# Brownian Motion and $\mathcal{F}_t$ brownian motion

I am trying to figure out why a $$\mathcal{F}_t^B$$ Brownian motion is also a Brownian motion in the regular sense. Here $$\mathcal F_t^B$$ is just the natural filtration.

I am getting confused with the definition of the $$\mathcal F_t^B$$-Brownian motion. The only definition I have is that for some bounded measurable $$f$$ $$E[B_t\vert \mathcal F_s]=P_{t-s}f(B_s).$$

• What exactly is $$P_{t-s}f(B_s)$$?
• What would be a smart approach? The usual definition with independent increments and normal distribution or rather showing it is a Gaussian process with correct covariance?
• What exactly is your definition of an $\mathcal{F}_t^B$ Brownian motion? A stochastic process satisfying $\mathbb{E}(f(B_t) \mid \mathcal{F}_s) = P_{t-s} f(B_s)$ does, in general, not need to be a Brownian motion (but just a Markov process). – saz Jan 14 '17 at 6:38
• The definition I have is a continuous time stochastic process $B_t$ such that $E[f(B_t)|\mathcal F_s^B]=P_{t-s} f(B_s)$ where we use the natural filtration. – Bennibenben Jan 14 '17 at 11:51
• @saz can you help me out with what the definition of $P_{t-s} f(B_s)$ is? – Bennibenben Jan 14 '17 at 11:57
• Typically, $P_t$ denotes the semigroup, i.e. $P_t f(x) = \mathbb{E}f(x+B_t)$ (...but why do you ask me? I'm pretty sure that this is explained somewhere in the notes/book/... you are using). – saz Jan 14 '17 at 12:28