Conditional Probability of two Standard Brownian Motions

Question

I am trying to prove this statement: if s < t, then $$P(B_{s}|B_{t} = z)$$ is also a Normal distribution with mean of $$\frac{zs}{t}$$ and a variance of $$\frac{s(t-s)}{t}$$ What I have been doing is by definition, I have $$P(B_{s}|B_{t} = z) = \frac {P(B(s) \wedge B(t) = z)}{P(B_{t}) = z} = \frac{P(B_{s} \in [y, y+\delta y] \wedge B_{t} \in [z,z+\delta z])}{P(B_{t} = z)} =\frac{P(B_{s} \in [y, y+\delta y] \wedge B_{t-s} \in [z-y,z-y+\delta (z-y)])}{P(B_{t} = z)} $$ How should I proceed to the next step? Thank you.

Answer 1

Try using Bayes' Theorem (reference: Wikipedia). You know the density functions of $ B_s $ and $ B_t $, and can use the property of Brownian Motion to easily find the conditional density of $ B_t $ given $ B_s $.