 Solution

Let $(B_t{)}_{0\le t\le 1}$ be standard Brownian motion with its continuous version, and defineEvery finite vector is a linear transform of a Gaussian Brownian vector, so $Z$ is a centred Gaussian process. Using $\mathbb{E}[B_s{B}_t]=\min (s,t)$,For $s\le t$, this is $s(1-t)$, the prescribed Brownian bridge covariance kernel. Its paths are continuous on the probability-one event of Brownian continuity, because $t\mapsto t{B}_1$ is continuous on every sample. The construction also gives $Z_0=Z_1=0$. Thus it realizes a Brownian bridge with almost surely continuous paths, and part (i) identifies its process law.