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Given that $\{B_t,t\ge0\}$ is a standard Brownian process. What is the conditional distribution of $B(s)$ given $B(t_1)=x_1$ and $B(t_2)=x_2$, for $0<t_1<s<t_2$?

My try: First i tried to write $B(s)=B(s)-B(t_1)+B(t_1)$ and then try to find the conditional distribution $\Pr(B(s),B(t_1)=x_1,B(t_2)=x_2)$ then we get $P(B(s)-x_1+x_1,B(t_1)=x_1,B(t_2)=x_2)$-but then i have no idea how to proceed. I am thinking of a 2nd apporach, using the moment generating function but seems that it is not easy to get the moment generating function of a condition random variable, as i don't even know how its distributed.

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Would you happen to know the fact that $(B_s)_{t_1\leq s\leq t_2}\mid B_{t_1}=x_1,\,B_{t_2}=x_2$ is a Brownian bridge? – Stefan Hansen Dec 21 '12 at 10:29
@StefanHansen I heard about the brownian bridge but don't know that the above example is a Brownian bridge – Mathematics Dec 21 '12 at 16:56

One approach uses joint distributions to calculate the conditional distribution:

Since $(B_t)_{t \geq 0}$ is a Brownian motion we know that: $$\begin{equation*} (B_{t_1},B_s,B_{t_2}) \sim N(0,C) \qquad \text{mit} \, \, C:= \begin{pmatrix} t_1 & t_1 & t_1 \\ t_1 & s & s \\ t_1 & s & t_2 \end{pmatrix} \end{equation*}$$ The density of the joint distribution is given by$$\begin{equation*} p_{t_1,s,t_2}(x) = \frac{1}{(2\pi)^{\frac{3}{2}}} \cdot\frac{1}{\sqrt{(t_2-s) \cdot (s-t_1) \cdot t_1}} \cdot \exp \left(- \frac{1}{2}\cdot\left( \frac{(x_3-x_2)^2}{t_2-s} + \frac{(x_2-x_1)^2}{s-t_1}+ \frac{x_1^2}{t_1} \right) \right) \end{equation*}$$ where $x=(x_1,x_2,x_3)$. Now we can calculate the conditional density $$\begin{align*} p_{s|t_1,t_2}(x) &= \frac{p_{t_1,s,t_2}(x_1,x_2,x_3)}{p_{t_1,t_2}(x_1,x_3)} \\ &= \frac{1}{\sqrt{2\pi}} \cdot \sqrt{\frac{t_2-t_1}{(t_2-s) \cdot (s-t_1)}} \cdot \exp \left( - \frac{1}{2} \cdot \left( \frac{(x_3-x_2)^2}{t_2-s} + \frac{(x_2-x_1)^2}{s-t_1} - \frac{(x_3-x_1)^2}{t_2-t_1} \right) \right) \end{align*}$$ where $p_{t_1,t_2}$ denotes the joint distribution if $(B_{t_1},B_{t_2})$. Hence $$\mathbb{P}[B_s \in B|B_{t_1}=y_1,B_{t_2}=y_3] = \int 1_B(x) \cdot p_{s|t_1,t_2}(y_1,x,y_3) \, dx$$

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