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Consider the covariance, evaluated at time $t$, between two call options written on two different but not independent underlyings $S_1$ and $S_2$ defined on the same (filtered) measure space $\left(\Omega,\mathbb{F},P,\mathbb{\bar{F}}\right)$: \begin{equation} E_t\left(\left(\left(S_{1,T}-k_1\right)^{+}-E_t\left(\left(S_{1,T}-k_1\right)^{+}|\mathbb{F}_t\right)\right)\left(\left(S_{2,T}-k_2\right)^{+}-E_t\left(\left(S_{2,T}-k_2\right)^{+}|\mathbb{F}_t\right)\right)|\mathbb{F}_t\right) \end{equation}
Is the correlation continuous with respect to $k_1$ and $k_2$? Consider each component of the covariance and let $\tilde{\Omega}$ be the space of events that make the payoff of both call options positive: \begin{equation} \begin{aligned} E_t\left(\left(S_{1,T}-k_1\right)^{+}\left(S_{2,T}-k_2\right)^{+}|\mathbb{F}_t\right)&=E_{t,\tilde{\Omega}}\left(\left(S_{1,T}-k_1\right)\left(S_{2,T}-k_2\right)|\mathbb{F}_t\right)=\\ &= E_{t,\tilde{\Omega}}\left(S_{1,T}S_{2,T}-k_1S_{2,T}-S_{1,T}k_2+k_1k_2|\mathbb{F}_t\right)=\\ &=E_{t,\tilde{\Omega}}\left(S_{1,T}S_{2,T}|\mathbb{F}_t\right)-E_{t,\tilde{\Omega}}\left(S_{2,T}|\mathbb{F}_t\right)k_1+\\ &-E_{t,\tilde{\Omega}}\left(S_{1,T}\right)k_2+k_1k_2P\left(\tilde{\Omega}\right) \end{aligned} \end{equation} \begin{equation} \begin{aligned} E_t\left(\left(S_{1,T}-k_1\right)^{+}|\mathbb{F}\right)E_t\left(\left(S_{2,T}-k_2\right)^{+}|\mathbb{F}\right)&=E_{t,\tilde{\Omega}}\left(\left(S_{1,T}-k_1\right)|\mathbb{F}\right)E_{t,\tilde{\Omega}}\left(\left(S_{2,T}-k_2\right)|\mathbb{F}\right)\\ &=E_{t,\tilde{\Omega}}\left(\left(S_{1,T}\right)|\mathbb{F}\right)E_{t,\tilde{\Omega}}\left(\left(S_{2,T}\right)|\mathbb{F}\right)-E_{t,\tilde{\Omega}}\left(\left(S_{1,T}\right)|\mathbb{F}\right)k_2-E_{t,\tilde{\Omega}}\left(\left(S_{2,T}\right)|\mathbb{F}\right)k_1+k_1k_2P\left(\tilde{\Omega}\right) \end{aligned} \end{equation} Therefore I'd conclude that the correlation is continuous in both $k_1$ and $k_2$. Is this correct?

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