Sorting out formulae for St

[nicolathompson]

Hi, I’ve been trying to sort out all the formulae / distributions in my head and have found the following 6 at various places in the notes. Am confused by the last two...

1. dSt = St(µdt + σdZt) under P (basic definition of Geometric Brownian motion)
2. dSt = St(rdt + σdZtildat) under Q (Chapter 15, p18)


3. St = Soexp[(µ-σ2/2)t + σZt] under P (basic definition of Geometric Brownian motion)
4. St = Soexp[(r-σ2/2)t + σZtildat] under Q (comes from using C-M-G theorem, chapter 14, p12)


Then why do these not match the same way (replace µ by r and Z by Ztilda)?

5. ln[(St+dt/St] has N(µdt, σ2dt) under P (Chapter 9)
6. ln[(St+dt/St] has N((r-σ2/2)dt, σ2dt) under Q (Calibrating binomial models, chapter 12)

 

[Hamilton]

Hi again Nicola

Em 5 ? where did that come from its not true sure its not , we can get 5 from 3 by divide across by S zero then taking logs. (and a bit of fooling around with the times t , t zero.)

likewise for 6 can be got from 4 in the same fashion.

So you can in fact always replace mu with r and z with zTilda.

But whats that i hear you ask , why oh why , is the variances both sigma squared dt in 5 and 6 ????

well this is cause we are changing the drift but not the volatility of our Brownian motion when we shift to the EMM q .
So our variance in both cases is the same sigma squared dt

how do we work out this variance
well it comes from the fact that

::EDIT :: integral sigma dBs

is a random variable that is normal distributed with mean zero and variance given by

integral sigma squared dt

this is the variance that appears in 5 and 6.

Hope this is understandable and the correct way to think about it cause I am by no means a pro at this .
Good luck in your exam.

 

[nicolathompson]

Thanks for reminding me that you can change the drift but not the volatility when we shift from P to Q (I had never really got what they were going on about when that was mentioned in the C-M-G theorem).

So are 5 and 6 both correct then?
I got 5 from:
log(Su) - log(St) is N with mean mu(u-t) and volatility sigma-squared*(u-t)
which I believe is the same thing as my no.5

I got 6 from the proof of calibrating binomial models.

Good luck too! I have a few long evenings of study to do!