Somewhat embarrassingly I'm stuck with something very elementary.

I want to find the conditional probability of a stock movement (GBM):

$$\mathbb{P} \big( S_t \geq b \vert S_s \leq b) $$

for $ t > s$. My main problem is to determine what $\mathbb P(S_t \geq b, S_s \leq b\big)$ equals.


3 Answers 3


By a certain algebraic manipulation, what you need is the probability $P(W_t \ge a, W_s \le c)$, which can be computed as below: \begin{align*} P(W_t \ge a, W_s \le c) &= P(W_t-W_s \ge a-W_s, W_s \le c)\\ &=E\big(E\left(1_{W_t-W_s \ge a-W_s} 1_{W_s \le c} \mid W_s \right)\big)\\ &=E\big(1_{W_s \le c}E\left(1_{W_t-W_s \ge a-W_s} \mid W_s \right)\big)\\ &=E\Big(1_{W_s \le c}\Big[1-\Phi\Big(\frac{a-W_s}{\sqrt{t-s}}\Big)\Big]\Big)\\ &=\Phi\Big(\frac{c}{\sqrt{s}} \Big)-\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\frac{a}{\sqrt{s}}}\Phi\Big(\frac{a-\sqrt{s}x}{\sqrt{t-s}} \Big)e^{-\frac{x^2}{2}} dx, \end{align*} where $\Phi$ is the cumulative distribution function of a standard normal random variable.

  • $\begingroup$ Can you explain the second equality (when going to indicators)? $\endgroup$
    – Nid
    Commented Nov 25, 2016 at 19:28
  • $\begingroup$ That is based on the tower law for conditional expectation. $\endgroup$
    – Gordon
    Commented Nov 25, 2016 at 19:40


Let $$dS_t=\mu S_tdt+\sigma S_t dW_t\,.\tag 1$$ Set $x_t=\ln S_t$, By application of Ito's lemma we have $$x_t=x_0+\left(\mu-\frac 12 \sigma^2\right)t+\sigma W_t\tag 2$$ thus

$$x_t\sim\mathcal {N}\left(x_0+\left(\mu-\frac 12 \sigma^2\right)t\,,\,\sigma^2 t\right)\tag 3$$ on the other hand $$\mathbb P(S_t \geq b\,, S_s \leq b\big)=\mathbb P(\ln S_t \geq \ln b\,, \ln S_s \leq \ln b\big)=\mathbb P(x_t\ge \ln b \,,\,x_s\le \ln b)\tag 4$$

Note $$z_t=\frac{x_t-x_0-\left(\mu-\frac 12 \sigma^2\right)t}{\sigma\sqrt{t}}\sim \mathcal{N}(0,1)$$

  • $\begingroup$ So this I knew, but what confuses me is that $S_t$ and $S_s$ are not independent. I understand that $\ln(t) - \ln (s)$ and $\ln (s)$ are independent, but how can I use this? I am not doing homework here so please give a straight answer if you have it :) $\endgroup$
    – Nid
    Commented Nov 25, 2016 at 18:55
  • $\begingroup$ Yes these are not independent, and you should calculate $$\text{Cov}(x_t,x_s)=\mathbb{E}[x_tx_s]-\mathbb{E}[x_t]\mathbb{E}[x_s]$$ $\endgroup$
    – user16651
    Commented Nov 25, 2016 at 19:00
  • $\begingroup$ Note $\mathbb{E}[W_t,W_s]=\min\{t,s\}=s$ $\endgroup$
    – user16651
    Commented Nov 25, 2016 at 19:01
  • $\begingroup$ finally , we have $$\rho=\frac{\text{Cov}(x_t,x_s)}{\sigma\sqrt{ts}}$$ $\endgroup$
    – user16651
    Commented Nov 25, 2016 at 19:03
  • $\begingroup$ look at it: en.wikipedia.org/wiki/Multivariate_normal_distribution $\endgroup$
    – user16651
    Commented Nov 25, 2016 at 19:06

Another way to obtain many multi-period risk-neutral probabilities related to geometric Brownian motion processes is to use the valuation function for higher-order binaries. These contracts are special cases of the multi-asset multi-period $\mathbb{M}$-binaries introduced by Skipper and Buchen (2003).

In contrast to the others answers, I don't provide the derivation but refer to the original paper. See also Appendix B my paper Zhang and Thul (2016) for a summary of the notation and key results (sorry for the plug).

2-nd Order Bond Binaries

The time $T_2$ terminal value of a $2$-nd order bond binary is given by

\begin{equation} \mathcal{B}_{\xi_1, \xi_2}^{s_1, s_2} \left( S_{T_1}, S_{T_2}, T_2 \right) = \mathrm{1} \left\{ s_1 S_{T_1} > s_1 \xi_1 \right\} \mathrm{1} \left\{ s_2 S_{T_2} > s_2 \xi_2 \right\}. \end{equation}

I.e. this contract has a unit payoff conditional on the asset prices at times $T_1$ and $T_2$ being above ($s_i = 1$) or below ($s_i = -1$) the levels $\xi_1$ and $\xi_2$, respectively. Its time $0 \leq t < T_1$ value is given by

\begin{equation} \mathcal{B}_{\xi_1, \xi_2}^{s_1, s_2} \left( S_t, t \right) = e^{-r \tau_2} \mathcal{N}_2 \left( \alpha_{0, 1}, \alpha_{0, 2}; \rho \right), \end{equation}

where $\tau_i = T_i - t$,

\begin{equation} \alpha_{0, i} = \frac{s_i}{\sigma \sqrt{\tau_i}} \left( \ln \left( \frac{S}{\xi_i} \right) + \left( r - \frac{1}{2} \sigma^2 \right) \tau_i \right) \end{equation}


\begin{equation} \rho = s_1 s_2 \sqrt{\frac{\tau_1}{\tau_2}}. \end{equation}

Here, $\mathcal{N}_2$ is the bivariate standard normal distribution function with the given correlation.

Joint Probability

Using these results, you obtain the probability of interest as

\begin{equation} \mathbb{P} \left\{ S_{T_1} < b, S_{T_2} > b \right\} = e^{r \tau_2} \mathcal{B}_{b, b}^{-, +} \left( S_0, 0 \right) \end{equation}

While you could just as well work out this particular case by hand, the above framework can often still be applied when you are interested in more complex probabilities. For example, barriers can be easily incorporated using the method of images; see e.g. Buchen (2001).


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

Skipper, Max and Peter W. Buchen (2003) "The Quintessiential Option Pricing Formula", Working Paper, School of Mathematics and Statistics, University of Sydney

Zhang, Ally Quan and Matthias Thul (2016) "How Much is the Gap? Efficient Jump Risk-Adjusted Valuation of Leveraged Certificates", Working Paper, available on SSRN


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