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I think the variance of the instantaneous shifts in the spread is meant: $V \left[ dX \right]=V \left[ dS_1-dS_2 \right]$ And the individual variances (in the conditional and local sense) are: $V \left[ dS_1 \right]= \sigma_1^2 S_1^2dt$ $V \left[ dS_2 \right]= \sigma_2^2 S_2^2dt$ And the covariance term is, assuming the two Brownians are correlated:...


@Quantuple had a good answer in the comments. Basically, I was not using a fine enough grid in my replication. Click here to see the original comment that answers the question.


For the first, where $|\beta| < 1.0$, you can write it using the lag operator. $x_t (1 - \beta L) = (1 + \theta L) u_t $ $X_t = \frac{(1 + \theta L) u_t}{(1- \beta L)} $ Since $|\beta| < 1.0 $, this is an infinite sum that converges: $X_t = \sum_{i=0}^\infty \beta^{i}( 1 + \theta L) u_{t-i}$ The $u_{t}$ are independent and normal with mean zero ...

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