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Conditional GBM Threshold Probabilities Using Brownian Increments

Article Quant Q&A · Author: Nid

Summary

The document asks for the probability that a stock modeled by geometric Brownian motion is at or above a threshold at a later time, given that it was at or below that threshold earlier. The responses explain that taking logarithms converts the stock process into a drifted Brownian process, so the event becomes a joint threshold event for the log prices at two times.

One derivation conditions on the earlier Brownian value and uses the independence and normal distribution of the subsequent increment. This yields an integral expression for the joint event, which can then be divided by the probability of the conditioning event to obtain the conditional probability. Another response points to a bivariate normal formulation through higher-order binary option valuations, which can extend to more complex joint events. The post supplies methods rather than a particular numerical probability; the result depends on model parameters, thresholds, and the chosen probability measure.

Key ideas

  • Taking logarithms turns geometric Brownian motion into a drifted Brownian process.
  • The joint threshold event can be evaluated by conditioning on the process value at the earlier time.
  • The later Brownian increment is independent of the earlier value and has a normal distribution.
  • A bivariate normal formulation offers another way to express joint threshold probabilities.
  • The post gives no numerical probability, which depends on the model inputs and probability measure.

Tags

Full text
# How to compute the conditional probability for a geometric Brownian process?


# How to compute the conditional probability for a geometric Brownian process?












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.

## Answer by Gordon (score 1, accepted)

https://quant.stackexchange.com/a/31201

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.

## Answer by user16651 (score 1)

https://quant.stackexchange.com/a/31200

Hint

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

## Answer by LocalVolatility (score 0)

https://quant.stackexchange.com/a/31219

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}

and

\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).

References

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

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.