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$(B_t)_{t\geq 0}$ is a brownian motion.

I can't seem to find le covariance for $X_t$ and that's what I need to show they have the same distribution since both are gaussian. Is there another way to find it?

alalala
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1 Answers1

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You should mention that $Y_t$ and $X_t$ are both Brownian bridges.

Integration by parts $$\tag1 d\left\{\frac{B_t}{1-t}\right\}=\frac{B_t\,dt}{(1-t)^2}+\frac{dB_t}{1-t} $$ implies $$\tag2 B_t=(1-t)\int_0^t\frac{B_s\,ds}{(1-s)^2}+(1-t)\int_0^t\frac{dB_s}{1-s} $$ so that we see that $X_t$ is the Gaussian $$\tag3 (1-t)\int_0^t\frac{dB_s}{1-s}\,. $$ Further straightforward calculations show that it has the covariance function $$\tag4 \operatorname{Cov}[X_t,X_s]=\min(t,s)-ts $$ of the Brownian bridge.

Kurt G.
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