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I'd like to price the following contingent claim using a copula model. $$V_T = (S_T-K)1_{\{S_T>K\}}1_{\{L\leq X_T\leq U\}}$$

where $S$ and $X$ are two stock price processes which follow a non-flat vol model. In particular, $S$ and $X$ are not standard GBM models with flat vols. Suppose you are given the distributions of $S_T$ and $X_T$ as a black box. Moreover, suppose you are given a copula density function, $f_{(S,X)}$, which is very good at approximating the joint distribution. I would like to derive a semi-analytic formula for the price, $V_t$. Below is my attempt. Does it look correct? What should I do next? Can I simplify further? \begin{align*} V_t & = e^{-r(T-t)} E[(S_T-K)1_{\{S_T>K\}}1_{\{L\leq X_T\leq U\}}] \\ &= e^{-r(T-t)}\int_{L}^{U}\int_{K}^{\infty}(a-K)f_{S,X}(a,b)dadb \\ &= e^{-r(T-t)}\int_{K}^{\infty}(a-K)\int_{L}^{U}f_{S,X}(a,b)dbda \\ &= e^{-r(T-t)}\int_{K}^{\infty}(a-K)\left(F_{S,X}(a,U) - F_{S,X}(a,L)\right)da \\ &= e^{-r(T-t)}\left( (a-K)\int_{K}^{\infty}F_{S,X}(a,U) - F_{S,X}(a, L) da - \int_{K}^{\infty} \int_{K}^{\infty}F_{S,X}(a,U) - F_{S,X}(a, L)dada \right) \end{align*}

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  • $\begingroup$ Oh nevermind, X =/= S! $\endgroup$ – stackoverblown Dec 15 '20 at 13:18
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I don't know whether you are studying in the same class with the author of this question Copula analytic formula for $max(S_T^1−K,0)1_{L<S_T^2<U}$.

The idea is to decompose the payoff into 2 parts. The first part is $1_{S_T>K}1_{L<X_T<U}$ and the second one is $S_T 1_{S_T>K}1_{L<X_T<U}$ .

The value of the first part is equal to $P(\{(W_T^1-W_t^1) > d_1 \}\cap \{ d_2<(W_T^2-W_t^2))<d_3 \}) $ and you can use the density function $f_{S,X}$ to have its closed form solution.

And the second part can be transformed into the first part form with the change of numéraire. After that, you use the same method to have its closed form solution.

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  • $\begingroup$ Thanks, but that assumes flat vols. So you are saying to just do the same thing with a copula instead? $\endgroup$ – John Doe Dec 14 '20 at 22:44
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    $\begingroup$ If the volatility is a deterministic but time-dependent functionk, for example $\sigma_S = \sigma_S(t)$, you can use the same argument. If the volatility is not a deterministic function, for example, the volatility is a function of $t$ and $S_t$ as following $\sigma_S = \sigma_S(S_t,t)$, I don't think you can have closed form solution, even if the payoff were just $(S_T-K)^+$. $\endgroup$ – NN2 Dec 14 '20 at 23:08

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