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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?

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  • $\begingroup$ For $t<T_1$, \begin{align} V_t &= e^{-r(T_2-t)}\mathbb{E}^{\mathbb Q}_t\left[ (S_{T_2}-K)^+\mathbb{1}_{\{S_{T_1}>B\}}\right] \\ &= e^{-r(T_2-t)}\mathbb{E}^{\mathbb Q}_t\left[\mathbb{E}^{\mathbb Q}_{T_1}\left[ (S_{T_2}-K)^+\mathbb{1}_{\{S_{T_1}>B\}}\right]\right] \\ &= e^{-r(T_1-t)}\mathbb{E}^{\mathbb Q}_t\left[ e^{-r(T_2-T_1)}\mathbb{E}^{\mathbb Q}_{T_1}\left[ (S_{T_2}-K)^+\right]\mathbb{1}_{\{S_{T_1}>B\}} \right] \\ &= e^{-r(T_1-t)}\mathbb{E}^{\mathbb Q}_t\left[C_{T_1}\mathbb{1}_{\{S_{T_1}>B\}} \right], \end{align} where $$C_{T_1}=S_{T_1}e^{-q(T_2-T_1)}N(d_1) - Ke^{-r(T_2-T_1)}N(d_2).$$ $\endgroup$
    – Kevin
    Commented Jun 4, 2021 at 10:14
  • $\begingroup$ Further, note that $$\mathbb{E}[f(X)\mathbb{1}_{\{X>B\}}]=\mathbb{E}\left[f(X)|\{X>B\}\right]\text{Pr}[\{X>B\}].$$ The probability can be easily computed $$\mathbb{Q}_t[\{S_{T_1}>B\}]=N\left(\frac{\ln\left(\frac{S_t}{B}\right)+\left(r-q-\frac{1}{2}\sigma^2\right)(T_1-t)}{\sigma\sqrt{T_1-t}}\right).$$ $\endgroup$
    – Kevin
    Commented Jun 4, 2021 at 10:24
  • $\begingroup$ @Kevin: I don't think this approche works because you have $S_{T_1}$ in $d_1$ and $d_2$. $\endgroup$
    – NN2
    Commented Jun 4, 2021 at 11:37
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    $\begingroup$ @NN2 Because $C_{T_1}$ depends on $S_{T_1}$ in a non-linear way, computing $\mathbb{E}_t^\mathbb{Q}\left[C_{T_1}\mathbb{1}_{\{S_{T_1}>B\}}\right]$ is not trivial. That's why the second comment illustrates that one needs to compute a conditional expectation, because of the covariance between the option price, $C_{T_1}$ and the indicator, $\mathbb{1}_{\{S_{T_1}>B\}}$. My comment left open how to compute $\mathbb{E}_t^\mathbb{Q}[C_{T_1}|\{S_{T_1}>B\}]$ (I don't have much time right now). Did I miss something where I went completely wrong? Your answer is arguably easier, hence I upvoted it! :) $\endgroup$
    – Kevin
    Commented Jun 4, 2021 at 11:43
  • $\begingroup$ @Kevin Thank for the upvoting. Indeed, what I meant is it's difficult to have closed form solution because it's not trivial to compute the conditional expectation while there is $S_{T_1}$ in the $N(...)$ as you said. $\endgroup$
    – NN2
    Commented Jun 4, 2021 at 11:47

1 Answer 1

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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

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