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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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    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$2011-09-11

2 Answers 2

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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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    I think that's not difficult. You can consider the Riemann Summation2011-09-12
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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.

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    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.)2011-10-11