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

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

  1. bobby sanger

    bobby sanger Made 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
     
    bobby sanger, Mar 18, 2020
    #1
  2. anees aslam

    anees aslam Member

    It is because dBt is a standard brownian motion which follows normal distributed with parameter mu = 0 and sigma =1
     
    anees aslam, Mar 18, 2020
    #2
  3. bobby sanger

    bobby sanger Made first post

    great thank you anees!
     
    bobby sanger, Mar 19, 2020
    #3
  4. John Potter

    John Potter ActEd Tutor Staff 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
     
    John Potter, Mar 25, 2020
    #4

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