Tiny question on Geometric Brownian Function

[NS206]

Hi,

I feel unsure about something very basic, so hoping this is a very quick question - I wanted some help to understand the difference in the geometric brownian motion function equation in Ch9 vs Ch10. I've included a snip of both from the summaries - is it just that one is presented as a differential and the other is not? If we are asked to define it, can we give either formula?

Thanks!

 

[Anna Bishop]

Hello

The answer is that it depends on the context of the question. If the question asks you how geometric BM relates to standard BM then I would go for the Chapter 9 formulation. If the question askes you to state the geometric BM SDE, I would go for the Chapter 10 formulation. Similarly, if the question asks you to state the distribution for the continuous time lognormal model, go for the Chapter 11 formulation!

Note though, that all three formulations are related and it's important to be able to move between them all.

Typically, the exam question will ask you to define the geometric BM SDE, in which case you would write down:

dSt = μSt dt + σSt dWt

However, you need to be completely familiar with the proof for solving this SDE for the exam, and the distribution, ie:

1) use Taylor's theorem / Ito's lemma to obtain the SDE for dlogSt:

dlogSt = (μ - 0.5σ^2) dt + σ dWt

2) integrate both sides between t and T to obtain the expression:

log ST - log St = (μ - 0.5σ^2)(T - t) + σ(WT - Wt)

3) use log laws to obtain:

ST/St = exp{(μ - 0.5σ^2)(T - t) + σ(WT - Wt)}

or, taking St over to the RHS:

ST = exp {log St + (μ - 0.5σ^2)(T - t) + σ(WT - Wt)}

4) be able to build up / quote the distribution of either of these above expressions, for example:

The distribution of ST/St = exp{(μ - 0.5σ^2)(T - t) + σ(WT - Wt)} is built up as follows:

WT - Wt has a N(0, T - t) distribution by definition
σ(WT - Wt) has a N(0, σ^2 (T - t)) distribution
(μ - 0.5σ^2)(T - t) + σ(WT - Wt) has a N((μ - 0.5σ^2)(T - t), σ^2 (T - t)) distribution

ST/St is logN((μ - 0.5σ^2)(T - t), σ^2 (T - t))

The distribution of ST = exp{log St + (μ - 0.5σ^2)(T - t) + σ(WT - Wt)} is built up as follows:

WT - Wt has a N(0, T - t) distribution by definition
σ(WT - Wt) has a N(0, σ^2 (T - t)) distribution
log St + (μ - 0.5σ^2)(T - t) + σ(WT - Wt) has a N(log St + (μ - 0.5σ^2)(T - t), σ^2 (T - t)) distribution

ST is logN(log St + (μ - 0.5σ^2)(T - t), σ^2 (T - t))

Note we have assumed that St is known here, so really we should be writing:

ST|Ft is logN(log St + (μ - 0.5σ^2)(T - t), σ^2 (T - t))

It is now worth thinking about how this relates to the formulation on the Summary page of Chapter 9. I reckon it is the same but with:

• T and t replaced with t and 0 respectively
• W0 = 0 (std BM starts at 0 at time 0)
• Z0 = log S0
• the parameter μ replaced with μ + 0.5σ^2

Let me know whether you agree and can follow the post, as it's an important corner of the course to understand.

Best wishes
Anna

 

[NS206]

Hello

The answer is that it depends on the context of the question. If the question asks you how geometric BM relates to standard BM then I would go for the Chapter 9 formulation. If the question askes you to state the geometric BM SDE, I would go for the Chapter 10 formulation. Similarly, if the question asks you to state the distribution for the continuous time lognormal model, go for the Chapter 11 formulation!

Note though, that all three formulations are related and it's important to be able to move between them all.

Typically, the exam question will ask you to define the geometric BM SDE, in which case you would write down:

dSt = μSt dt + σSt dWt

However, you need to be completely familiar with the proof for solving this SDE for the exam, and the distribution, ie:

1) use Taylor's theorem / Ito's lemma to obtain the SDE for dlogSt:

dlogSt = (μ - 0.5σ^2) dt + σ dWt

2) integrate both sides between t and T to obtain the expression:

log ST - log St = (μ - 0.5σ^2)(T - t) + σ(WT - Wt)

3) use log laws to obtain:

ST/St = exp{(μ - 0.5σ^2)(T - t) + σ(WT - Wt)}

or, taking St over to the RHS:

ST = exp {log St + (μ - 0.5σ^2)(T - t) + σ(WT - Wt)}

4) be able to build up / quote the distribution of either of these above expressions, for example:

The distribution of ST/St = exp{(μ - 0.5σ^2)(T - t) + σ(WT - Wt)} is built up as follows:

WT - Wt has a N(0, T - t) distribution by definition
σ(WT - Wt) has a N(0, σ^2 (T - t)) distribution
(μ - 0.5σ^2)(T - t) + σ(WT - Wt) has a N((μ - 0.5σ^2)(T - t), σ^2 (T - t)) distribution

ST/St is logN((μ - 0.5σ^2)(T - t), σ^2 (T - t))

The distribution of ST = exp{log St + (μ - 0.5σ^2)(T - t) + σ(WT - Wt)} is built up as follows:

WT - Wt has a N(0, T - t) distribution by definition
σ(WT - Wt) has a N(0, σ^2 (T - t)) distribution
log St + (μ - 0.5σ^2)(T - t) + σ(WT - Wt) has a N(log St + (μ - 0.5σ^2)(T - t), σ^2 (T - t)) distribution

ST is logN(log St + (μ - 0.5σ^2)(T - t), σ^2 (T - t))

Note we have assumed that St is known here, so really we should be writing:

ST|Ft is logN(log St + (μ - 0.5σ^2)(T - t), σ^2 (T - t))

It is now worth thinking about how this relates to the formulation on the Summary page of Chapter 9. I reckon it is the same but with:

• T and t replaced with t and 0 respectively
• W0 = 0 (std BM starts at 0 at time 0)
• Z0 = log S0
• the parameter μ replaced with μ + 0.5σ^2

Let me know whether you agree and can follow the post, as it's an important corner of the course to understand.

Best wishes
Anna

Thanks Anna, that is really helpful. I can follow the post if I think through each step. It seems like the SDE is the most common starting point for most questions and that makes sense why it's that one that we use now.

 

[Anna Bishop]

smashing, thank you and good luck