September 2006, q8

Discussion in 'CT8' started by 12345, Apr 13, 2009.

  1. 12345

    12345 Member

    Looking at the ASET solution for this (p34), slightly confused as to how they jump from the definition of log(St) = log(So) + etc to the conclusion that log(St) must have a normal distribution with those parameters.

    Can anyone put me out of my misery?
     
  2. Graham Aylott

    Graham Aylott Member

    Hi.

    Here we're working under the set of risk-neutral probabilities, or equivalent martginale measure (EMM) Q. By definition, this is the set of probabilities that ensure that the expected rate of (continuously compounded) return/drift of the underlying risky share is equal to the risk-free force of interest, r.

    So, under Q, the stochastic differential equation for S(t) must be:

    dS(t) = S(t).[r.dt + sigma.dZ(t)~]

    where Z(t)~ is a standard Brownian motion under the Q probabilities.

    (This is what you have already stated in part (i) of the question.)

    So, if we apply Ito's Lemma to the function:

    G(S(t) = ln S(t)

    as per the example on p22 of Chapter 7 of the Course Notes, then we can see that the corresponding formula for S(t) itself is:

    S(t) = S(0). exp[(r - sigma/2)t + sigma.Z(t)~]

    {Note that whenever you have a formula of this type, if you ignore the -sigma/2 term, then whatever else is in the coefficient on t (ie r here), tells you what the drift of S(t) is. The -sigma/2 term appears because of the 2nd-order term in Ito's Lemma.}

    Also, if we log the equation for S(t), then we can show that lnS(t) has a Normal distribution, as shown on p34 of the ASET, and hence that S(t) itself has a logNormal distribution.

    Hope this helps.

    Graham
     
  3. 12345

    12345 Member

    Thanks Graham - I get the bulk of the question, but just confused as to the last part, i.e. "we can show that lnS(t) has a Normal distribution", exactly how do we know that lnS(t) has a Normal distribution?
     
  4. capitalH

    capitalH Member

    The key is the equation that Graham presented:
    S(t) = S(0). exp[(r - sigma/2)t + sigma.Z(t)~]

    Rewrite as
    ln(S(t)/S(0)) = [(r - sigma/2)t + sigma.Z(t)~]

    Now we now Z(t) is N[0,t] since it is standard BM
    so ln(S(t)/S(0)) or (lnS(t) if S(0)=1)
    must be N[(r - sigma/2)t,sigma^2t]
     
  5. 12345

    12345 Member

    Got it, being slow, thanks for the replies
     
  6. barney

    barney Member

    what about part (v) of that question?! I'm lost!

    i don't have ASET, only the Revision notes, so maybe that's my problem but I can't understand beyond the basic fee of 0.1%!
     

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