If we have sampled a Brownian motion at $t_i$, how can we get samples at the midpoints of $[t_{i-1},t_i]$ using a Brownian bridge? Suppose we have sampled a Brownian motion $(B_t)_{t\ge0}$ at $0=t_0<t_1<\cdots$. How can we obtain a sample at the midpoints of $[t_{i-1},t_i]$ from those samples? I've read that this is possible by a Brownian bridge, but wasn't able to figure out how exactly.


Could you please tell me how we prove that $(B_{t_1},B_{t_2},B_{t_3})$ is normally distributed?

The joint normal distribution stems from two aspects of Brownian motion:

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*The distribution of the increment/distance over a certain time is normal distributed (described by the diffusion equation). $$f(x,\Delta t) = \frac{1}{\sqrt{4\pi D}} \frac{e^{-\frac{(x - x_0)^2}{4 D \Delta t}}}{\sqrt{\Delta t}}$$ Einstein, Albert. "Über die von der molekularkinetischen Theorie der Wärme geforderte Bewegung von in ruhenden Flüssigkeiten suspendierten Teilchen." Annalen der physik 4 (1905).


*The displacements in non-overlapping time intervals is independent (Markov property, the future changes are only dependent on the current state but and not on past states/history of the system)...
So the you can describe the changes in two subsequent intervals $B_{t_2}-B_{t_1} = \epsilon_a$ and $B_{t_3}-B_{t_2} = \epsilon_b$ as independent normal distributed variables (with variance depending on the length of the time interval and the diffusion constant). And the positions can be described as $B_{t_2} = B_{t_1} + \epsilon_a$ and $B_{t_3} = B_{t_1} + \epsilon_a + \epsilon_b$, which defines the joint normal distribution of $B_{t_2}$ and $B_{t_3}$ from which you can derive the conditional distribution of $B_{t_2}$ given $B_{t_3}$.

Above we simulated 200 Brownian motions as cumulative sums of Gaussian white noise. On the left you see the paths. On the right you see the joint distribution of the position at two different time points. Highlighted are paths for conditional on the position of $B_{t_3}$. The position of $B_{t_2}$ given the position of $B_{t_3}$ can be derived based on the joint distribution of the two.
