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}
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 ${(t-s)^3}/{12}$.