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Let $B_t$ be a Wiener Process, then $U_t=B_t-tB_1,~0\le t \le 1$ is a Brownian bridge.

Show that $X_t=(1+t)U_{{t}/({1+t})}$ is a Wiener Process. I'm not quite sure how to start this off.

Any help would be greatly appreciated.

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What is the expectation and variance of $X_t-X_s$? Are $X_t-X_s , X_s$ independent for $s<t$? Is $X_t$ continuous? Levy's Characterization shall do the rest: almostsure.wordpress.com/2010/04/13/… Alternately try Ito's Lemma directly. –  Alex R. Feb 1 '13 at 0:11

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The process $(X_t)_{t\geqslant0}$ is gaussian centered and starts from $X_0=U_0=B_0=0$. For $t\leqslant s$, the covariance of $X_t$ and $X_s$ is $(1+t)(1+s)$ times $$ \mathbb E((B_\tau-\tau B_1)(B_\sigma-\sigma B_1))=\mathbb E(B_\tau B_\sigma)-\tau \mathbb E(B_\sigma B_1)-\sigma \mathbb E(B_\tau B_1)+\tau\sigma\mathbb E(B_1^2), $$ where $\tau=t/(1+t)$ and $\sigma=s/(1+s)$, hence $\tau\leqslant\sigma$. Applying $\mathbb E(B_uB_v)=\min(u,v)$ for every $(u,v)$ to each term yields $$ \mathbb E((B_\tau-\tau B_1)(B_\sigma-\sigma B_1))=\tau-\tau\sigma-\tau\sigma+\tau\sigma=\tau(1-\sigma), $$ hence $$ \mathbb E(X_tX_s)=(1+t)(1+s)\tau(1-\sigma)=t. $$ This proves that the process $(X_t)_{t\geqslant0}$ is a standard Brownian motion.

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