# Acted notes – page 39 – Q 10.1 (ii) (e)

Discussion in 'CM2' started by bobby sanger, Mar 18, 2020.

1. ### bobby sangerMade first post

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

2. ### anees aslamMember

It is because dBt is a standard brownian motion which follows normal distributed with parameter mu = 0 and sigma =1

3. ### bobby sangerMade first post

great thank you anees!

4. ### John PotterActEd TutorStaff Member

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