Deriving the SDE under black scholes using Ito's lemma

[vegan]

Hi

Core reading 9 of the Black Scholes chapter states an explicit formula for the share price, S_t , which it defines as a function as below:

\[g(t,z)=S_oexp((μ - 0.5σ)t+σz)\]

It then says, therefore:

\[S_t=g(t,Z_t)\]

1) Why is there a small z in the first function; and then a large Zt in the other?

It then says (core reading 9) Ito's lemma will be applied but instead, shows Taylor's second order to theorem formula, as below

\[dg(t,Z_t)= \frac{\partial g}{\partial t}dt+\frac{\partial g}{\partial z}dZ_t +\frac{1}{2}\frac{\partial^2 g}{\partial z^2}dt\]

2) Why does it just not say it will use Taylor's second order formula. Are Ito's formula and Taylor's formula used interchangeably; or is there a reason why Ito's lemma cannot be applied in this case here?

The workings produce this solution:

\[S_t*((μ-\frac{1}{2}σ^2)dt +σdZ_t\frac{1}{2}σ^2dt)\]

By applying the derivatives of ito's/taylor's formula to g(t,Z_t), I am unable to see how the following occurs:
3a) how the exponential function gets removed in the final solution
3b) how S_o becomes S_t in the final solution
Please can someone explain why this happens?

Thanks

 

[Graham Aylott]

NB This query is about material on page 8 in Chapter 14 of the Course Notes.

(1) Don't worry about the Core Reading switching z and Zt. As long as you get the correct end answer in the exam, this is unlikely to matter.

(2) The Core Reading here is just being inconsistent when saying that it will apply "Ito's formula" and then using a Taylor Series. However, in past Examiners' Meetings, they have said that they don't mind whether you use a Taylor Series approach or Ito's Lemma, which is just a special case of a Taylor Series.

Using a Taylor Series:

dg = dg/dt * dt + dg/dZ * dZ + 1/2 * d^2g/dZ^2 * (dZ)^2

From the multiplication table for increments we have (dZ)^2 = dt, so the line above becomes:

dg = dg/dt * dt + dg/dZ * dZ + 1/2 * d^2g/dZ^2 * dt (*)

Next we have:

dg/dt = (mu - 1/2*sigma^2) * S0*exp[(mu - 1/2*sigma^2)*t +sigma*Z]

As the bit in bold in just St, we can simply write this as:

dg/dt = (mu - 1/2*sigma^2) * St

Likewise:

dg/dZ = sigma*St

and:

d^2g/dZ^2 = sigma^2 * St

Substituting these into (*) above and doing a bit of cancelling then gives the required answer:

dg = dSt = St * [mu*dt + sigma*dZ]

Hope this helps.

Graham