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I know that given a Brownian motion $W(t)$, its quadratic variation is $$[W,W](t) = t$$

Then for a brownian bridge, $X(t) := W(t) - \frac{t}{T} W(T) $, what is its quadratic variation?

athos
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  • Use the fact that quadratic variation is bilinear and that the process $t\to \frac{t}{T}W_T$ has finite variation which kills all covariations with that term. You should end up with something eerily familiar. – James Yang Oct 06 '18 at 16:15
  • @JamesYang is it also $t$? – athos Oct 14 '18 at 04:50
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    Yes! In general if you have $X_t = Y_t + A_t$ where $A_t$ is a finite variation process, $\langle X \rangle_t = \langle Y \rangle_t$, where $\langle \cdot \rangle_t$ is quadratic variation, by the same argument. – James Yang Oct 14 '18 at 16:39
  • thanks so much@JamesYang – athos Oct 15 '18 at 04:13

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