The continuous time lognormal model

[Harashima Senju]

How do you derive the following distribution
\[ ln S_u - ln S_t \sim N\big( \mu (u-t), \sigma^2(u-t)\big) \]

 

[Mark Mitchell]

It is obtained by solving the SDE for geometric Brownian motion:

dSt = St ( (mu + 0.5*sigma^2) dt + sigma dBt)

where Bt is standard Brownian motion.

Note that this SDE (and hence distribution) is parameterised differently to the SDE for geometric Brownian motion used in conjunction with the Black-Scholes model:

dSt = St (mu dt + sigma dBt)

 

[Harashima Senju]

It is obtained by solving the SDE for geometric Brownian motion:

dSt = St ( (mu + 0.5*sigma^2) dt + sigma dBt)

where Bt is standard Brownian motion.

Note that this SDE (and hence distribution) is parameterised differently to the SDE for geometric Brownian motion used in conjunction with the Black-Scholes model:

dSt = St (mu dt + sigma dBt)

Why do we have 2 different SDEs, are they also equivalent?

 

[Mark Mitchell]

The two SDEs relate to the same basic model (geometric Brownian motion) just with the parameters defined differently. They both give rise to a lognormal distribution, but with different parameters, so I can't quite say they're "equivalent".

The SDE:

dSt = St ( (mu + 0.5*sigma^2) dt + sigma dBt)

appears rarely in CT8 - I quoted it as it answers the question you asked - how that particular form of the lognormal distribution is derived.

 

[Harashima Senju]

The two SDEs relate to the same basic model (geometric Brownian motion) just with the parameters defined differently. They both give rise to a lognormal distribution, but with different parameters, so I can't quite say they're "equivalent".

The SDE:

dSt = St ( (mu + 0.5*sigma^2) dt + sigma dBt)

appears rarely in CT8 - I quoted it as it answers the question you asked - how that particular form of the lognormal distribution is derived.

Thanks, I was able to derive it with the equation you quoted