April 2014 Question 3

[Molly]

Hi all,

Am really confused about this question, in particular part iib.

The ASET solutions in the revision notes have quoted dWt=1dWt+0dt. where has this come from?
I am comfortable that f(t,W^2_t)=W^2_t-t, I'm just really unsure as to where this SDE has come from.

I also had a look at the IFOA solutions on the past papers site, and am even more lost as to what they have done! Where has their S_t=S_0* exp(\mu_t+\sigmaW_t) come from?

Please can someone help?

Thanks

 

[vidhya36]

Does this help? I used Taylor's formula.

Let f(B_t) = B_t^2 - t
Taylor’s formula: df(B_t) = f’(B_t) dB_t + f’’(B_t)/2 [dB_t]^2 + f’(t) dt
f’(B_t) = 2B_t ; f’’(B_t) = 2; f’(t) = -1
df(B_t) = 2B_t.dB_t + (2/2) dt – 1.dt
the dt terms cancel out to give => df(B_t) = 2B_t.dB_t

I don't have ASET. But, the process you stated at start with zero drift indicates the dt terms would be zero. Which is what we are getting above. So, f(t, B_t^2) is indeed a process with zero drift.

(In above, Notation B_t = W_t)

 

[vidhya36]

Regarding the second query - Question asks to state the continuous-time log-normal model in part (i)

 

[Molly]

Does this help? I used Taylor's formula.

Let f(B_t) = B_t^2 - t
Taylor’s formula: df(B_t) = f’(B_t) dB_t + f’’(B_t)/2 [dB_t]^2 + f’(t) dt
f’(B_t) = 2B_t ; f’’(B_t) = 2; f’(t) = -1
df(B_t) = 2B_t.dB_t + (2/2) dt – 1.dt
the dt terms cancel out to give => df(B_t) = 2B_t.dB_t

I don't have ASET. But, the process you stated at start with zero drift indicates the dt terms would be zero. Which is what we are getting above. So, f(t, B_t^2) is indeed a process with zero drift.

(In above, Notation B_t = W_t)

Hi,

Thank you so much!! Makes much more sense to do it that way, thats really helpful!

With the answer to the second query, i cant seem to find S_t=S_0* exp(\mu_t+\sigmaW_t) is the continuous-time log-normal model anywhere in the book i would have said that the answer to this was
log(S_u)-log(S_t)-N(\mu(u-t),\sigma^2(u-t)). Am i missing some sort of theory here, or did you derive the S_t=S_0* exp(\mu_t+\sigmaW_t) from scratch?

Thank you!

 

[Molly]

Hi,

Thank you so much!! Makes much more sense to do it that way, thats really helpful!

With the answer to the second query, i cant seem to find S_t=S_0* exp(\mu_t+\sigmaW_t) is the continuous-time log-normal model anywhere in the book i would have said that the answer to this was
log(S_u)-log(S_t)-N(\mu(u-t),\sigma^2(u-t)). Am i missing some sort of theory here, or did you derive the S_t=S_0* exp(\mu_t+\sigmaW_t) from scratch?

Thank you!

Also, if anyone knows where the dWt=1dWt+0dt has come from in the ASET solutions, would be really intrigued to find out why this was used. Thank you!

 

[John Potter]

dWt=1dWt+0dt is merely an observation. It's just saying that, trivially the drift = 0 and the volatility = 1.

Imagine that all fields consist of cows and horses, so that the field may be specified by a and b, where a = the number of horses and b = the number of cows. If we had a field with just 1 horse and that's it, we could say "Oh look, in that field a = 1 and b = 0"

It's the same here except we're looking for drift and volatility, not cows and horses. We're saying "Oh look, dWt has drift = 0 and volatility = 1. It is equal to itself."

 

[Molly]

dWt=1dWt+0dt is merely an observation. It's just saying that, trivially the drift = 0 and the volatility = 1.

Imagine that all fields consist of cows and horses, so that the field may be specified by a and b, where a = the number of horses and b = the number of cows. If we had a field with just 1 horse and that's it, we could say "Oh look, in that field a = 1 and b = 0"

It's the same here except we're looking for drift and volatility, not cows and horses. We're saying "Oh look, dWt has drift = 0 and volatility = 1. It is equal to itself."

Ah, makes much more sense when you put it like that - thank you!!

Sorry, do you know why they have used S_t=S_0* exp(\mu_t+\sigmaW_t) as the continuous-time log-normal model in the ifoa solutions rather than log(S_u)-log(S_t)-N(\mu(u-t),\sigma^2(u-t)). i assume they must be equivalent, somehow?

 

[vidhya36]

Hey Molly, continuous time lognormal model is same as geometric Brownian motion
You can refer to Pg 20, CM2 Ch 8 in CMP course notes where you would see a version of the same equation with
S_t = S_0. exp [(mu - 0.5sigma^2)t + sigma.W_t]
In examiner report, they presented a simplified verison of same equation and either should be valid to fetch full marks?
S_t = S_0. e^(mu t + sigma W_t)
where -0.5 sigma sq is the correction term which gets adjusted within during calculation of mu
Here is what I found in wiki -

The correction term of − σ^2/2 corresponds to the difference between the median and mean of the log-normal distribution, or equivalently for this distribution, the geometric mean and arithmetic mean, with the median (geometric mean) being lower. ​

Citation: Wikipedia - Itô's lemma - Wikipedia (refer the geometric Brownian motion section)
This is also discussed here - Log N model and GBM | Actuarial Education (acted.co.uk)
May be @John Potter can explain it better.

 

[John Potter]

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]

Hi both, that makes so much more sense now, thank you both so much!!