What is a martingale and how it compares with a random walk in the context of the Efficient Market Hypothesis?


Samuelson suggested in 1965 that the stock prices follow a martingale (see P. Samuelson “Proof That Properly Anticipated Prices Fluctuate Randomly”).

Assume there is a security with a random payoff $X_T$ at date $T$. Let $..., P_{t–1}, P_t, P_{t+1},...$ be the time series of prices of a security with this payoff. Finally, define the price change $\Delta P_{t+1}=P_{t+1} – P_{t}$ for any pair of successive dates $t$ and $t + 1$. Samuelson begins by defining “properly anticipated prices” as prices that are equal to the expected value of $X_T$ at every date $t \leq T$, based on the information $\Phi_t$ available at date t (which, in particular, includes the present and all past price realizations for that security, $...,P_{t–2}, P_{t–1}, P_t$). That is, for all $t \leq T$: $$P_t = \mathbb E(X_T|\Phi_t).$$

In particular, $P_T = X_T$. He then proves that the “prices fluctuate randomly” since it follows that for all $t \leq T$, $P_t = \mathbb E(P_{t+1}|\Phi_t)$ or alternatively that $\mathbb E(\Delta P_{t+1}|\Phi_t) = 0$, and $$\mathbb E(\Delta P_{t+1}\Delta P_{t+2}...\Delta P_T|\Phi_t) = \mathbb E(\Delta P_{t+1}|\Phi_t) \mathbb E(\Delta P_{t+2}|\Phi_t)...\mathbb E(\Delta P_T|\Phi_t)=0.$$ In words, prices follow a martingale, and successive price changes are mutually uncorrelated.

This implies that if “prices are properly anticipated,” all the information in the past price series that is useful for forecasting next period’s expected price is contained in the current price. Note that this is a much weaker statement than to say that all information in the past price series that is useful for forecasting the probability distribution of next period’s price is contained in the current price (which is the random walk hypothesis suggested by Fama in his thesis).

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    $\begingroup$ To the user who has downvoted the answer: could you possibly explain your reasons for doing so? I would be glad to learn how it can be improved. $\endgroup$
    – olaker
    Feb 9 '11 at 21:49
  • $\begingroup$ Is this theory "frequency agnostic"? I mean, does it equaly applies to daily, monthly and tick data? $\endgroup$
    – user40
    Apr 18 '11 at 13:03

A martingale is a random process $X(t)$ which has the following properties:

$ E[X(T)|\mathcal{F}_t] = X(t) $

for $T > t$ and

$ E[|X(T)|] < \infty $

where $\mathcal{F}_t$ is the filtration at time $t$.

A martingale is a random walk, but not every random walk is a martingale. A Brownian random walk is a martingale if it does not have drift.

Also, a martingale does not have to be a Markov process.

EMH is not directly related to martingales.

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    $\begingroup$ As is, the definition is incorrect. You have to condition with respect to information up to time $t$. $\endgroup$
    – gappy
    Feb 9 '11 at 12:08
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    $\begingroup$ Yes, that's an important detail one is likely to forget about. $\endgroup$
    – quant_dev
    Feb 9 '11 at 14:02
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    $\begingroup$ This doesn't address the question w.r.t. EMH. $\endgroup$
    – Shane
    Feb 9 '11 at 14:34
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    $\begingroup$ Added some more. $\endgroup$
    – quant_dev
    Feb 10 '11 at 12:42
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    $\begingroup$ Which definition of a random walk are you using here? $\endgroup$
    – Ilya
    May 21 '13 at 11:22

Often one will find the argument that a random walk of price changes would be a proof of the efficient market hypothesis, but this is (IMO) a logical fallacy: Only because the EMH does imply random walks in the price changes, the finding of random walks does not imply automagically that the EMH is true.


A martingale can be viewed as a fair game (a game in which there is no arbitrage strategy)

A (centered) random walk is a martingale (think of it as the total Gain of the fair game)

If EFH is in order, then you can think that all information is in the current price, I think this more comparable to Markov Property than to Martingale property.

Hope that this helps a bit


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    $\begingroup$ A martingale does not have to be a Markov process. $\endgroup$
    – quant_dev
    Feb 9 '11 at 9:29
  • $\begingroup$ I have never said that it had to be one $\endgroup$
    – TheBridge
    Feb 9 '11 at 9:32
  • $\begingroup$ A small precision : a centered random walk is a martingale :) $\endgroup$
    – egoroff
    Feb 9 '11 at 10:14
  • $\begingroup$ "you can think that all information is in the current price, I think this more comparable to Markov Property" -- I think rather that EMH is saying that the price is an adapted process -- Markov or not. $\endgroup$
    – quant_dev
    Feb 10 '11 at 12:42

There are many good answers already, but I give this one just to provide some additional intuition:

The simplest random walk is tossing a coin several times: heads means one up, tails means one down. Because of the symmetry of this process the sum of those tosses adds up to zero, on average: it is a martingale!

Intuitively a martingale means that, on average, the expected value of your cumulative stochastic process stays the same, no matter how many coin tosses in the future.

If you add a drift to your random walk by e.g. saying that the up-move is not one but two it is no longer a martingale because, on average, the expected value will go higher and higher.

At the moment I see no direct connection with the Efficient Market Hypothesis (EMH).


I probably will answer your question in a simple fashion before getting to a much mathematical term because I may suspect you are not familiar with stochastic terms/jargons yet.

A variable could be called a martingale if the expectation of the variable at $t+1$ equals to the expectations of the variable at $t$ (basically we joke that if we don't learn anything it means that we are getting a martingale - we are not better off).

So let's look at the mathematical jargon, according to

A basic definition of a discrete-time martingale is a discrete-time stochastic process (i.e., a sequence of random variables) $X_1, X_2, X_3, \dots$ that satisfies for any time $n$, $$ \begin{aligned} &\mathbf{E} ( \vert X_n \vert ) < \infty \\ &\mathbf{E} (X_{n+1}\mid X_1,\ldots,X_n) = X_n. \end{aligned} $$ That is, the conditional expected value of the next observation, given all the past observations, is equal to the last observation. Due to the linearity of expectation, this second requirement is equivalent to: $$ \mathbf{E} (X_{n+1} - X_n \mid X_1,\ldots,X_n)=0 $$ or $$ \mathbf{E} (X_{n+1} \mid X_1,\ldots,X_n)- X_n=0 $$ which states that the average "winnings" from observation $n$ to observation $n+1$ are $0$.

  • $\begingroup$ Your equations aren't showing up.... $\endgroup$
    – SmallChess
    Sep 29 '15 at 4:37

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