2006 FRM Practice Exams - Geometric Brown Motion

[dennis_cmpe]

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.

 

[David Harper CFA FRM]

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

 

[David Harper CFA FRM]

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

 

[Krivetka]

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?

 

[Aleksander Hansen]

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