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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 $\{\tau_B\leq T\}= \{X_T^*\geq B\}$ 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

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  • 1
    $\begingroup$ Check the scaling. You should probably have $\lambda = \bar{\mu} / \sigma$. $\endgroup$ – Ben Derrett Nov 9 '14 at 12:21

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