I recently read an article and they provide a formula for the first-passage probability as

$$Z = {1 \over \sigma }\left[ {\log S/{S_t} + (r - {1 \over 2}{\sigma ^2})t} \right]$$

${{S_t}}$ value of the stock at time t , $r$ ror on the stock, $\sigma $ standard deviation.

The authors ref. Feller (1971) "An Introduction to Probability Theory and Its Applications, Vol 2. " But i have been unable to find the formula , does anyone know where this comes from and any literature where this formula is presented?

Update: Is it derived from this ? $$Z(t) = \exp \left\{ {\sigma X(t) - (\sigma \mu + {1 \over 2}{\sigma ^2})t} \right\}$$

where $$X(t) = \log S/Sc$$

update: is there a typo in their formula? after reading some literature should it not be the case that the first passage probability should be

$$Z = {1 \over \sigma \sqrt{t} }\left[ {\log S/{S_t} + (r - {1 \over 2}{\sigma ^2})t} \right]$$

Link : Feller pdf

Link Article : https://forum.wilmott.com/download/file.php?id=7750


1 Answer 1


Their formula looks correct. As is usually the case, there are multi ways to derive this result. I will outline two of them here.

Reflection Principle & Measure Change

The solution to the risk-neutral dynamics of $S$ is

\begin{equation} S_t = S_0 \exp \left\{ \left( r - \frac{1}{2} \sigma^2 \right) t + \sigma W_t^* \right\}, \end{equation}

where $W^*$ is a $\mathbb{P}^*$-Brownian motion. We have $S_t = B$ when

\begin{equation} W_t^* + \frac{1}{\sigma} \left( r - \frac{1}{2} \sigma^2 \right) t = \frac{1}{\sigma} \ln \left( \frac{B}{S_0} \right). \end{equation}

Now let

\begin{equation} \lambda = \frac{1}{\sigma} \left( r - \frac{1}{2} \sigma^2 \right), \qquad \alpha = \frac{1}{\sigma} \ln \left( \frac{B}{S_0} \right) \end{equation}

and define a new probability measure $\hat{\mathbb{P}}$ equivalent to $\mathbb{P}^*$ through the Radon-Nikodym derivative process

\begin{equation} \xi_t \left( \mathbb{P}^*, \hat{\mathbb{P}} \right) = \left. \frac{\mathrm{d} \hat{\mathbb{P}}}{\mathrm{d} \mathbb{P}^*} \right| \mathfrak{F}_t = \mathcal{E}_t \left( - \int_0^\cdot \lambda \mathrm{d}W_u^* \right) \qquad \mathbb{P}^*\text{-a.s.}, \end{equation}

where $\mathcal{E}$ is the Doleans-Dade exponential martingale. It follows by Girsanov's theorem that the process $\hat{W}$ defined by

\begin{equation} \hat{W}_t = W_t^* + \lambda t \end{equation}

is a standard Brownian motion under $\hat{\mathbb{P}}$. Next, by the reflection principle for Brownian motion, PDF of the first passage time $\nu$ of $\hat{W}$ to a level $\alpha$ is given by

\begin{equation} \hat{\mathbb{P}} \left\{ \nu \in \mathrm{d}t \right\} = \frac{\vert \alpha \vert}{t \sqrt{2 \pi t}} \exp \left\{ -\frac{\alpha^2}{2 t} \right\} \mathrm{d}t; \end{equation}

see for example Equation (II.6.3) in Karatzas and Shreve (1991), p. 80 or Theorem 3.7.1. in Shreve (2004), p. 113. I take this result as given and you can find details on its derivation in the references. Using the abstract Bayes rule; see for example Lemma A.1.4 in Musiela and Rutkowski (2005), p. 615, we get

\begin{eqnarray} \mathbb{P}^* \left\{ \nu \in \mathrm{d}t \right\} & = & \mathbb{E}_{\mathbb{P}^*} \left[ \mathrm{1} \{ \nu \in \mathrm{d}t \} \right]\\ & = & \mathbb{E}_{\hat{\mathbb{P}}} \left[ \xi_t^{-1} \left( \mathbb{P}^*, \hat{\mathbb{P}} \right) \mathrm{1} \{ \nu \in \mathrm{d}t \} \right]\\ & = & \mathbb{E}_{\hat{\mathbb{P}}} \left[ \exp \left\{ \lambda \hat{W}_t - \frac{1}{2} \lambda^2 t \right\} \mathrm{1} \{ \nu \in \mathrm{d}t \} \right]. \end{eqnarray}

Now, when $\nu = t$ then $\hat{W}_t = \alpha$ and thus

\begin{eqnarray} \mathbb{P}^* \{ \nu \in \mathrm{d}t \} & = & \exp \left\{ \lambda \alpha - \frac{1}{2} \lambda^2 t \right\} \hat{\mathbb{P}} \{ \nu \in \mathrm{d}t \}\\ & = & \frac{\vert \alpha \vert}{t \sqrt{2 \pi t}} \exp \left\{ -\frac{(\alpha - \lambda t)^2}{2 t} \right\} \mathrm{d}t. \end{eqnarray}

Note that the $\alpha - \lambda t$ term is equal to $Z$ in your reference.

