September 2016 Question 8 part v

[Molly]

Hi all,

I have just finished this question, and although the solutions make sense i am stumped as to why my approach isn't correct.

we are trying to find the value of an option v_0=exp(-0.02)*[50*P(S_1/S_0 <0.8) *P(R_1/R_0 <0.6))
we are given r=0.02 and sigma_s=0.32 and sigma_r=sqrt(0.15)
my approach would be to use that
S_1/S_0-lognormal(muT, \sigma^2T)
and since under black scholes mu is = r, we know mu=0.02
S_1/S_0-lognormal(0.02, 0.32^2)
R_1/R_0-lognormal(0.02, 0.15)

so P(S_1/S_0 <0.8)=P(Z<ln0.8-0.02/0.32)=0.22386
and P(R_1/R_0<0.6)=P(Z<ln0.6-0.02/sqrt(0.15))=0.08525
so v_0=exp(-0.02*)*[50*0.22386*0.08525)
=0.9346 which is not equal to the 1.61 which is the correct answer.

If anyone is able to let me know why we cant use this method that would be amazing please - is my assumption that under black scholes mu=r wrong?

Thank you

 

[John Potter]

I think your approach is going to work, you just need to adjust your drifts. mu = r, good but then this drift needs adjusting to r - 0.5*sigma^2 = 0.02 - 0.5*0.32^2 =-0.0312
P(S_1/S_0 <0.8)=P(Z<(ln0.8+0.0312)/0.32)=0.2743
and 0.02 - 0.5*0.15 =-0.055
P(R_1/R_0<0.6)=P(Z<(ln0.6+0.055)/sqrt(0.15))=0.1196

John

 

[Molly]

I think your approach is going to work, you just need to adjust your drifts. mu = r, good but then this drift needs adjusting to r - 0.5*sigma^2 = 0.02 - 0.5*0.32^2 =-0.0312
P(S_1/S_0 <0.8)=P(Z<(ln0.8+0.0312)/0.32)=0.2743
and 0.02 - 0.5*0.15 =-0.055
P(R_1/R_0<0.6)=P(Z<(ln0.6+0.055)/sqrt(0.15))=0.1196

John

Hi John, thank you for this! that makes complete sense, and has actually made me realise that i was getting a little confused with the whole \mu=r but \mu also being the drift thing.
for example E(S_t)=S_0*exp(\mu t+1/2\sigma^2) (call this formula 1)
Im also saying that \mu=r (call this formula 2)
I cant necessarily say that E(S_t)=S_0*exp(rt+1/2\sigma^2) because the mu in formula 1 and mu in formula 2 aren't the same thing (i think?).
If I go with the mindset of mu being the drift (aka the coefficient of t in the log normal model S_t), and if not given mu directly but instead given r then mu(aka drift)=r-0.5, then that should be ok i think?
Thank you!

 

[John Potter]

Yes, that sounds sensible but careful not to fixate on mu being a thing. I don't think you are but, just in case, recall this thread...
https://www.acted.co.uk/forums/index.php?threads/april-2014-question-3.19981/
and in particular the bit where I write...

"Yes, GBM and cts time logN model are the same model, we just need to be careful what we mean by "mu".
If we start with dSt = St (mu dt + sigma dZt ) and solve, we will get log St = log S0 + (mu - 0.5sigma^2) t + sigma Zt, which is the same as saying St ~ logN [log S0 + (mu - 0.5sigma^2) t , sigma^2 t ]

If we start with log St = log S0 + mu t + sigma Zt, which is the same as saying St ~ logN [log S0 + mu t , sigma^2 t ], this would trace back to dSt = St ((mu +0.5 sigma^2) dt + sigma dZt )

Try not to fixate on mu as a thing. It's just a letter of the Greek alphabet. I could say that mu - 0.5sigma^2 = alpha and then hand my solution to GBM to a friend as log St = log S0 + alpha t + sigma Zt. This friend might then decide that they don't like alpha and they're going to change it to mu before handing it to a third person - that third person might get confused about why the mu in the first representation is not the same mu as the one in the second representation. That's all that's going on here, but it's still the SAME model,

John"

 

[Molly]

Yes, that sounds sensible but careful not to fixate on mu being a thing. I don't think you are but, just in case, recall this thread...
https://www.acted.co.uk/forums/index.php?threads/april-2014-question-3.19981/
and in particular the bit where I write...

"Yes, GBM and cts time logN model are the same model, we just need to be careful what we mean by "mu".
If we start with dSt = St (mu dt + sigma dZt ) and solve, we will get log St = log S0 + (mu - 0.5sigma^2) t + sigma Zt, which is the same as saying St ~ logN [log S0 + (mu - 0.5sigma^2) t , sigma^2 t ]

If we start with log St = log S0 + mu t + sigma Zt, which is the same as saying St ~ logN [log S0 + mu t , sigma^2 t ], this would trace back to dSt = St ((mu +0.5 sigma^2) dt + sigma dZt )

Try not to fixate on mu as a thing. It's just a letter of the Greek alphabet. I could say that mu - 0.5sigma^2 = alpha and then hand my solution to GBM to a friend as log St = log S0 + alpha t + sigma Zt. This friend might then decide that they don't like alpha and they're going to change it to mu before handing it to a third person - that third person might get confused about why the mu in the first representation is not the same mu as the one in the second representation. That's all that's going on here, but it's still the SAME model,

John"

Yes that part makes a lot of sense now, thank you!!