We know that the Brownian motion has independent increment. So if we have the following: $$E[tB_t-B_s|\mathcal F_s]$$ which are the problem that arise? can I say that $E[tB_t-B_s|\mathcal F_s]= E[tB_t-B_s]$ and what is the distribution of $tB_t-B_s$?
Brownian motion and independent increment
brownian motionindependencerandom variables
Best Answer
No, it is not true. Let us transform left-hand side a bit: $$\mathbb{E}[tB_{t}-B_{s}|\mathcal{F}_{s}]=\mathbb{E}[tB_{t}-(t-t+1)B_{s}|\mathcal{F}_{s}]$$ $$=t\cdot \mathbb{E}[B_{t}-B_{s}|\mathcal{F_{s}}]+\mathbb{E}[(t-1)B_{s}|\mathcal{F_{s}}]$$ $$=t\cdot \mathbb{E}[B_{t}-B_{s}]+(t-1)B_{s}.$$
Note that, $E[tB_t-B_s]$ from Your proposal is a constant and $$t\cdot \mathbb{E}[B_{t}-B_{s}]+(t-1)B_{s}$$ is not - because of $(t-1)B_{s}$ part.
We know that $B_{s}\sim \mathcal{N}(0,s)$ and hence:
$$t\cdot \mathbb{E}[B_{t}-B_{s}]+(t-1)B_{s} \sim \mathcal{N}(t\cdot \mathbb{E}[B_{t}-B_{s}], (t-1)^2\cdot s).$$
Simillary, we have: $$tB_t-B_s=t(B_{t}-B_{s}) + (t-1)B_{s},$$ $$t(B_{t}-B_{s})\sim \mathcal{N}(0, t^{2}(t-s)),$$ $$(t-1)B_{s}\sim \mathcal{N}(0,(t-1)^{2}s),$$ and $$t(B_{t}-B_{s})\perp (t-1)B_{s}$$ so we can simply add up parameters.