3
$\begingroup$

I have read that in the derivation of the Black-Scholes PDE, we assume that the return of a stock $S$ is given by $$\frac{dS}{S}=\mu dt+\sigma dB$$ where $\mu$ is the average growth of $S$, $\sigma$ is the volatility of $S$ (assumed constant) and $dB$ is an infinitesimal increment of a Brownian motion.

My question is, is the $\sigma dB$ trying to represent the 'random' fluctuations of $S$ from the rate people are trading the stock or is it trying to capture the movements seen in $S$ as a result of external information such as company earnings or some geopolitical news.

$\endgroup$
  • 4
    $\begingroup$ Both interpretations are valid. It represents the random price changes from trading the stock. For people who support the efficient markets hypothesis, those changes are caused by reaction to external information and therefore the second interpretation also holds. $\endgroup$ – Alex C Jan 20 at 22:56
  • $\begingroup$ +1 I like this question, not entirely sure why we have 2 close votes on it. It is getting at something of central focus in the derivatives pricing field that is indeed quite confusing. Although it is possible that what you are looking for was covered here somewhat: quant.stackexchange.com/questions/14755/… If not, hopefully my answer cleared up what you were looking for, and if not please specify. $\endgroup$ – Theodore Weld Jan 26 at 0:52
1
$\begingroup$

Yes.

It may prove useful to see why / how Brownian motion plays a role in the growth of a stock in general, and then the role it plays in pricing derivatives as the latter is fairly complex. The following stochastic differential equation represents how the price of a stock follows a geometric Brownian motion:

$dS(t) = \mu S(t)dt + \sigma S(t)\ dW(t, w)$

  • where $S(t)$ is the price of the stock at a time $t$ greater than or equal to 0 (and the quantity that is $S(t)$ is a random variable.

  • $\mu > 0$ is the rate at which the price of the stock grows (and it is assumed to be constant, i.e., independent of the state of the system).

  • $\sigma > 0$ is the volatility of the stock, which is also assumed to be constant.

  • $W(t)$ is a standard Brownian motion (also known as a Weiner Process—same thing, Brown discovered it but he was a botanist thus couldn't explain it, then Weiner came along and was able to explain it).

The mean of this SDE would be

$\mathbf{E}[S(t)] = S_{0}e^{\mu t}$

and the variance would be described as

$Var [S(t)] = S_{0}^2e^{2 \mu t}(e^{\sigma ^2t-1}$)

Now regarding:

is the $σdB$ trying to represent the 'random' fluctuations of S from the rate people are trading the stock or is it trying to capture the movements seen in S as a result of external information such as company earnings or some geopolitical news.

As Alex C stated in the comments, both interpretations are valid as geometric Brownian motion sets to account for the random fluctuations assets experience, and given the fact that geometric Brownian motion is considered to be what is known as a Markov process, it assumes that the past behavior / fluctuations / prices / whatever are already incorporated. This is also how Efficient market hypothesis plays into this as has already been stated: EMH is the idea that the prices of assets are fully reflective of all the information available, i.e., going back to this idea of all prior fluctuations in price being incorporated. This also means that all future fluctuations / movements in the price of the asset are going to be conditionally independent.

We can also represent this with

$\frac{\Delta}{\Delta S} = \mu \Delta t + \sigma \epsilon \sqrt{\Delta t}$

again where $S$ is the stock price, $\mu$ is the expcted return, $\sigma$ is the standard deviation of the returns, $t$ is time, and $\epsilon$ is a random variable.

This can be rearranged trivially so we can solve for just the change in stock price $S$.

$\Delta S = S(\mu \Delta t + \sigma \epsilon \sqrt{\Delta T})$

although we see a couple things have changed a bit, mainly we now have a first term representing drift and the second being shock, i.e., the price is going to drift upwards as per the expected return - and this is where shock comes in as it (depending on what the drift does) is going to be added to or subtracted from the drift.

I think this may be a good way to visualize this if all the probability theory is too much; thinking of an asset to be following these incremental steps, and where each of these steps are a drift with shock added to or subtracted from it.

Geometric Brownian motion has a couple properties that make it attractive for being used to model certain parts of derivatives pricing models.

  1. One of the reasons is that $S(t) > 0$ for all $t \in [0, T]$ which should make sense intuitively in this context of pricing options.
  2. Normality: Increments of Brownian motion are defined to be the time $t$ between two other moments, say $s$ and $s + t$ is going to be $B_{s+t} - B_{s}$ also with $N(0, t)$... meaning it is normally distributed with zero mean and variance $t$.
  3. Being continuous; $B_{t}$ has the continuous path, as well as $B(t ,0) = 0$.

if it seems as if some things are not adding up / making sense then you’re not alone nor have you stumbled upon anything profound as it is well known that there are problems with the current options pricing models and how geometric Brownian motion plays into this. I’d recommend reading Why should we expect geometric Brownian motion to model asset prices? for more information on why it is used and the limitations.

$\endgroup$
0
$\begingroup$

In the equation $$ \frac{dS}{S}=\mu dt + \sigma dB$$ ,where $\mu$ is the constant drift (expected return) of the security price $S_t$, while $\sigma$ is the constant volatility, and $dW_t$ is the standard wiener process with zero mean and unit rate, some books write the wiener process as $dB_t$.

Simply put, $dB_t$ captures the random movement of the stochastic process.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.