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Brownian Motion and Geometric Brownian Motion in Option Pricing

Article Quant Q&A · Author: S_Star

Summary

The document explains the difference between arithmetic Brownian motion, where price changes are modeled in absolute terms, and geometric Brownian motion, where proportional price changes are modeled. It shows how the GBM equation can be transformed with Itô’s lemma: the log price has drift reduced by half the variance rate and retains the Brownian shock. Integrating that equation makes log returns normally distributed and prices lognormally distributed.

The discussion gives the equation-solving steps but is informal and contains imprecise wording. In particular, a Brownian increment integrated over a time interval has variance equal to the interval length, rather than unit variance in general. It also does not develop option valuation itself or explain how to estimate the model parameters. Its value is as a conceptual derivation of the lognormal price assumption and the role of the volatility correction in log-price drift.

Key ideas

  • Arithmetic Brownian motion models additive price changes, while geometric Brownian motion models proportional changes.
  • Applying Itô’s lemma to the logarithm of a geometric Brownian price introduces a negative half-variance adjustment to drift.
  • Integrated log returns are normally distributed under the stated model, so the corresponding price is lognormally distributed.
  • The variance of the Brownian shock over a period depends on the period’s length.

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Full text
# Ito's lemma and Lognormal Property


# Ito's lemma and Lognormal Property












What would be the difference between:

\begin{align} dS = udt + \sigma dz \end{align}

and

\begin{align} dS=u*S*dt + \sigma*S*dzdS \end{align} Is that the former is in absolute terms and the latter is in relative terms with the stock price?

Therefore, if I want to derive the lognormal property for ( $G =\ln S => dG = (u-\sigma^2/2)*dt + \sigma*dz$) pricing an option, can the first equation be used and how? In John Hull book is done by using the second one.

Thank you.

## Answer by StackG (score 0, accepted)

https://quant.stackexchange.com/a/57511

The answer to

\begin{align} dS = \mu dt + \sigma dW_t \end{align} is simply

\begin{align} S(t) - S(0) = \mu t + \sigma W_t \end{align}

(as discussed here in the first page, for example)

## Answer by gte (score 2)

https://quant.stackexchange.com/a/57661

Perhaps it might help if we define the difference between Brownian Motion (BM) and Geometric Brownian Motion (GBM). BM has independent, identically distributed increments while GBM has independent, identically distributed ratios between successive factors. The definition is inherited from that of arithmetic random walks, which are modelled as sums of random terms, and geometric random walks, modelled as products of random factors.

Let's look at them a bit more in detail.

The BM differential equation is:

$dS_{t} = \mu dt + \sigma dW_{t}$

where the first term, $\mu dt$, is the drift term and the second term $ \sigma dW_{t}$ is the diffusion term characterised by the Wiener process $W_{t}$.

To resolve it, we add integrals on both sides:

$\int_{t=0}^T dS_{t} =\mu \int_{t=0}^T dt + \sigma \int_{t=0}^T dW_{t}$

Here, the last term $\int_{t=0}^T dW_{t}$ is your random variable, i.e. shock.

Let us now look at the GBM. As we said earlier, the GBM is characterised by i.i.d ratios between successive factors. We define it as

$ \frac{dS_{t}}{S_{t}} = \mu dt + \sigma dW_{t}$

Here, $ \frac{dS_{t}}{S_{t}}$ is the finite-time price. To resolve, we take logs and after applying ito's lemma we obtain

$d(logS_{t}) = (\mu - \frac{1}{2} \sigma^2)dt+\sigma dW_{t}$

now we can add integrals, as we have a normal diffusion:

$\int_{t=0}^Td(logS_{t}) = (\mu - \frac{1}{2} \sigma^2)\int_{t=0}^Tdt+\sigma \int_{t=0}^TdW_{t}$

hence

$log S_{T}-log S_{0} = (\mu - \frac{1}{2} \sigma^2)T + \sigma \int_{t=0}^TdW_{t}$

Here, $(\mu - \frac{1}{2} \sigma^2)T$ is the mean of the log price after T years, and $\sigma \int_{t=0}^TdW_{t}$ is the shock, i.e. the variance after T years (normally distributed with mean 0 and variance 1).

Finally, we have the rate of return equal to

$\frac{S_{t}}{S_{0}} = exp((\mu - \frac{1}{2} \sigma^2)T + \sigma \int_{t=0}^TdW_{t})$

which is log-normally distributed.

I hope this helps!

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.