Conditional distribution in Brownian motion

Question

I need to prove the following:

Let $X$ be a Brownian motion with drift $\mu$ and volatility $\sigma$. Pick three time points $s < u < t$. Then, the conditional distribution of $X_u$ given $X_s = x$ and $X_t = y$ is normal; in fact $$(X_u\mid X_s = x, X_t = y)\sim \mathcal{N}\left(\frac{t-u}{t-s}x+\frac{u-s}{t-s}y,\sigma^2\frac{(t-u)(u-s)}{t-s}\right)$$

I know the conditional distributions of Gaussian are also Gaussian. I also have formulae to find $\mathbb E(X\mid Y=y)$ and $\operatorname{Var}(X\mid Y=y)$ when $X$ and $Y$ are jointly Gaussian. However I cannot find a way to apply those results here. Could someone help me please?

Answer 1

You need to apply this theorem about how on conditioning a Gaussian vectors.

You are in the situation where you have this 3-d Gaussian vector $V=(X_s,X_u,X_t)$.

You will have to compute the mean vector and covariance matrix of $V$, and then applying the theorem to your situation.

Regards

Answer 2

Consider this theorem on conditional expectation of multivariate Gaussian distribution. Let $x=X_u$ and $Y=\left[X_s \quad X_t\right]^\top$. Let $X_0=0$, so that $\mu_X=0$ and $\mu_Y=\left[0 \quad 0\right]^\top$.

\begin{align} \Sigma_{XX}&=Cov(X_u,X_u)=\sigma^2u\\ \Sigma_{XY}&=E(XY^\top)=[E(X_uX_s)\quad E(X_uX_t)]=\sigma^2[s\quad u]\\ \Sigma_{YY}&=E(YY^\top)=\left[\begin{matrix}E(X_s^2) & E(X_sX_t) \\ E(X_tX_s) & E(X_t^2)\end{matrix}\right]=\sigma^2\left[\begin{matrix}s & s \\ s & t\end{matrix}\right] \end{align}

Plugging the formula, the random variable $X_u\mid (X_s=x,X_t=y)$ is also Gaussian with

• mean: $\Sigma_{XY}\Sigma_{YY}^{-1}\left[\begin{matrix}x \\ y\end{matrix}\right]=\frac{t-u}{t-s}x+\frac{u-s}{t-s}y$
• variance:$\Sigma_{XX}+\Sigma_{XY}\Sigma_{YY}^{-1}\Sigma_{YX}=\sigma^2u+\sigma^2[s\quad u]\left[\begin{matrix}s & s \\ s & t\end{matrix}\right]^{-1}\left[\begin{matrix}s \\ u \end{matrix}\right]=\sigma^2\frac{(t-u)(u-s)}{t-s}$