I am not sure about the answer to this question. For a Brownian motion $B_t$ and a process $M$ defined by $M_t=B_{t-s}$ if $t>s$ and 0 else, what is the Quadratic Covariation $[B,M]_t$ ?

I find $(t-s)$ if $t>s$ and $0$ else, but the right answer shall be 0 and I can't figure out where I missed something.

Best regards


Well, we'd like to show that the following limit

$$\lim_{n\to\infty} \sum_{k=0}^{\lfloor 2^n t \rfloor}(M_{(k+1)2^{-n}}-M_{k2^{-n}})(B_{(k+1)2^{-n}}-B_{k2^{-n}})$$ exists a.s., since this would then equal $[B, M]_t$. Rewrite the limit as $$ \lim_{n\to\infty} \sum_{k=\lceil 2^ns\rceil}^{\lfloor 2^n t \rfloor}(B_{(k+1)2^{-n}-s}-B_{k2^{-n}-s})(B_{(k+1)2^{-n}}-B_{k2^{-n}})\\ $$

and see that it's the limit of random variables $X_n$, with $\mathbb{E}(X_n) = 0$ and $\text{Var}(X_n)\to 0$. (Then show convergence in the usual way, with the first Borel-Cantelli lemma.)

When I first tried this question, I didn't do this - I tried to show that $MB$ was a continuous local martingale. The trouble with this - as far as I can see - is that you have to find a filtration with respect to which both $M$ and $B$ are semimartingales. $M$ doesn't seem to be a semimartingale w.r.t. the natural filtration of $B$.


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