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We know the payment function of Spread option is $$\max\{X_T - Y_T-K,0\}$$ here $$d X_t = (\mu_x - D_x)X_t dt + \sigma_xX_td W^x_t$$ $$d Y_t = (\mu_y - D_y)Y_t dt + \sigma_yY_td W^y_t$$ $$d W^x_td W^y_t = \rho dt$$ and we know the the increasing of $\rho$ will reduce the value of spread option, but how to explain this result without deducing the mathematical formula?

I know the Kirk’s approximation formula, $\rho$ only contribute to the equivalent volatility $$\sigma^2 = \sigma^2_x - 2\rho\sigma_x\sigma_z + \sigma^2_z$$ here $\sigma_z$ is a transformation of $\sigma_y$ which is fixed. Then increasing of $\rho$ will reduce the $\sigma,$ and will reduce the value of option.

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the payoff is max(X-Y-K,0). so this option pays you the most if X goes up and Y goes down. So you need X and Y to move in opposite directions. The more X and Y move in the same direction (high rho) the less you get paid.

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