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Chp 11 the binomial model comparison logN

W

withoutapaddle

Member
I don't understand pg 37...

E[S(t +dt)/S(t)] = exp(r.dt) ... ok that makes sense but..

E[S(t +dt)/S(t)]^2 only involves terms of higher order than (dt) ???

my basic maths says...

E[S(t +dt)/S(t)]^2 = exp(2.r.dt)~= 1 +2.r.dt +o{(dt)^2}

maybe i'm missing the plot?

also

E[{S(t +dt)/S(t)}^2] = q.u^2 +(1-q).d^2 where did th ln's come from.. i.e.

Var{S(t +dt)/S(t)}=q.(lnu)^2 +(1-q).(-lnu)^2 -E[S(t +dt)/S(t)]^2 ???
 
i think i answered my own question

1.) Var{S(t+dt)/S(t)}=q*(1-q)(u^2 +d^2 +2*u*d) -binomial assumption
2.)Var{S(t+dt)/S(t)}=(dt)*sigma^2 -lognormal assumption

equating (1) & (2) then we have a quaratic in u^2 (as d=1/u)

using simplifying assumptions:

since only over dt, q*(1-q)~=0.25
also {1+o(dt)}^0.5 ~= 1+0.5*o(dt)

then u~=1+dt*sigma^2 +sigma*(dt)^0.5

and since sigma*(dt)^0.5 >> dt*sigma^2

u~=exp(sigma*(dt)^0.5)

but this is nowhere near how the book derived it?
i don't care, i like my way better because it makes sense (or does it? )
 
Last edited by a moderator:
i do find this whole "derive the SDE" for St kind of strange too.

Given dSt = S_0 [mu.dt + sigma.dZt]

you then let f = ln(St),

go thru the whole Ito thingy, then finish with a line that looks like this:

df = (mu - sigma^2/2).dt + sigma.dZt

which looks something like the one given in the first place, but this is called the "solution" and from this we say that ln(St) follow a normal distribution.

However, the first line that's given, also looks like it follows a normal distribution (if you let f = ln(St), then df = dSt/S_0) with different parameters.

Perhaps I have focused so much on exercises, I forgotten the theory behind :confused:
 
S_0 doesn't appear in the SDE
should be:
dSt = St [mu.dt + sigma.dBt]
note Bt~N(0,t) which isn't the same as Zt~N(0,1)
(dSt)^2=(St*mu*dt+St*sigma*dBt)=(dBt*St*sigma)^2=dt*(St*sigma)^2
then:
dln(St)+dt/2*sigma^2 = dSt/St = mu.dt + sigma.dBt
hence:
ln(St)=ln(S_0) + (mu-1/2*sigma^2)*t + sigma*Bt
so ln(St)~N{ln(S_0)+(mu-1/2*sigma^2)*t,t*sigma^2}
i.e. St is logNormally distributed

I'm not sure what "the first line that's given" is but:
dBt~N(0,dt)
dSt/St=mu*dt+sigma*dBt~N(mu*dt,dt*sigma^2)
so dSt/St is Normally Distributed

so it all makes sense, just doesn't read well
 
withoutapaddle said:
I'm not sure what "the first line that's given" is but:
dBt~N(0,dt)
Click to expand...

without being given "dSt = St [mu.dt + sigma.dBt]", what would you substitute your dSt and (dSt)^2 with?

If you differentiate:
ln(St)=ln(S_0) + (mu-1/2*sigma^2)*t + sigma*Bt,

you'll probably get:
dSt/St = (mu-1/2*sigma^2)*dt + sigma*dBt -----(B)

so if you compare this (B), with "dSt = St [mu.dt + sigma.dBt]", why can't we say this St has a lognormal distribution with parameters mu and sigma? Why do we need to go thru the whole Ito's Lemma thing just to get to (B) and say that St follows a lognormal distribution with parameters mu-1/2*sigma^2 and sigma?
 
Could any totor from ActEd clarify how the notes calculated the variance for S(t+dt)/St?
 
you'll probably get:

I don't understand your question?

St is a random variable, it can't be differentiated..

going from ln(St) to the intergrated form we cannot ignore the efferct of (dSt)^2

i.e. d{ln(st)} = dSt/St -0.5*(dSt/St)^2

I'm not too sound on my probablistic proofs, but "you'll probably get" probably means you you might be probably wrong
 
That section of the course notes (including part of the Core Reading) is wrong. The variances and other quadratic bits need to be in terms of the log share price ratio, not just the ratio itself. There's an erratum about this in the ActEd corrections.

For some reason, this section isn't in the revision notes booklet - maybe it's new this time. Anyway, I reckon that all you need to take away from this section is that d = 1/u unless you're told anything different, and that there are these formulae giving u and d in terms of the parameters of a log normal model - these formulae are in the gold tables. I'll be unpleasantly surprised if we have to derive these formulae!
 
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