2006 FRM Practice Exams - Geometric Brown Motion

Discussion in 'P2.T5. Market Risk (25%)' started by dennis_cmpe, Nov 6, 2008.

  1. dennis_cmpe

    dennis_cmpe New Member

    40. In the Geometric Brown Motion process for a variable S,

    I. S is normally distributed
    II. d ln(S) is normally distributed
    III. dS/S is normally distributed
    IV. S is lognormally distributed

    a. I only
    b. II, III and IV
    c. IV only
    d. III and IV

    To answer this question. I noted that:

    1) Price levels are lognormally distributed
    2) Price returns are normally distributed
    3) If the log of a variable is normally distributed, then the variable is lognormally distributed

    So this helps me determined that lll and lV are part of the answer. But I don't understand why ll is part of the answer too. The answer explanation below mentions that dS/S is equal to dln(S). How is this?

    ANSWER: B
    In the Geometric Brownian Motion (GBM) process for variable S:

    dS = ยต S dt + s S dz

    From the above relation it follows that dS/S, which is equal to d ln(S), is normally distributed, whereas S is lognormally distributed.
  2. Hi Dennis,

    To be candid, I would have given your answer because it doesn't say "approximately normal;" i.e., I thought dS/S was only approximately normal and technically lognormal.

    I don't have time to do the research i'd like here, so i posted it to Wilmott:
    http://www.wilmott.com/messageview.cfm?catid=8&threadid=66557

    Let's see what they say. Thanks, David
  3. Oops, the question is right. I forgot to take the derivative:d lnS = dS/S, so II and III are the same - David
  4. Krivetka

    Krivetka New Member

    If a little change question:
    In the Geometric Brown Motion process for a variable S,

    I. S is normally distributed
    II. ln(S) is normally distributed
    III. dS/S is normally distributed
    IV. S is lognormally distributed

    What is correct answer to question?
  5. Aleksander Hansen

    Aleksander Hansen Well-Known Member

    If S is log-normally distributed, then ln(S) is normally distributed,
    dlnS/dS = 1/S => dS/S = dlnS, is normally distributed
    S = exp(X) => S lognormal if and only if X is normal

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