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 = µ 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.

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

Oops, the question is right. I forgot to take the derivative:d lnS = dS/S, so II and III are the same - David

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?

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