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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)

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$X(t) \sim N(0,t)$ 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.

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