Acted notes – page 39 – Q 10.1 (ii) (e) When applying ito’s lemma, we note that dBt = 1*dBt + 0*dt Equivalently, mu(t)=0 and sigma(t)=1 Why/how is this so? Is this due to Bt is normally distributed with mean=0 and sigma=1? thanks bobby
It is because dBt is a standard brownian motion which follows normal distributed with parameter mu = 0 and sigma =1
Hi Bobby, It is 'so' by inspection. dBt = something * dt + something else * dBt How do we make the LHS = RHS? Well something = 0 and something else = 1 Standard Brownian motion has a drift of 0 and a volatility of 1. We can instead say this sentence by writing Bt - Bs ~ N(0,t-s) but we don't need to use this fact to inspect the above. Just look at it, something = 0 and something else = 1 John