Consider the Brownian $W_u$. Suppose you are only considering realizations of this brownian that verify both $W_0=0$ and, for a specific (given) $t$, $W_t=a$.
Under these specific conditions, what is the conditional mean and variance of $I=\int_0^tW_udu$ ?
Edit : after a bit more search, I have learned that $W_u$ is then called a brownian bridge and managed to produce some results, although I am basically a beginner in stochastic calculus (don't have and never had any kind of course on the subject).
I have (I believe) no trouble proving that $I$ is gaussian.
For the mean, I obtain
$$E[I|W_t]=\int\limits_0^tE[W_u|W_t]du=\int\limits_0^t\frac{uW_t}{t}du=\frac{tW_t}{2}$$
For the variance, I obtain
$$var(I|W_t)=cov\left(\int\limits_0^tW_udu|W_t,\int\limits_0^tW_vdv|W_t\right)$$ $$=\int\limits_0^t\int\limits_0^tcov(W_udu|W_t,W_vdv|W_t)$$ $$=\int\limits_0^t\int\limits_0^t\left(min(u,v)-\frac{uv}{t}\right)dudv$$ $$=\frac{t^3}{12}$$
Is this correct ?