# How can I prove this random process to be Standard Brownian Motion?

$B_t,t\ge 0$ 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

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.
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.
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