# Conditional distribution of integral of brownian motion

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}

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