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[Short answer] No closed-form formula in general. You need to resort to numerical methods. Monte Carlo is preferred by most practitioners but you could also use Finite Difference schemes (and sometimes even Fourier inversion techniques depending on the model used and the instruments to be priced). [Long answer] One usually distinguishes between 2 classes ...

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The answer is yes. In fact, there always exist a 'Black Scholes like' formula. Easy to show too. If the risk neutral distribution of the price has cumulative density $P$ and probability density $p$, then $$E(S-K)^+=E((S-K)\ 1_{S>K})=E(S\ 1_{S>K})-K\ E(1_{S>K})$$ The second expectation is just $P(K)$, ie the probability that the option ends up in ...

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Measure change is still the most natural approach for such problems. We assume that, under the measure $P$, \begin{align*} dX_t &= \mu X_t dt + \sigma X_t dW_t^1,\\ dY_t &= \mu Y_t dt + \sigma Y_t \left(\rho dW_t^1 + \sqrt{1-\rho^2} dW_t^2 \right), \end{align*} based on the Cholesky decomposition, where $\{W_t^1, t \ge 0\}$ and $\{W_t^2, t \ge 0\}$ ...

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For an American option, you have the right to exercise at any intermediate time. Then, at time $T-1$, if you exercise your option, you obtain the payoff $X_{T-1}$. However, if you wait to exercise at the maturity $T$, your value is $\frac{1}{1+r}\mathbb{E}^Q\left(X_T \mid \mathscr{F}_{T-1} \right)$. Your option value at time $T-1$ is the maximum of these ...

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Do these work for you? P34 of http://web.mit.edu/junpan/www/SVJ.pdf P1360 of http://www.darrellduffie.com/uploads/pubs/DuffiePanSingleton2000.pdf P2045 of http://www.math.ku.dk/~rolf/bakshi.pdf

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Relatively quick Solution If $U$ and $V$ be normally distributed with means $\mu_u\,,\,\mu_v$, variances $\sigma^2_u\,,\,\sigma^2_v$ and correlation $\rho$ then we can show ( by definition of expectation and apply joint density function ) \mathbb{E}\left[\left(e^U-e^V\right)^+\right]={\large{e^{\mu_u+\frac{1}{2}\sigma_u^2}}}\Phi\left(d_1\right)-{\large{e^...

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