x*****i
发帖数: 287 | 1 W_1 and W_2 are independent Brownian motions,
prove \int^T_0 h*dW_1 + \int^T_0 \sqrt(1-h^2) * dW_2 is a Brownian motion.
Of course, if h is a small constant, straightforward.
how about h is also a adapted random process ? Assume -1 < h < 1 .
Thanks a lot! | L**********u
发帖数: 194 | 2 let
W(t):=h*W_1 + \sqrt(1-h^2) * W_2
1, W(0)=0,
2, W(t) is continuous with respect to almost surely.
3, It is a martingale obviously
4, quadratic variation is t since
dW(t)=h*dW1+\sqrt(1-h^2)dW2,
dW*dW=h^2dt+(1-h^2)dt=dt.
By Levy theorem, W(t) is a Brownian motion
【 在 xiguapi (guagua) 的大作中提到: 】 | x*****i
发帖数: 287 | 3 Thanks a lot. I forgot the Levy theorem, just struggling to prove the
increment is normally distributed. | L**********u
发帖数: 194 | 4 I forget the integral in the definition, but it does not matter.
the proof is still true | x*****i
发帖数: 287 | 5 It is the same. Prove W^2- t is a martingale. | L**********u
发帖数: 194 | 6 这个就是直接求导而已，屁都没有
【在 x*****i 的大作中提到】
: It is the same. Prove W^2- t is a martingale.
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