I posted this question before on MSE

I need to use it in a small step in the middle of a simulation and I think I'm not getting correct results to this probabilities and so for my all subsequent simulation. Could someone please check my results and my proof below?

I would like to obtain the law of the first hitting time of a geometric Brownian motion.

More precisely, to compute $\mathbb P [\tau_B \leq T]$ where $$\tau_B := \inf\{0 \leq t \leq T: X_t \leq B \}$$

and $X = (X_t)_{t \geq 0}$ is a geometric brownian motion of drift $\mu$ and volatility $\sigma$ with initial condition $X_0> B$.

It's known that it can be obtained in terms of the law of the minimum/maximum of a drifted Brownian motion.

Indeed, since $X_t = X_0 \exp ((\mu -\frac{\sigma^2}{2})t + \sigma W_t )$ and $\{\tau_B\leq T\}= \{X_T^*\leq B\}$ where $X^* = (X_t^*)_{t \geq 0}$ defined by $X^*_t =\min_{t \in [0,T]} X_t$, one can write

\begin{align} \mathbb P [\tau_B \leq T]&= \mathbb P [X_T^*\leq B]\\ &= \mathbb P \left[\xi_T^{(\lambda)}\leq \ln \left(\frac{B}{X_0}\right)\right] \end{align}

where $\xi_t^{(\lambda)}=\min_{s \in [0,t] }\{\bar{\mu} t+\sigma W_t \}= \min_{s \in [0,t] }\{\lambda t + W_t \}$ (last equality comes by scaling) with $\bar{\mu}=(\mu -\frac{\sigma^2}{2}) $ and $\lambda=\frac{\bar{\mu}}{\sigma^2}$.

Then one can just use the formula (see here for reference, the proof is a straightforward application of Girsanov theorem)

$$\mathbb P \left[\Lambda_T^{(\eta)}\geq x\right]=e^{2\eta x}\mathcal N\left(\frac{-x-\eta t}{\sqrt t}\right)+\mathcal N\left(\frac{-x+\eta t}{\sqrt t}\right)$$

for the law of the maximum of the drifted Brownian motion $\Lambda_t^{(\eta)} := \max_{s \in [0,t] }\{\eta t+W_t \}$. Therefore since by symmetry $\Lambda_t^{(\eta)}\overset{\mathcal L} {=} -\xi_t^{(-\eta)}$ where $\xi_t^{(\eta)} :=\min_{s \in [0,t] }\{\eta t+ W_t \}$ is the minimum of the drifted brownian motion, one can can obtain the following formula

\begin{align} \mathbb P [\tau_B \leq T]&= \mathbb P [X_T^*\leq B]\\ &= \mathbb P \left[\xi_T^{(\lambda)}\leq \ln \left(\frac{B}{X_0}\right) \right] \\ &= \mathbb P \left[-\Lambda_T^{(-\lambda)}\leq \ln \left(\frac{B}{X_0}\right) \right] \\ &=e^{2\lambda \ln \left(\frac{B}{X_0}\right)}\mathcal N\left(\frac{ \ln \left(\frac{B}{X_0}\right)+\sigma^2 \lambda T}{\sigma \sqrt T}\right)+\mathcal N\left(\frac{ \ln \left(\frac{B}{X_0}\right)-\sigma^2 \lambda T}{\sigma \sqrt T}\right)\\ &=\left(\frac{B}{X_0}\right)^{2\lambda}\mathcal N\left(\frac{ \ln \left(\frac{B}{X_0}\right)+\bar{\mu} T}{\sigma \sqrt T}\right)+\mathcal N\left(\frac{ \ln \left(\frac{B}{X_0}\right)-\bar{\mu} T}{\sigma \sqrt T}\right) \end{align}

Please let me know if you notice any mistake.

I'm having values bigger than one and an unexpected shape for the function $f(B) := \mathbb P [\tau_B \leq T]$ as you can see below in the graphic (Axes $y =f(B)$ and $x =B$; here X_0= 68).

I forced it to be 1 for barrier values equal or higher than the initial value. Nonetheless I think there is still a problem with the formula not only it should be automatically 1 for $B \leq X_0$. The probabilities I have calculating it by Monte Carlo are higher than the values obtained by this formula as you can see in the second graphic bellow.

Many thanks

enter image description here enter image description here

  • $\begingroup$ An interesting reference using similar arguments seems to be "How Likely is it to Hit a Barrier? Thoretical and Emperical Estimates" by Lujing Su and Marc Oliver Rieger. $\endgroup$ – philippe Sep 30 '14 at 11:21
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    $\begingroup$ You do not post your implementation, but I am guessing that you check the values of drifted Brownian motion at some prespecified time points $\delta t, 2 \delta t, ..., N \delta t$. If so, you will be overestimating the probability that $\tau_b<T$ because you do not take into account the situations where $ n \delta t > B$, $(n+1) \delta t > B$ but $\tau \in [ n \delta t, (n+1) \delta t]$. You can check if my guess is correct by reducing $\delta t$ $\endgroup$ – Yulia V Sep 30 '14 at 12:08
  • $\begingroup$ @YuliaV: Thanks a lot for your hint. You suppose you meant $X_{n\delta t}> B$ and $X_{(n+1) \delta t}> B$, right? In this case I would be underestimating the probability of hitting the barrier before $T$ instead of overestimating it since I'm excluding a lot of paths in with the condition is satisfied. Am I right ? $\endgroup$ – Paul Sep 30 '14 at 12:48
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    $\begingroup$ @James: Thanks for your comment. That is not the case. However I just found a mistake in the transcription of the closed formula my code. I've written $\left(\frac{B}{X0} \right)^{2 \bar{\mu}}$ instead of $\left(\frac{B}{X0} \right)^{2 {\lambda}}$ which explain the scaling difference. Now the match is perfect! $\endgroup$ – Paul Sep 30 '14 at 14:47
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    $\begingroup$ Hi Paul, welcome to Quant.SE! Good to hear you found your solution. Can you put it as an answer? Then it's clear this question is solved. $\endgroup$ – Bob Jansen Sep 30 '14 at 15:23

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