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If the value of an option at Maturity is

what is the off-setting position you take for X and Y, if you are

i)Long Call of the option ii)Short Call of the option iii)Long Put of the option iv)Short Put of the option

when the correlation between X & Y >1, & it's a European option

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  • $\begingroup$ what do you he correlation between X & Y >1? $\endgroup$ – Gordon Jan 7 '16 at 13:38
  • $\begingroup$ Note that, if both $X$ and $Y$ are log-normal, then $X/Y$ is also log-normal, and you can use the Black-Scholes' formula. $\endgroup$ – Gordon Jan 7 '16 at 14:18
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We assume that \begin{align*} dX_t &= X_t(rdt + \sigma_x dW^x_t)\\ dY_t &= Y_t\Big[rdt + \sigma_y\Big(\rho dW^x_t + \sqrt{1-\rho^2} dW^y_t \Big)\Big], \end{align*} where $\rho$ is the correlation, which we assume to be less than 1, and $W^x$ and $W^y$ are two independent standard Brownian motions. Then \begin{align*} d(X_t/Y_t) &= \frac{1}{Y_t}dX_t - \frac{X_t}{Y_t^2}dY_t +\frac{X_t}{Y_t^3}d\langle Y, \, Y\rangle_t - \frac{1}{Y_t^2}d\langle X, \, Y\rangle_t\\ &=\frac{X_t}{Y_t}\Big[\sigma_x dW^x_t - \sigma_y\Big(\rho dW^x_t + \sqrt{1-\rho^2} dW^y_t \Big) +\big(\sigma_y^2-\rho\sigma_x\sigma_y\big)dt\Big]\\ &=\frac{X_t}{Y_t}\Big[ \big(\sigma_y^2-\rho\sigma_x\sigma_y\big)dt +\sigma dW_t\Big], \end{align*} where $$\sigma = \sqrt{\sigma_x^2+\sigma^2_y-2\rho\sigma_x\sigma_y},$$ and $W$ is standard Brownian motion. The option payoff $$(X_T/Y_T-K)^+$$ at maturity $T$ has the value given by \begin{align*} C &\equiv e^{-rT}E\big((X_T/Y_T-K)^+ \big)\\ &=e^{-rT}\left(e^{(\sigma_y^2-\rho\sigma_x\sigma_y)T} \frac{X_0}{Y_0}N(d_1)-KN(d_2)\right)\\ &= e^{-rT +(\sigma_y^2-\rho\sigma_x\sigma_y)T}\left(\frac{X_0}{Y_0}N(d_1)-e^{-(\sigma_y^2-\rho\sigma_x\sigma_y)T} KN(d_2)\right), \end{align*} where \begin{align*} d_1 &=\frac{\ln (X_0/Y_0) + (\sigma_y^2-\rho\sigma_x\sigma_y)T + \frac{1}{2}\sigma^2 T }{\sigma \sqrt{T}}, \end{align*} and $$d_2 = d_1 - \sigma \sqrt{T}.$$

Consequently, the delta hedge ratios with respect to $X_0$ and $Y_0$ can be computed correspondingly. Specifically, \begin{align*} \frac{\partial C}{\partial X_0} &= \frac{1}{Y_0}\frac{\partial C}{\partial (X_0/Y_0)}\\ &=\frac{1}{Y_0}e^{-rT +(\sigma_y^2-\rho\sigma_x\sigma_y)T}N(d_1), \end{align*} and \begin{align*} \frac{\partial C}{\partial Y_0} &= -\frac{X_0}{Y_0^2}\frac{\partial C}{\partial (X_0/Y_0)}\\ &=-\frac{X_0}{Y_0^2}e^{-rT +(\sigma_y^2-\rho\sigma_x\sigma_y)T}N(d_1). \end{align*}

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