Expectation for a Brownian Motion E(W(s)|W(t)] when s<t

Question

I was wondering what is the Expectation of a Brownian motion W(t) in the following case:

E[W(s)|W(t)] with s < t

My guess is to use Brownian Bridge but I have no clue at all of the answer. Does someone have any idea?

Thanks in advance,

JC

Answer 1

We know $$f_{W(t)}(y)=\frac{1}{2\sqrt{\pi\,t}}\text{exp}\left(-\frac{y^2}{2t}\right)$$ and $$f_{W(s),W(t)}(x,y)=\frac{1}{2\pi\sqrt{s(t-s)}}\text{exp}\left(-\frac{x^2}{2s}-\frac{(y-x)^2}{2(t-s)}\right)$$ therefore $$f_{W(s)|W(t)}(y)=\frac{f_{W(s),W(t)}(x,y)}{f_{W(t)}(y)}=\frac{1}{\sqrt{2\pi}}\frac{\sqrt{t}}{\sqrt{s(t-s)}}\text{exp}\left(-\frac{x^2}{2s}-\frac{(y-x)^2}{2(t-s)}+\frac{y^2}{2t}\right)$$ In other words $$f_{W(s)|W(t)}(y)=\frac{1}{\sqrt{2\pi}}\frac{1}{\sqrt{s-\frac{s}{t}}}\text{exp}\left(-\frac{\left(x-\frac{s}{t}y\right)^2}{2\left(s-\frac{s}{t}\right)}\right)$$ thus $$\mathbb{E}[W_s|W_t=y]=\frac{s}{t}y$$ moreover $$\mathbb{E}[W_t|W_s=x]=x$$ Other way $$\mathbb{E}\left[W_s\middle | W_t\right]=\mathbb{E}\left[W_s-\frac{s}{t}W_t\middle |\, W_t\right]+\frac{s}{t}W_t=\frac{s}{t}W_t$$