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?
  • $\begingroup$ 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). $\endgroup$ – saz Jan 14 '17 at 6:38
  • $\begingroup$ 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. $\endgroup$ – Bennibenben Jan 14 '17 at 11:51
  • $\begingroup$ @saz can you help me out with what the definition of $P_{t-s} f(B_s)$ is? $\endgroup$ – Bennibenben Jan 14 '17 at 11:57
  • $\begingroup$ 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). $\endgroup$ – saz Jan 14 '17 at 12:28

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