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Consider the payoff function $$ V_T = max(S_T^1 - K, 0) 1_{\{L<S_T^2<U\}} = (S_T^1 - K)1_{\{S_T^1 > K\}}1_{\{L<S_T^2<U\}}$$

where $S_T^1$ and $S_T^2$ are two GBM distributed stocks with correlation, $\rho$. How would you find $V_t$ without the use of MC simulation (hint: use a Gaussian copula)?

My attempt: $$V_t = e^{-r(T-t)}E[(S_T^1 - K)1_{\{S_T^1 > K\}}1_{\{L<S_T^2<U\}}]$$ $$V_t = e^{-r(T-t)}\int_{L}^{U}\int_{K}^{\infty}(x-K)f_{x,y}(x,y)dxdy$$

where $f_{x,y}$ is the joint pdf coming from the Guassian copula. How do I go further now? If I can't, how would I implement the double integral on a computer?

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For simplicity, we suppose $r = 0$. $$V_t = E^Q((S_T^1 - K)1_{\{S_T^1 > K\}}1_{\{L<S_T^2<U\}}) = E^Q(S_T^11_{\{S_T^1 > K\}}1_{\{L<S_T^2<U\}}) - KE^Q(1_{\{S_T^1 > K\}}1_{\{L<S_T^2<U\}})$$ For the second term: $$E^Q(1_{\{S_T^1 > K\}}1_{\{L<S_T^2<U\}})=P(\{S_T^1 > K\}\cap \{ L<S_T^2<U \}) =P(\{(W_T^1-W_t^1) > d_1 \}\cap \{ d_2<(W_T^2-W_t^2))<d_3 \}) $$ with $$d_1 = \frac{\ln (K/S_t^1)+(1/2\sigma^1)(T-t))}{\sigma^1}$$ $$d_2 = \frac{\ln (L/S_t^2)+(1/2\sigma^2)(T-t))}{\sigma^1}$$ $$d_3 = \frac{\ln (U/S_t^2)+(1/2\sigma^2)(T-t))}{\sigma^1}$$ Because $(W_T^1-W_t^1,W_T^2-W_t^2)$ follows the bidimensional gaussian distribution, you can easily obtain the analytic formula for $E^Q(1_{\{S_T^1 > K\}}1_{\{L<S_T^2<U\}})$.

Now, for the first term $E^Q(S_T^11_{\{S_T^1 > K\}}1_{\{L<S_T^2<U\}})$, we change the mesure to the $S_t^1$ as numéraire. $$E^Q(S_T^1 1_{\{S_T^1 > K\}}1_{\{L<S_T^2<U\}}) = S_t^1 E^{Q_{S^1}}(1_{\{S_T^{1} > K\}}1_{\{L<S_T^{2}<U\}})$$ In the new mesure, we have $$dW_t^{1'} = dW_t^1 -\sigma^1dt$$ $$dW_t^{2'} = dW_t^2 -\sigma^1\rho dt$$ with $$\frac{dS_t^1}{S_t^1} = \sigma^1 dW_t^1 = \sigma^1 (dW_t^{1'} + \sigma^1dt)$$ $$\frac{dS_t^2}{S_t^2} = \sigma^2 dW_t^2 = \sigma^2 (dW_t^{2'} + \sigma^2 \rho dt)$$ By the same argument we did with the first term, we obtain the analytic formula for $E^{Q_{S^1}}(1_{\{S_T^{1} > K\}}1_{\{L<S_T^{2}<U\}})$.

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