 Solution

By the Markov property, all factors in the stationary likelihood function that involve $X_2$ are the transitions from $X_1$ to $X_2$ and from $X_2$ to $X_3$. Hence conditioning on later observations adds no information once the two neighbors are known. Writing $u=X_2-\mu$, $u_1=x_1-\mu$ and $u_3=x_3-\mu$, its conditional density is proportional toThe quadratic coefficient of $u$ is $1+{\varphi }^2$ and the linear coefficient is $-2\varphi (u_1+u_3)$. Completing the square therefore gives the Gaussian AR1 bridgeHere ${\sigma }^2$ is the innovation variance, not the marginal stationary variance.