# Distribution of Conditional Brownian Motion

Let $$\ X(t),t \ge 0$$ be a Brownian motion process.

That is, $$\ X(t)$$ is a process with independent increments such that:

$$\ X(t) - X(s) \sim N(0,t-s), 0\le s \lt t$$
and $$\ X(0)= 0$$.

Derive the conditional distribution of $$\ X(s), s \lt t$$ conditional on $$\ X(t) = B$$ and state its mean and variance.

(I am pretty sure from looking online that the mean = $$\ {Bs\over t}$$ and variance = $$\ {s(t-s) \over t}$$ but I cannot derive the distribution to show this)

$$X(t) \sim N(0,t)$$ and $$X(s) \sim N(0,s)$$. $$\operatorname{Var}(X(t) - X(s)) = \operatorname{Var}(X(t))+\operatorname{Var}(X(s))-2\operatorname{Cov}(X(t),X(s)) = t - s$$ ==> $$\operatorname{Cov}(X(t),X(s))=s$$.
So $$\left( \begin{matrix} X(s)\\X(t) \end {matrix}\right) \sim N\left[\left( \begin{matrix} 0\\0 \end {matrix}\right),\left( \begin{matrix} s&s\\s&t \end {matrix}\right)\right]$$
Following $$Y|X=x∼N\left(μ_Y+ρ\frac {σ_Y}{σ_X}(x−μ_X),σ_Y^2(1−ρ^2)\right)$$, you get your results.