CT8 - April 2018 - Q3

Discussion in 'CM2' started by Niall1234, Mar 24, 2023.

  1. Niall1234

    Niall1234 Made first post

    I'm having trouble understanding the solution to part (ii).

    It's only worth 2 marks so should be straight forward.

    I have to find E[St] = E[exp^B(5)], where B(5) is standard brownian motion.

    From the notes, the expected value of a security price at time t using Geometric brownian motion is :

    E[St] = exp((Zo + mu*t) + 0.5 *sigma^2*t)

    As B(5) - N(0, 5), I would expect that, E[S(5)] = exp((0+0*5)+0.5*5*5)= exp(12.5),

    Where, mu=0, sigma^2=5, t=5, Zo=0

    However in the solutions they end up with a much different answer, and E[S(5)] = exp(0.5*5*1^2), so I'm not sure where the 1^2 is coming from.

    Any help would be greatly appreciated.
     
    Bill SD likes this.
  2. CapitalActuary

    CapitalActuary Ton up Member

    You’re confusing different notation here. The mu and sigma in the notes’ formula for E[St] are the drift and volatility parameters for a Brownian motion.

    For standard Brownian motion, drift=0 and volatility=1, so the equation says E[St]=exp(0.5*t). Here we’re at t=5, so the answer is exp(0.5*5).

    Equivalently you could note that B(5)~N{0,5}, so exp(B(5)) is lognormal with parameters mu=0 and sigma=sqrt(5). Here mu and sigma are different concepts; they’re the parameters of a lognormal distribution, not the drift and volatility of Brownian motion. Then using the formula for expected value of a lognormal variable, again we get exp(0.5*5).

    When you included the “5” term twice in your formula you were double counting by thinking of sigma as a volatility parameter=5, which is not correct.

    The 1^2 term comes from sigma=volatility parameter=1, t=5, and E[St]=exp(0.5*t*sigma^2).
     
    Bill SD, Niall1234 and Steve Hales like this.

Share This Page