Differentiating the Barrier Option Price

Let $\psi = -1$ ($\psi = +1$) indicate an upper (lower) barrier. Consider a contract with the terminal payoff $V_T = \mathrm{1} \{ \nu > T \}$. Using the method of images, it can be shown that its valuation function in terms of the time-to-maturity $\tau = T - t$ is given by

\begin{equation} \tilde{V}(S, \tau) = \mathcal{B}_B^\psi(S, \tau) - \mathcal{I} \left\{ \mathcal{B}_B^\psi(S, \tau) \right\}, \end{equation}


\begin{eqnarray} \mathcal{B}_B^\psi & = & e^{-r \tau} \mathcal{N} \left( \psi d_-(S, B) \right),\\ d_-(S, B) & = & \frac{1}{\sigma \sqrt{\tau}} \left( \ln \left( \frac{S}{B} \right) + \left( r - \frac{1}{2} \sigma^2 \right) \tau \right),\\ \mathcal{I} \left\{ \tilde{V}(S, \tau) \right\} & = & \left( \frac{S}{B} \right)^{2 \alpha} \tilde{V} \left( \frac{B^2}{S}, \tau \right)\\ \alpha & = & \frac{1}{2} - \frac{r}{\sigma^2};\\ \end{eqnarray}

see e.g. Buchen (2001) or Wilmott et al. (1995). Note that the option price is linked to the first passage time CDF through

\begin{eqnarray} \mathbb{P}^* \{ \nu > \tau \} & = & e^{r \tau} \tilde{V}(S, \tau)\\ & = & \int_\tau^\infty \mathbb{P}^* \{ \nu \in \mathrm{d}\tau \} \end{eqnarray}

and thus

\begin{equation} \frac{1}{\mathrm{d} \tau} \mathbb{P}^* \{ \nu \in \mathrm{d}\tau \} = -\frac{\partial}{\partial \tau} \left\{ e^{r \tau} \tilde{V}(S, \tau) \right\}. \end{equation}

Carefully differentiating the two terms in $\tilde{V}(S, \tau)$ yields

\begin{eqnarray} \frac{\partial}{\partial \tau} e^{r \tau} \mathcal{B}_B^\psi(S, \tau) & = & -\psi \mathcal{N}' \left( \psi d_-(S, B) \right) \frac{1}{2 \sigma \tau \sqrt{\tau}} \left( \ln \left( \frac{S}{B} \right) - \left( r - \frac{1}{2} \sigma^2 \right) \tau \right) \end{eqnarray}


\begin{eqnarray} \frac{\partial}{\partial \tau} e^{r \tau} \mathcal{I} \left\{ \mathcal{B}_B^\psi(S, \tau) \right\} & = & -\psi \left( \frac{S}{B} \right)^{2 \alpha} \mathcal{N}' \left( \psi d_-(B, S) \right) \frac{1}{2 \sigma \tau \sqrt{\tau}} \left( \ln \left( \frac{B}{S} \right) - \left( r - \frac{1}{2} \sigma^2 \right) \tau \right). \end{eqnarray}

Through some tedious algebra, we can show that

\begin{eqnarray} \left( \frac{S}{B} \right)^{2 \alpha} \mathcal{N}' \left( \psi d_-(B, S) \right) & = & \mathcal{N}' \left( \psi d_-(S, B) \right)\\ & = & \mathcal{N}' \left( d_-(S, B) \right). \end{eqnarray}


\begin{eqnarray} \frac{1}{\mathrm{d} \tau} \mathbb{P}^* \{ \nu \in \mathrm{d}\tau \} & = & \frac{-\psi \ln (B / S)}{\sigma \tau \sqrt{\tau}} \mathcal{N}' \left( d_-(S, B) \right). \end{eqnarray}

It is easy to check that this is the same expression that we obtained before.


Buchen, Peter W. (2001) "Image Options and the Road to Barriers," Risk Magazine, Vol. 14, No. 9, pp. 127-130

Karatzas, Ioannis and Steven E. Shreve (1991) Brownian Motion and Stochastic Calculus: Springer, 2nd edition.

Musiela, Marek and Marek Rutkowski (2005): Martingale Methods in Financial Modelling: Springer.

Shreve, Steven E. (2004) Stochastic Calculus for Finance II - Continuous Time Models: Springer.

Wilmott, Paul, Sam Howison and Jeff Dewynne (1995) The Mathematics of Financial Derivatives: Cambridge University Press

  • 1
    $\begingroup$ +1, very nice explanation (I suspect there might be a slight notational problem with $\Bbb{P}^*$ and $\hat{\Bbb{P}}$ when you write abstract Bayes though - same when you give the distribution of the first passage time under $\hat{\Bbb{P}}$) $\endgroup$
    – Quantuple
    Apr 24, 2017 at 10:07
  • $\begingroup$ @Quantuple - thanks for pointing this out! $\endgroup$ Apr 24, 2017 at 10:11

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