I am confused by this question. We all know that Brownian Bridge can also be expressed as:
$Y_t=bt+(1−t)\int_a^b \! \frac{1}{1-s} \, \mathrm{d} B_s $
Where the Brownian motion will end at b at $t = 1$ almost surely. Hence I can write it as:
$Y_t = bt + I(t)$
where $I(t)$ is a stochastic integral, and in this case it is a martingale. Since it is a martingale, the co-variance can be calculated as:
\begin{array} {lcl} E[Y_t Y_s] & = & b^2 ts + E(I(t)I(s)] \\ & = & b^2 ts + E\{(I(t)-I(s))* I(s) \} + E [I(s)^2] \\& =&b^2 ts + Var[I(s)] + b^2s^2 \\ & = & b^2 ts + b^2 s^2 + s(1-s) \end{array}
Hence the variance is just $ b^2 s^2 + s(1-s)$. However I read online, the co-variance of the Brownian Bridge should be $s(1-t)$. I am relaly confused. Please advise. Thanks so much!