1
$\begingroup$

The rate of return of a stock is denoted $\frac{dS}{S dt}$ where $S$ is the solution to the SDE modeling the price of a stock. Can someone give an explanation of the rate of return and what it is supposed to represent in this context.

$\endgroup$
1
$\begingroup$

In discrete time the (annualized) logarithmic return is defined as $\frac{\Delta \ln(S)}{\Delta t}=\frac{\ln(S_{t+\Delta t})-\ln(S_t)}{\Delta t}$

In continuous time this becomes $\frac{d \ln(S)}{d t}=\frac{1}{S}\frac{d S}{d t}$

Note that $\frac{d S}{dt}$ is the instantaneous rate of change in price, dividing it by $S$ turns it into a percentage change in price. And that is how we usually assess stock price changes, in percentage terms.

$\endgroup$
  • $\begingroup$ What does annualized mean in this context? it dosnt seem to be "annualized return" by which I mean a sort of backtracked of the yearly return that would add up to a total return. $\endgroup$ – user1 Sep 24 '19 at 5:05
  • 1
    $\begingroup$ It just means "per unit time" and the unit is whatever you want, such as (in finance) 1 year. Whatever unit t is expressed in. $\endgroup$ – Alex C Sep 24 '19 at 15:13
2
$\begingroup$

A fundamental premise of the BS model is that equity prices move according to a Weiner process. Moreover, when there are a series of many small random movements in the share price the track that it is tracing can be assumed to be geometric Brownian motion.

This process is then symbolically defined as (see Ito's lemma)

$$dS = \mu S dt + \sigma dz$$

Where

  1. $\mu$ is constant and represents the return on the share reported as an annual rate
  2. $\sigma$ is constant and represents the share's volatility also reported as an annual rate
  3. $dt$ is an infinitesmal passage of time
  4. $dz$ represents a term which generates randomness into the movement of the Share price, $S$.

In this process, if there is no randomness, $dz=0$, thus \begin{align} dS &= \mu S dt \\ \implies \frac{dS}{S dt} &= \mu \end{align}

$\endgroup$

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.