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First this is not a full answer, but it might help you. You probably hit $B$ quickly with $(1)$ than with $(2)$. Hint of previous assertion I might reformulate your question. I suppose your pricing condition is $$\left\langle X^{(2)}\right\rangle_t=\left\langle X^{(1)}\right\rangle_t $$ so you get : $$X^{(1)}_t = ...


Using $$X^i_t = (X^i_0 + \int_0^t\sigma_i e^{a_i u} dB^i_u)e^{-a_it} $$ and $$ X^i_t-\mathbb{E}[X^i_t] = e^{-a_it} \int_0^t\sigma_i e^{a_i u} dB^i_u $$ and thus : $$\text{Cov}(X^1_t,X^2_t)=\mathbb{E}\left[e^{-a_1t} \int_0^t\sigma_1 e^{a_1 u} dB^1_u e^{-a_2t} \int_0^t\sigma_2 ...


I dont know what exactly you want but have a look at : you calibrate the first one in stand alone, then the second one in stand alone, endly you can compute correlation on the residuals of the increments knowing your parameters


let define $$ \text{RP}_t = \sum_{u< t} \frac{dP_u}{P_u}$$ $$ \text{RQ}_t =\sum_{u<t} \frac{dP_u}{P_u}$$ $X$ is a mean reverting process so : $$ dX = \alpha (\mu - X)dt + \sigma dB $$ where $B$ is a brownian motion meanwhile using your relationship you get : $$ X_t = \text{RP}_t - b \text{RQ}_t - a t $$ you use $X$ dynamics with this and you get: ...

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