# Standard Brownian Motion

Let $\{X_t,t\ge 0\}$ be a standard Brownian motion. Compute the density of $X_t$ conditioned by $X_{t_1}$ and $X_{t_2}$ assuming that $t_1 <t<t_2$.

Can anyone give me some hint to start the question?

• $(X_{t_1},X_t,X_{t_2})$ is a Gaussian random variable, with values in $\mathbb{R}^3$; you know its mean vector and variance matrix. You want the conditional distribution of $X_t | X_{t_1}, X_{t_2}$. – Vincent Zoonekynd May 6 '13 at 8:20

Let $f_{t}(x) = \frac{1}{\sqrt{2\pi t}}e^{-x^2/2t}$ be the $N(0,t)$ density, and let $f_{s,t}$ be the joint density of $(X_s, X_t)$. If $s<t$, then since the bijection $(X_s,X_t - X_s) \mapsto (X_s,X_t)$ has inverse $(u,v) \mapsto (u, v-u)$ with jacobian determinant $1$, $$f_{s,t}(x,y) = f_{s,t-s}(x,y-x) = f_s(x)f_{t-s}(y-x).$$ Then $$f_{t|s}(x|y) = \frac{f_{s,t}(x,y)}{f_{s}(x)} = f_{t - s}(y-x) = \frac{1}{\sqrt{2\pi(t-s)}}e^{-(y-x)^2/2(t-s)}.$$

Apply this to $(X_{t_1},X_t,X_{t_2})$: the bijective map $(x,y,z) \mapsto (x,y-x,z-y)$ has inverse $(u,v,w) \mapsto (u,u+v,u+v+w)$ with jacobian determinant $1$, and so for $s < t < u$ $$f_{s,t,u}(x,y,z) = f_{s,t-s,u-t}(x,y-x,z-y) = f_s(x)f_{t-s}(y-x)f_{u-t}(z-y).$$ Then $$f_{t|s,u}(y|x,z) = \frac{f_{s,t,u}(x,y,z)}{f_{s,u}(x,y)} = \frac{f_s(x)f_{t-s}(y-x)f_{u-t}(z-y)}{f_{s,u}(x,y)} = \frac{f_{t-s}(y-x)f_{u-t}(z-y)}{f_{u-s}(y-x)}.$$

There is no need to delve into the specifics of gaussian densities here since, when restricted to some gaussian family, conditionings amount to linear algebra computations (one reason of the success of gaussian models in appplications).

In the present case, the triplet $(X_t,X_{t_1},X_{t_2})$ is centered normal hence, conditionally on $(X_{t_1},X_{t_2})$, $X_t$ is distributed like $E[X_t\mid X_{t_1},X_{t_2}]+\sigma Z$ for some $\sigma$, where $Z$ is standard normal independent of $(X_{t_1},X_{t_2})$.

One knows that $E[X_t\mid X_{t_1},X_{t_2}]=a_1X_{t_1}+a_2X_{t_2}$ for some $(a_1,a_2)$ and that the variance of $a_1X_{t_1}+a_2X_{t_2}+\sigma Z$ should equal the variance $t$ of $X_t$.

All this plus the fact that $E[X_uX_v]=\min(u,v)$ for every $(u,v)$, allows to compute $(a_1,a_2,\sigma)$. Conditionally on $(X_{t_1},X_{t_2})$, $X_t$ is normal with mean $a_1X_{t_1}+a_2X_{t_2}$ and variance $\sigma^2$. Numerically, $$a_1=\frac{t_2-t}{t_2-t_1},\qquad a_2=\frac{t-t_1}{t_2-t_1},\qquad \sigma^2=\frac{(t_2-t)(t-t_1)}{t_2-t_1}.$$ Edit: This is how Paul Lévy constructed Brownian motion $B$ on the time interval $[0,1]$, starting from $B_0=0$, from $B_1$ standard normal, then drawing $B_{1/2}$ as $\frac12(B_0+B_1)$ plus half an independent standard normal, then $B_{1/4}$ and $B_{3/4}$ conditionally on $(B_0,B_{1/2},B_1)$, and so on for every dyadic time in $[0,1]$.

• Shouldn't it read $\sigma^2 = \frac{(t_2-t)(t-t_1)}{t_2-t_1}$...? – saz May 13 '13 at 11:49
• @saz Indeed, if only for dimensional analysis reasons... Thanks. – Did May 13 '13 at 13:43