TITLE: Conditional distribution of integral of brownian motion
QUESTION [8 upvotes]: I am trying to calculate the conditional distribution of
$$\biggl( \int_s^t W_u du \; \biggl| \; W_s = x, W_t= y\biggl) $$
where $W$ is a Standard Brownian Motion and $s\leq u \leq t$.
Any help would be greatly appreciated :)

My approach is the following: Using this  helpful answer I can show that
$$ \mathcal{L}\biggl( W_u \; \biggl| \; W_s = x, W_t= y\biggl) \sim \mathcal{N}\biggl( \frac{t-u}{t-s}x+\frac{u-s}{t-s}y, \frac{(t-u)(u-s)}{t-s} \biggl) $$
and thus
\begin{align}
\mathbb{E}\biggl[\int_s^t W_u du \; \biggl| \; W_s = x, W_t= y \biggl]
&= \int_s^t \mathbb{E}[W_u \; | \; W_s = x, W_t= y] du \\
&= \int_s^t \biggl(\frac{t-u}{t-s}x+\frac{u-s}{t-s}y\biggl) du \\
&= \frac{t-s}{2}x+ \frac{t-s}{2}y
\end{align}
However, I struggle with calculating the variance... Is the following correct so far?
\begin{align}
Var\biggl[\int_s^t W_u du \; \biggl| \; W_s = x, W_t= y \biggl] 
&= \mathbb{E}\biggl[\biggl(\int_s^t W_u du \biggl)^2\; \biggl| \; W_s = x, W_t= y \biggl]  - \biggl(\mathbb{E}\biggl[\int_s^t W_u du \; \biggl| \; W_s = x, W_t= y \biggl]\biggl)^2 \\
&= \mathbb{E}\biggl[\int_s^t \int_s^t W_v W_u du dv\; \biggl| \; W_s = x, W_t= y \biggl]  - \biggl(\frac{t-s}{2}x+ \frac{t-s}{2}y\biggl)^2 \\
&= \int_s^t \int_s^t \mathbb{E}[W_v W_u | \; W_s = x, W_t= y]du dv  - \biggl(\frac{t-s}{2}x+ \frac{t-s}{2}y\biggl)^2 
\end{align}

REPLY [3 votes]: As mentioned in the comments, the variance is independent of $x$ and $y$, so we can assume $x=y=0$. Also set for simlicity $s=0, t=1$. Then, conditionally on $W_0=W_1=0$, $W$ is a standard Brownian bridge, so
$$
\mathrm{Var}\left(\int_0^1 W_u du\,\middle|\, W_0=W_1=0 \right) = \mathrm{E}\left(\int_0^1\int_0^1 W_z W_u du\,dz\,\middle|\, W_0=W_1=0 \right) \\
=2 \int_0^1\int_0^z u (1-z) du\,dz = \int_0^1 z^2(1-z)dz = \frac1{12}.
$$
Thanks to self-similarity and homogeneity of increments, for general $s,t$ the answer is ${(t-s)^3}/{12}$.