Geometric Brownian Motion vs. Lognormal Model

I have seen in a number of places that Geometric Brownian Motion is an alternative way of presenting the Lognormal Model.

Under the lognormal model:-
ln St - ln Ss ~ N[mu(t-s),sigma(t-s)]

But when I start with the Geometric Brownian Motion model:-
dSt = St (mu dt + sigma dZ)

and derive the distribution I get:-
ln St - ln Ss ~ N[(mu - 1/2 sigma^2)(t-s),sigma(t-s)]

So if they really are equivlent ways of expressing the same thing, then why do I get the -1/2 sigma^2 term in the drift when I use Ito's lemma to derive ln St - ln Ss?

 

It was slightly confusing for me too at first but I think this explanation is correct:

The Lognormal model describes share price returns as following the normal distribution with mean parameter mu and variance parameter sigma^2, both proportional to the time period in question (as you've discussed).

However geometric brownian motion is a special type of model where the SDE (I.e. Movement in infinitesimally small increments) can be written as the SDE you've written. What it means is that the instantaneous return (dSt) is the current share price, St, multiplied by a deterministic drift factor, mu, with some randomness around that increasing factor (the volatility factor). Ignore the Lognormal model up to this point to avoid confusion.

That's all well and good, but what we can do solve the SDE to give an equation that looks remarkably like the Lognormal model we've already defined. In fact, it IS the Lognormal Lognormal model, but with new parameters for the mean and standard deviation!

 

Oops didn't mean to press Post!

The distribution you have is correct. The distribution of the log return that satisfies the SDE is the second one you wrote.

Remember: mu and mu - 1/2 sigma^2 are just numbers. The first mu is because we are used to using mu as the mean of a normal distribution. The second one is because we have used mu as the average return in the SDE model. Maybe it would be less confusing if the mean of the Lognormal model was alpha instead of mu.

 

You will realize later that it won't matter for derivative valuation, all you'll need to do is change the drift so that the present value of the discounted process is a martingale under risk neutral measure.

For : ln St - ln Ss ~ N[mu(t-s),sigma(t-s)]
you will set Mu = r - [sigma]^2/2 and
Wt bar = Wt + [ r - [sigma]^2/2]*dt

For : ln St - ln Ss ~ N[(mu - 1/2 sigma^2)(t-s),sigma(t-s)]
you will set Mu = r
and Wt bar = Wt + [ r ]*dt

 

The -1/2sigma^2 term appears or disappears when you use Ito's Lemma to move between the formula for St itself and the formula for dSt - the change in St over a very short instant of time of length dt. It arises due to the second-order partial derivative (with respect to dZt) term in Ito's Lemma.

So, under the real-world P probabilities (with no dividends):

St = S0*exp[(mu-0.5*sigma^2)*t + sigma*Zt]

and differentiating this using Ito gives:

dSt = St*[mu*dt + sigma*dZt]

Likewise, under the risk-neutral Q probabilities (with no dividends):

St = S0*exp[(r - 0.5*sigma^2)*t + sigma*Zt]

and differentiating this using Ito gives:

dSt = St*[r*dt + sigma*dZt]

I think that part of the confusion arises because of the notation used in Chapter 9 when the lognormal model is introduced using the equation:

logSu - logSt ~N[mu*(u-t), sigma^2*(u-t)]

instead of

logSu - logSt ~N[(mu-0.5*sigma^2)*(u-t), sigma^2*(u-t)]

which is equally valid and is consistent with the way the SDE for geometric Brownian motion is normally presented, i.e.:

dSt = St*[mu*dt + sigma*dZt]

Hope the exam goes / went well.