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 $t$..
 A: For every nonnegative $t$, let $Z_t=B_{1-\mathrm e^{-t}}=\displaystyle\int_0^{1-\mathrm e^{-t}}\mathrm dB_s$. 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 $Z_t=\displaystyle\int_0^t\mathrm e^{-s/2}\mathrm d\beta_s$ for every nonnegative $t$.
In particular, $X_t=\displaystyle\mathrm e^{t/2}\int_0^t\mathrm e^{-s/2}\mathrm d\beta_s$ and
$$
\int_0^tX_u\mathrm du=\int_0^t\mathrm e^{u/2}\int\limits_0^u\mathrm e^{-s/2}\mathrm d\beta_s\mathrm du=\int_0^t\mathrm e^{-s/2}\int_s^t\mathrm e^{u/2}\mathrm du\mathrm d\beta_s,
$$
hence
$$
\int_0^tX_u\mathrm du=\int_0^t\mathrm e^{-s/2}2(\mathrm e^{t/2}-\mathrm e^{s/2})\mathrm d\beta_s=2\mathrm e^{t/2}\int_0^t\mathrm e^{-s/2}\mathrm d\beta_s-2\beta_t=2X_t-2\beta_t.
$$
This proves that $Y_t=X_t-\displaystyle\frac12\int\limits_0^tX_u\mathrm du=\beta_t$ and that $(Y_t)_{t\geqslant0}$ is a standard Brownian motion.
A: 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.
