Let $B = (B_t : t \geq 0)$ be a standard Brownian Motion. Fix $0 \leq s \leq t$.
How can I prove that, conditionally on $\{B_s = x, B_t = z\}$, the intermediate value $$B_{\frac{t+s}{2}}$$ has Gaussian distribution with mean $\left(\frac{x+z}{2}\right)$ and variance $\left(\frac{t-s}{4}\right)$ ?.
It looks like it is a bit more tricky than it actually seems at first sight.
Thanks in advance for your answers.
We introduce the random variable
$$ Z=B_{(t+s)/2}+\frac{x+z}{2}-\frac12(B_t+B_s), $$
Then, $Z$ is a Gaussian random variable since $(B_t)_{t\ge0}$ is a Gaussian process. Additionally, $Z$ is independent from $B_t$ and $B_s$ since $\mathbb E[ZB_s]=E[ZB_t]=0=E[Z]E[B_s]=E[Z]E[B_t]$. This implies that $Z$ has the same distribution as $B_{(t+s)/2}$ conditional on $B_t=x$ and $B_s=z$.
It remains to check that $E[Z]=\frac{x+z}{2}$ and \begin{align*} \mathrm{Var}\,{Z}&=E\left[\left(B_{(t+s)/2}-\frac12(B_t+B_s)\right)^2\right]\\ &=E\left[\left(B_{(t+s)/2}\right)^2\right]+\frac14E\left[B_t^2+B_s^2+2B_tB_s\right]-E\left[\left(B_{(t+s)/2}\right)(B_t+B_s)\right]\\ &=\frac{t+s}2+\frac14(t+s+2s)-\frac{t+s}2-s\\ &=\frac{t-s}{4}. \end{align*}