Covariance of Brownian Bridge? 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!

I think the given representation of the Brownian Bridge is not correct. It should read
$$Y_t = a \cdot (1-t) + b \cdot t + (1-t) \cdot \underbrace{\int_0^t \frac{1}{1-s} \, dB_s}_{=:I_t} \tag{1}$$
instead. Moreover, the covariance is defined as $\mathbb{E}((Y_t-\mathbb{E}Y_t) \cdot (Y_s-\mathbb{E}Y_s))$, so you forgot to subtract the expectation value of $Y$ (note that $\mathbb{E}Y_t \not= 0$).
Here is a proof using the representation given in $(1)$:
$$\begin{align} \mathbb{E}Y_t &= a \cdot (1-t) + b \cdot t \\
\Rightarrow \text{cov}(Y_s,Y_t) &= \mathbb{E}((Y_t-\mathbb{E}Y_t) \cdot (Y_s-\mathbb{E}Y_s)) = (1-t) \cdot (1-s) \cdot \mathbb{E}(I_t \cdot I_s) \\ &= (1-t) \cdot (1-s) \cdot \underbrace{\mathbb{E}((I_t-I_s) \cdot I_s)}_{\mathbb{E}(I_t-I_s) \cdot \mathbb{E}I_s = 0} + (1-t) \cdot (1-s) \mathbb{E}(I_s^2) \tag{2} \end{align}$$
for $s \leq t$ where we used the independence of $I_t-I_s$ and $I_s$. By Itô's isometry, we obtain
$$\mathbb{E}(I_s^2) = \int_0^s \frac{1}{(1-r)^2} \, dr = \frac{1}{1-s} -1.$$
Thus we conclude from $(2)$:
$$\text{cov}(Y_t,Y_s) = (1-t) \cdot (1-s) \cdot \left( \frac{1}{1-s}-1 \right) = s-t \cdot s = s \cdot (1-t).$$
