I am a bit confused about how to formulate a problem where I have to price an option on a stock. Many papers say that stock prices are best modeled using a geometric Brownian motion (GBM), and I understand that GBM is a Markov process. But i have recently read about Martingales and I am confused. Are stock prices modeled as GBM or martingales? Is a martingale a special type of GBM where drift is zero?

Can you please explain to me in detail and in basic language the relation and differences between Markov, GBM, and Martingale processes. And can you please provide me with an example, starting with a GBM, and how to derive a Martingale process from that?

Thank you very much!!

  • $\begingroup$ Exactly, there is no relation between Markov process and Martingale. By definition, a Markov process is a stochastic process that next state has no dependencies on previous state. While a martingale is a process that the drift zero, which does not require the dependencies between different states. see physicsforums.com/showthread.php?t=477725 $\endgroup$ Nov 14, 2013 at 13:27
  • $\begingroup$ The way I think about it a Markov process has a discrete and finite memory, and a martingale has no memory. Hence the conditional expectation of the future is the the same as the current state. A unbiassed brownian motion ($W_t$) or Wiener process is a martingale but a geometric brownian motion with nonzero drift $\mu$ (SDE: $dS_t=\mu S_t dt + \sigma S_t dW_t$) is not. $\endgroup$
    – Walter
    Nov 14, 2013 at 15:47
  • $\begingroup$ "Many papers say that stock prices are best modeled using a geometric Brownian motion (GBM)" Actually I doubt many papers say this, as stock prices aren't best modeled with GBM. $\endgroup$ Nov 25, 2013 at 6:49

3 Answers 3


Roughly speaking, we can express the difference between a Markov process and a martingale as follows:

  • A Markov process is one for which conditioning its future value on its history is the same as conditioning its future value on its present value, so that $E(h(X_t)\,|\,X_u,\,u\leq s)=E(h(X_t)\,|\,X_s)$, for any appropriate function $h$;

  • A martingale is a process whose expected future value equals its present value, when conditioned on its history, so that $E(X_t\,|\,X_u,\,u\leq s)=X_s$,

for all $s\leq t$. (Note that I am taking huge liberties by ignoring integrability conditions and other important caveats.)

In words, we might say that Markov processes have the property that their histories provide no information in excess of the information contained in their present values. Similarly, we might say that martingales are processes for which the best estimate of their future value is their current value.

These two concepts have an interesting history in Financial Economics. For example, the Efficient Market Hypothesis effectively asserts that asset price processes are Markov. On the other hand, much of Asset Pricing Theory characterises fair value for risky securities in terms of martingales, in one way or another.

To answer your question, although both the Markov condition and the martingale condition are expressed in terms of conditional expectations, they are in fact quite different notions. In particular, processes can be (1) Markov processes and martingales; (2) Markov processes but not martingales; (3) martingales but not Markov processes; and (4) neither martingales nor Markov process. A good exercise is to construct examples of all four types of process listed above (focus on discrete-time, rather than continuous-time).

Geometric Brownian motion is a process $X$ characterised by the stochastic differential equation $$d X_t=\mu X_t\,dt+\sigma X_t\,d B_t,$$ for all $t\geq 0$, where $B$ is a standard Brownian motion. It is always Markov (incidentally, this explains why the price of an option written on a security that follows a geometric Brownian motion is a function of the current price of the security, and not its price history). However, $X$ is only a martingale when $\mu=0$ (in which case we refer to it as driftless geometric Brownian motion.

  • $\begingroup$ when and only when $\mu = 0$ ? $\endgroup$
    – BCLC
    Sep 11, 2015 at 16:19
  • $\begingroup$ @BCLC When you can change the probability measure to say neutral under which it's zero. $\endgroup$
    – SmallChess
    Sep 8, 2016 at 5:56
  • $\begingroup$ @StudentT What do you mean? I was just wondering why 'only when' rather than 'when and only when' $\endgroup$
    – BCLC
    Sep 9, 2016 at 7:26

Just to give you two examples. Note that

  • $dX_t =a \; dt + dW_t$ is Markov but is not a martingale.

  • $dX_t=(\int_0^t X_s ds) \; dW_t$ is a martingale but is not Markov.


You have been given good answers above. Basically, a stochastic process ${X_t}$ is a Markov process if $P(\{X_{t} \leq x\} | \mathcal{F}_{s}) = P(\{X_{t} \leq x\} | X_{s})$, for $s \leq t$. Here $\mathcal{F}_{s}$ is a $\sigma$-algebra, a special collection of subsets of the underlying sample space $\Omega$, containing all information about the process $\{X_t\}$ up to $s$. In words this means that the distribution of the future values of $\{X_t\}$ does not depend on the path taken up to time $s$, but only on the value of $X$ at time $s$, i.e. $X_s$.

A martingale must be defined in terms of conditional expectations. A process $\{Y_t\}$ is a martingale with respect to the filtation $(\mathcal{F}_{t})_{t \geq 0}$ if $E\{Y_{t+h} | \mathcal{F}_{t}\} = Y_{t}$, for all $h \geq 0$. The best prediction of $Y_{t+h}$ given all information up to time $t$ is $Y_t$.

To connect this to your question on GBMs, if $dX_{t} = \mu X_t dt + \sigma X_t dB_t$, then $X_{t} = X_{0}e^{(\mu - \sigma^{2}/2)t + \sigma B_t}$. Solutions to stochastic differential equations are Markov Processes.

The martingale property depends on the probability measure. Discounted stock prices are martingales under the risk-neutral measure (using the money-market account as numérarie), but discounted stock prices are not martingales under the real-world probability measure.

  • $\begingroup$ 'Solutions to stochastic differential equations are Markov Processes.' ? wsw seems to disagree with 2nd example $\endgroup$
    – BCLC
    Sep 11, 2015 at 16:21

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