Credit de Servigny, Chapter 3 video - lognormal Future Firm Value

[EK]

Hi David,
Can you please help me out with the lognormal distribution, I think i’m missing some very basic concept here
Slide #12 says “Exp Future Firm Value (lognormal)=$13.34” - which is S*exp(mu-vol^2/2) - why is that? - I’ve always thought that, given the return follows GBM, the expected value is S*exp(mu) - see http://en.wikipedia.org/wiki/Geometric_Brownian_motion#Properties
Can you please clarify?

Thank you in advance,

 

[ShaktiRathore]

Hi,[take sign of d/dx as a sign of partial derivative]
The geometric brownian motion represents the change in stock price dSt as,
dSt=drift(=mean)*St*dt+sigma*dz*St in a small time 't' stock moves by dSt according to GBM
dz=e*sqrt(dt)=>dSt=drift(=mean)*St*dt+sigma*e*sqrt(dt)*St...1
From taylors series for changes by dt' and dx' in variables x and t the change in variable G which depends on x and t is given by according to taylors rule,
dG=(dG/dx)*dx'+(dG/dt)*dt'+.5*d^2G/dx^2*dx'^2
let G=lnS and x=St and t=t small change in stock price from St is dSt' and small change in time is dt'
d(lnSt)=(dG/dSt)*dSt'+(dG/dt)*dt'+.5*d^2G/dSt^2*dSt'^2
=>dG/dt=0,dG/dSt=1/St,d^2G/dSt^2=-(1/St)^2
=>(1/St)d(St)=(1/St)*dSt'+(0)*dt'+.5*(-1/St)^2*dSt'^2
=>d(St)=dSt'+.5*(-1/St)*dSt'^2...A
from for a small change dSt' GBM implies, dSt'=mean*St*dt+sigma*e*sqrt(dt)*St substituting dSt' in A=>,
d(St)=mean*St*dt+sigma*e*sqrt(dt)*St+.5*(-1/St)*(mean*St*dt+sigma*e*sqrt(dt)*St)^2.
d(St)=mean*St*dt+sigma*e*sqrt(dt)*St+.5*(-1/St)*(sigma*e*sqrt(dt)*St)^2.[ignoring higher powers of dt in expression ]
d(St)=mean*St*dt+sigma*dz*St+.5*(-1)*(sigma^2*(dt)*St ) we assume e^2=1 and after some simplification
d(St)/St=mean*dt+sigma*dz+.5*(-1)*(sigma^2*(dt) ) dividing by St both sides
d(St)/St=[mean-.5*sigma^2]dt+sigma*dz
now returns d(St)/St are assumed to be normally distributed with a drift/mean of returns of mean-.5*sigma^2 and volatility of sigma
ln(St/S0)~N(mean-.5*sigma^2,sigma)
=> ln(St/S0)=mean-.5*sigma^2
=>St=exp(mean-.5*sigma^2)*S0

thanks

 

[EK]

Hi,[take sign of d/dx as a sign of partial derivative]
...
d(St)/St=[mean-.5*sigma^2]dt+sigma*dz
now returns d(St)/St are assumed to be normally distributed with a drift/mean of returns of mean-.5*sigma^2 and volatility of sigma
ln(St/S0)~N(mean-.5*sigma^2,sigma)
=> ln(St/S0)=mean-.5*sigma^2
=>St=exp(mean-.5*sigma^2)*S0

thanks

Hi ShaktiRathore,

Thank you very much for your time! It's understood that Ln(St/S0) follows normal with a mean of mu-0.5*sigma^2, however, I don't quite get the last step in that conclude mean(St) = expected(St) = S0* exp(mu-0.5*sigma^2); To my knowledge, it's similar to stating that E(ln(x)) = ln(E(x)) which is incorrect! Rather, I believe that the expected return follows normal with mean = mu-0.5*sigma^2, while the future firm value follows lognormal with mean = S0*Exp(mu)

Please advise!

 

[ShaktiRathore]

Hi,
you cans that Ln(St/S0) follows normal with a mean of mu-0.5*sigma^2 a distribution N(mean-.5*sigma^2,sigma).
its similar to stating that
Ln(St/S0) has mean return of mu-0.5*sigma^2
or that exp(ln(St/S0)) has mean return of exp(mu-0.5*sigma^2)....raising both sides to the power e ie. taking taking exponents function both sides
or that St/S0 has mean return of exp(mu-0.5*sigma^2)
thus future firms value has a value S0*exp(mu-0.5*sigma^2)....dividing both sides with S0
The log returns of firm values returns are following normal distribution N(mean,sigma) , when log of something x follows a normal distribution then that x is said to follow a lognormal distribution, hence firm values follows lognormal distribution. Then the mean return of this x is exp(mean) but mean in our case is mean-.5*sigma^2 so that the mean return of x a.k.a firm value is exp(mean-.5*sigma^2). I hope you understood.
thanks

 

[EK]

Hi ShaktiRathore, the following statement of yours is incorrect:
========
Ln(St/S0) has mean return of mu-0.5*sigma^2
or that exp(ln(St/S0)) has mean return of exp(mu-0.5*sigma^2)....raising both sides to the power e ie. taking taking exponents function both sides
========
If you looked at any lognormal definition (eg http://en.wikipedia.org/wiki/Log-normal_distribution), you'd see that for any X=ln N(mu, sigma^2) - mean equals exp(mu + sigma^2/2), which in our case equals exp(mu-sigma^2/2 + sigma^2/2) = exp(mu)

 

[ShaktiRathore]

Hi
I think you have not understood how we get to the formula please see how its derived above carefully,
from the last step that i got on the derivation
d(St)/St=[mean-.5*sigma^2]dt+sigma*dz
=>d(lnSt)=mean-.5*sigma^2dt+sigma*dz is a GBM for a variable ln(St) with normal distribution N(mean-.5*sigma^2,sigma)[it follows that St follows a lognormal distribution as log of St follows normal distribution]
d(lnSt)=mean-.5*sigma^2dt+sigma*e*sqrt(dt)
Integrating above LHS So to St and Rhs from t=0 to t=1 i.e. dt=1
ln(St)-ln(S0)=(mean-.5*sigma^2)+sigma*e
ln(St)-ln(S0)~N((mean-.5*sigma^2),sigma)
ln(St/S0)=~N((mean-.5*sigma^2),sigma)
mean return of disribution over time t=1 is mean-.5*sigma^2
hence, ln(St/S0)=mean-.5*sigma^2
=>(St/S0)=e^(mean-.5*sigma^2)
St=S0*e^(mean-.5*sigma^2)
thanks

 

[ShaktiRathore]

also see http://en.wikipedia.org/wiki/Itō's_lemma#Geometric_Brownian_motion
thanks

 

[EK]

Hi ShaktiRathore,
Thank you very much for your patience! Finally, got it! The keyword is "geometric", which I've completely ignored up until this moment. What a relief!