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Valuing a Two-Date Barrier-Conditioned European Call under GBM

Article Quant Q&A · Author: Bravo
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Summary

The document considers a call whose payoff is paid at a later date only if the stock exceeded a barrier at an earlier date. Under geometric Brownian motion, the payoff can be split into two expectations: the probability-weighted strike component and the expected stock-value component, each restricted by the two date-specific price conditions.

The answer expresses the joint event probability using a bivariate normal cumulative distribution, reflecting the correlation between Brownian motion at the two dates. It proposes handling the stock-value expectation with a change of measure using the stock as numeraire, then evaluating a corresponding joint probability under the transformed measure. This gives a route to a closed-form valuation using bivariate normal probabilities. However, the displayed derivation appears to contain notation or algebra inconsistencies in its thresholds and dates, and it does not clearly state discounting or the risk-neutral drift assumptions. Those details should be verified before using the formula in practice.

Key ideas

  • The payoff can be decomposed into a joint-event probability term and a stock-value term.
  • The barrier and strike conditions form a two-time joint event under the stock’s lognormal dynamics.
  • A bivariate normal distribution represents the dependence between Brownian motion at the two dates.
  • A stock-numeraire change of measure can simplify the expectation involving the later stock price.
  • The displayed derivation has ambiguities that require checking before implementation.

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# How to compute the Present Value of this path-dependent option?


# How to compute the Present Value of this path-dependent option?












I have an option whose payoff depends on its value at two times $T_1$ and $T_2$ as follows.

$$V(t) = \mathbb{E}^{Q}[\mathbb{1}_{S(T_1)>B} (S(T_2)-K)^+)],$$

where the stock price follows the GBM dynamics $\mathrm{d}S_t=\mu S_t \mathrm{d}t+\sigma S_t\mathrm{d}W_t$. How do I compute its value using BS approach?

## Answer by NN2 (score 4, accepted)

https://quant.stackexchange.com/a/65389

We have

$$ \begin{align} V(t) &= \mathbb{E}^{Q}[\mathbb{1}_{S(T_1)>B} (S(T_2)-K)^+)] \\ &= \mathbb{E}^{Q}[\mathbb{1}_{S(T_1)>B}\mathbb{1}_{S(T_2)>K} (S(T_2)-K))] \\ &= \mathbb{E}^{Q}[\mathbb{1}_{S(T_1)>B}\mathbb{1}_{S(T_2)>K} S(T_2)]-K\mathbb{E}^{Q}[\mathbb{1}_{S(T_1)>B}\mathbb{1}_{S(T_2)>K}] \\ \end{align} $$

The second term is equal to $$ \begin{align} \mathbb{E}^{Q}[\mathbb{1}_{S(T_1)>B}\mathbb{1}_{S(T_2)>K}] &= P(S(T_1)>B,S(T_2)>K)\\ &=P( W_{T_1}>\frac{\ln(\frac{B}{S_0})+\frac{\mu^2}{2}T_1}{\sigma},W_{T_2}>\frac{\ln(\frac{K}{S_0})+\frac{\mu^2}{2}T_2}{\sigma}) \\ &=P( -W_{T_1}<-\frac{\ln(\frac{B}{S_0})+\frac{\mu^2}{2}T_1}{\sigma},-W_{T_2}<\frac{\ln(\frac{K}{S_0})+\frac{\mu^2}{2}T_2}{\sigma}) \\ &=\Phi_2 ((-d_1,-d_2);(0,0);\mathbf{\Sigma}) \end{align} $$ where

- $\Phi_2(\mathbf{x};\mathbf{\mu},\mathbf{\Sigma})$ is the cumulative probability function of $(X_1,X_2)$ following the bivariate normal distribution $\mathcal{N}_2(\mathbf{\mu},\mathbf{\Sigma})$

- $\mathbf{\Sigma}$ is the covariance matrix of $(-W_1,-W_2)$ and $$d_i =\frac{\ln(\frac{B}{S_0})+\frac{\mu^2}{2}T_i}{\sigma} $$

For the first term, make a change of measure with $S_t$ as the numeraire, you can transform it to $$\mathbb{E}^{Q}[\mathbb{1}_{S(T_1)>B}\mathbb{1}_{S(T_2)>K} S(T_2)] = \mathbb{E}^{Q_S}[\mathbb{1}_{S'(T_1)>B}\mathbb{1}_{S'(T_2)>K}]$$ after that, applying the same method used for the second term. Q.E.D

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.