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0

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 = ...


3

Using https://en.wikipedia.org/wiki/Ornstein%E2%80%93Uhlenbeck_process#Solution $$X^i_t = (X^i_0 + \int_0^t\sigma_i e^{a_i u} dB^i_u)e^{-a_it} $$ 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 ...


0

I dont know what exactly you want but have a look at : http://www.sitmo.com/article/calibrating-the-ornstein-uhlenbeck-model/ 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


1

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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