Brownian motion

[Benjamin]

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')?

 

[Steve Hales]

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
\[
dB_t\sim N(0,dt).
\]