)]$ as the probability that the Brownian motion with drift $\mu$ hits the upper barrier before the lower barrier.

I want to find conditions on $u,l$ such that $p$ is continuous (or differentiable).


I encourage yourself to verify below computations. Please tell me in comments if I am wrong, so I can edit the errors.

Let $T>0$, by Girsanov, $$p_T(\mu) = \mathbb{P}\left[\tau_u(\mu)\leq \min(T,\pi(\mu))\right]=\mathbb{E}\left[e^{\mu W_T-\frac{\mu^2}{2}T}\mathbb{1}_{\tau_u(0)\leq \min(T,\pi(0))}\right]$$

assume (prove) $p_T\to_{T\to\infty} p$


$p_T$ is differentiable with respect to $\mu$, and again by Girsanov and using independence of increments:

$$\frac{dp_T}{d\mu}=\mathbb{E}\left[\left(W_T-\mu T\right)e^{\mu W_T-\frac{\mu^2}{2}T}\mathbb{1}_{\tau_u(0)\leq \min(T,\pi(0))}\right]=\mathbb{E}\left[W_T\mathbb{1}_{\tau_u(\mu)\leq \min(T,\pi(\mu))}\right]=\mathbb{E}\left[W_{\min(T,\tau_u(\mu))}\mathbb{1}_{\tau_u(\mu)\leq \min(T,\pi(\mu))}\right]$$

and you need to find a condition to have convergence when $T\to\infty$

for example, $l,u$ bounded and $u$ continuous should give by dominated convergence theorem:

$$\frac{dp_T}{d\mu}\to_{T\to\infty}\mathbb{E}\left[W_{\tau_u(\mu)}\mathbb{1}_{\tau_u(\mu)\leq \pi(\mu)}\right]=\mathbb{E}\left[\left(u(\tau_u(\mu))-\mu\tau_u(\mu)\right)\mathbb{1}_{\tau_u(\mu)\leq \pi(\mu)}\right]$$

but these are strong assumptions

  • $\begingroup$ Thanks a lot for your quick answer, could you explain the application of Grisanov a little bit more. Here $\pi(\mu)=\tau_l(\mu)$ right? $\endgroup$ – Peter Jul 5 '16 at 9:50
  • $\begingroup$ quant.stackexchange.com/questions/26362/… might help you $\endgroup$ – MJ73550 Jul 6 '16 at 7:00
  • $\begingroup$ This is confusing to me and I think there might be a mistake in the above argument. Consider the case where $u$ is constant and equal to one. In this case the above argument yields that $$p'(\mu) = \mathbb{E} [u(\tau_u(\mu)) 1_{{\tau_u}\leq \pi(u)}] = \mathbb{E}[1_{{\tau_u}\leq \pi(u)}] = \mathbb{P}[{\tau_u}\leq \pi(u)] = p(\mu).$$ But this implies that $p(\mu) = p(0) e^{\mu}$ which is not in $[0,1]$ for large values of $\mu$. $\endgroup$ – Peter Jul 7 '16 at 9:56
  • $\begingroup$ thank you, you are right, last line was incorrect, I made the correction. $\endgroup$ – MJ73550 Jul 7 '16 at 13:59

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