Let $W_t$ be a standard Brownian motion and let $$Z(t) = \left\{ \begin{array}{ll} (1-t)B\left(\frac{t}{1-t}\right) & t \in [0,1) \\ 1 & \, t=1 \\ \end{array} \right. $$ I don't get why $Z$ is a Brownian Bridge. I have to prove that $Cov(Z(t),Z(s))=\min\{s,t\}-st$ for all $s,t\in(0,1)$, and $E(Z(t))=0$ and that $Z(t)$ is a continuous Gauß process. The first two thinks a trivial but I don't get why $Z(t)$ is continuous for $t=1$.
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Does this answer your question? Proving $ X_t = W_t - \int_0^t \frac{X_s}{1 - s} , ds$ is the Brownian Bridge – Kurt G. Dec 13 '23 at 14:30
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According to the law of large numbers for Brownian motion $$ \lim_{s\to\infty}\frac{B_s}{s}=0\,. $$ Therefore $$ \lim_{t\to 1}\left(t\frac{B_\frac t{1-t}}{\frac t{1-t}}\right)=\lim_{t\to 1}(1-t)B_\frac t{1-t}=0\,. $$
Kurt G.
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