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I agree with vanguard2k's comment: A few more details on the notation would be helpful. But, as far as I can tell, the second equality is a simple expansion. First, $\mathbf{1}'\mathbf{1} = T$ (assuming the vectors are elements of $\mathbb{R}^T$). The expression $\mathbf{1} (\mathbf{1}'\mathbf{1})^{-1} \mathbf{1}'$ is therefore nothing else than a $T\times ...


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let $\frac{\partial C}{\partial S}=\delta_c$ let $\frac{\partial^2 C}{\partial S^2}=\Gamma_c$ let $\frac{\partial C_0}{\partial S}=\delta_0$ let $\frac{\partial^2 C_0}{\partial S^2}=\Gamma_0$ we want $\frac{\partial V}{\partial S}=\frac{\partial C}{\partial S}=\delta_c$ and $\frac{\partial^2 V}{\partial S^2}=\frac{\partial^2 C}{\partial S^2}=\Gamma_c$ ...



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