2015 Q&A Bank 2.10 vs 2.14 Confused about Bt and dBt notation: It appears that the E[Bt] and E[dBt] are treated in the same way - I'm confused about to what I can apply the fact that Bt - Bs ~ N(0,t-s). Is the comment in the solution that "the dBt's are random quantities with mean 0 and variance dt" and application of the above (i.e. in this scenario, dt is the change in time which is the 't-s')?
Looks like you've got it to me! Let's start with the definition of \(dB_t\): \[ dB_t=B_{t+dt} - B_{t} \] which is just an increment of standard Brownian motion. Therefore we have: \[ B_{t+dt} - B_{t} \sim N(0,t+dt-t) \] ie \[ dB_t\sim N(0,dt). \]
