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$B_t,t\ge 0$ is a standard Brownian Motion. Then define $X(t)=e^{t/2}B_{1-e^{-t}}$ and $Y_t=X_t-\frac{1}{2}\int_0^t X_u du$. The question is to show that $Y_t, t\ge 0$ is a standard Brownian Motion.

I tried to calculate the variance of $Y_t$ for given $t$, but failed to get $t$..

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Is this homework? If so, it should be tagged as such. :) –  cardinal Sep 10 '11 at 17:23
    
I concur with @cardinal's suggestion. Having dealt with it, you could indicate what you tried to compute $E(X(t)X(s))$ and then $E(Y_tY_s)$. –  Did Sep 10 '11 at 19:37
    
Right. So now, what is $E(X(t)X(s))$? Before that (maybe), what are $E(X(t))$ and $E(X(t)^2)$? –  Did Sep 11 '11 at 9:56
    
yes I calculated $E(X(t))$ which is $\mathbb{E}[e^{t/2}B_{1-e^{-t}}]=e^{t/2}\mathbb{E}[B_{1-e^{-t}}]=0$ and \begin{align*}\mbox{Cov}(X_s,X_t)=\mathbb{E}(X_sX_t)&=\mathbb{E}[e^{s/2}B_{1-e^{‌​-s}}\cdot e^{t/2}B_{1-e^{-t}}]\\ &=e^{(s+t)/2}\mathbb{E}[B_{1-e^{-s}}B_{1-e^{-t}}]\\ &=e^{(s+t)/2}\min(1-e^{-s},1-e^{-t})\\ &=e^{(s+t)/2}(1-e^{-\min(s,t)}).\end{align*} But when I go to the $EY_t^2$, I dont know how to make use of the relation between $X_t$ and $Y_t$ –  Julie Sep 11 '11 at 13:40

2 Answers 2

For every nonnegative $t$, let $Z_t=B_{1-\mathrm e^{-t}}=\int\limits_0^{1-\mathrm e^{-t}}\mathrm dB_s$. Then $(Z_t)_{t\geqslant0}$ is a Brownian martingale and $\mathrm d\langle Z\rangle_t=\mathrm e^{-t}\mathrm dt$ hence there exists a Brownian motion $(\beta_t)_{t\geqslant0}$ starting from $\beta_0=0$ such that $Z_t=\int\limits_0^t\mathrm e^{-s/2}\mathrm d\beta_s$ for every nonnegative $t$. In particular, $X_t=\mathrm e^{t/2}\int\limits_0^t\mathrm e^{-s/2}\mathrm d\beta_s$ and $$ \int\limits_0^tX_u\mathrm du=\int\limits_0^t\mathrm e^{u/2}\int\limits_0^u\mathrm e^{-s/2}\mathrm d\beta_s\mathrm du=\int\limits_0^t\mathrm e^{-s/2}\int\limits_s^t\mathrm e^{u/2}\mathrm du\mathrm d\beta_s, $$ hence $$ \int\limits_0^tX_u\mathrm du=\int\limits_0^t\mathrm e^{-s/2}2(\mathrm e^{t/2}-\mathrm e^{s/2})\mathrm d\beta_s=2\mathrm e^{t/2}\int\limits_0^t\mathrm e^{-s/2}\mathrm d\beta_s-2\beta_t=2X_t-2\beta_t. $$ This proves that $Y_t=X_t-\frac12\int\limits_0^tX_u\mathrm du=\beta_t$ and that $(Y_t)_{t\geqslant0}$ is a standard Brownian motion.

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It seems that this is more advanced than what I've learnt. Is this stochastic integrals? –  Julie Oct 11 '11 at 22:42
    
Dear Zoe: how could I know what you have learnt and what you have not since you chose to remain completely and absolutely silent on this? (Anyway, the appearance of my post has had the side effect of making you accept your own answer to your question.) –  Did Oct 11 '11 at 23:34
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Calculate the covariance $E(Y_s,Y_t)$, and it is $min(s,t)$. But the algebra is really tedious, I wonder whether there is other simpler way to show it.

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Moreover, you'd still have to show that $Y_t$ is a Gaussian process, which doesn't seem obvious. –  Nate Eldredge Sep 11 '11 at 22:29
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I think that's not difficult. You can consider the Riemann Summation –  Julie Sep 12 '11 at 3:58

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