Why is $X_t=\sum\limits_{i=1}^n \int _0 ^t \frac{B_s^i}{\sqrt{(B_s^1)^2+\cdots + (B_s^n)^2}}dB_s^i$ a Brownian motion?

39 Views Asked by At

If $B_t^1,...,B_t^n$ are Brownian motions, why is $$X=\sum_{i=1}^n \int _0 ^t \frac{B_s^i}{\sqrt{(B_s^1)^2+\cdots + (B_s^n)^2}}dB_s^i$$ also a Brownian motion?

By the product rule, $X=\prod_1^n\frac{(B_s^i)^2}{(B_s^1)^2+\cdots + (B_s^n)^2}$. The product of Brownian motions is also a Brownian motion, but I don't think these fractions in the product are Brownian motions.