Martingale, random walk and rational expectations

Question

What is the link between these concepts?

For example let's take a process $Z_n$ which follows a random walk, I would say that:

1. This is a martingale, because my expectations of tomorrow, i.e. n+1 depends only on today
2. If we assume an individual with rational expectation, he will assume that the forecast of tomorrow is today

So in other words, the forecast of a rational individual follows a martingale?

I am not sure of the difference since one concept is used in finance while the other in economics, but they seem basically the same to me.

Last, I have seen in the example that a random walk implies a martingale, is this always true? Does a martingale implies a random walk?

Answer 1

All concepts are used in Economics. Definitions (not stated in a fully rigorous manner):

Martingale : A stochastic process $\{X_t\}$ is called "martingale" if and only if it holds that

$$E(X_{t+1} \mid X_t,X_{t-1},...) = X_t \tag{1}$$

There are extensions like "sub-martingale", "super-martingale" but the basic definition is the above

Random walk : A stochastic process $\{X_t\}$ is called "random walk" if and only if

$$X_{t+1} = X_t + u_{t+1}, \;\;\;u_t \sim \text{White Noise} \tag{2}$$

and you can look up the definition of "White Noise".

Here too there are spin-offs ("random walk with drift" etc).

Comment: it follows that a random walk is (one example of) a martingale, but a martingale does not imply a random walk, the former being a much broader concept.

Rational Expectations : Originally, the Rational Expectations Hypothesis stated that the aggregate expectation on some unknown (usually future) value of an aggregate random variable (see also this post on the matter), equals the conditional expectation (in the rigorous mathematical sense) of this variable "given all relevant information at the time of forming the expectation

$$X^e_{t+k|t} = E(X_{t+k}\mid I_t) \tag{3}$$

Comment: note that the conditioning set in the case of a martingale contains only past values of the process. The conditioning set in the case of the REH contains "whatever is available and is deemed relevant".

In a representative agent model, we are forced, for internal methodological consistency, to apply the Rational Expectations Hypothesis at the individual level (something that has raised all sorts of valid objections as regards the availability of information and the information processing constraints of an individual).

So do the forecasts of an individual follow a martingale? Let's examine it for one-step-ahead forecasts only: our stochastic process is

$$\{X^e_{t+1|t},\,X^e_{t+2|t+1},...\} = \{E(X_{t+1}\mid I_t),\,E(X_{t+2}\mid I_{t+1}),...\} \tag{4}$$

To obtain the martingale property it must hold that

$$E[X^e_{t+1|t} \mid X^e_{t|t-1}, X^e_{t-1|t-2},...] = ?\; X^e_{t|t-1} \tag{5}$$ (I will name in the end what expectations does eq. $(5)$ describes)

$$E[X^e_{t+1|t} \mid X^e_{t|t-1}, X^e_{t-1|t-2},...] = E[E(X_{t+1}\mid I_t) \mid \{E(X_{t}\mid I_{t-1}), E(X_{t-1}\mid I_{t-2}),...\}]$$

As already commented, the outer conditioning set is smaller than the inner conditioning set. By the Law of Iterated Expectations (the Tower property) of Conditional Expectation, we then obtain

$$E[E(X_{t+1}\mid I_t) \mid \{E(X_{t}\mid I_{t-1}), E(X_{t-1}\mid I_{t-2}),...\}] = E[X_{t+1} \mid \{E(X_{t}\mid I_{t-1}), E(X_{t-1}\mid I_{t-2}),...\}] \tag{6}$$

The right-hand-side of $(6)$ is not necessarily equal to the right-hand-side of $(5)$, so we cannot say that the stochastic process of one-step-ahead forecasts under the Rational Expectations Hypothesis is a martingale.

Equation $(5)$ describes the following situation: standing at period $t-1$ we form the expectation $X^e_{t|t-1}$, and "the best we can say" (best in the mean-squared-error sense) about our subsequent expectation $X^e_{t+1|t}$ is that it will be equal to $X^e_{t|t-1}$. This is (sometimes) called "static expectations" but beware because the term has two totally different meanings in the literature: for some authors, eq. $(5)$ represents "static" expectations in the sense that the expectation itself remains (or is expected to remain) unchanged in value. But you will often find authors who write the term "static expectations" and mean something totally different, namely $E(X_{t+1}\mid I_t) = X_t$ ("what is, will be"). This in turn looks like the martingale property but it is at best an extended concept of it (because the conditioning set is bigger), and in any case, it is a martingale-like property as regards the actual process $\{X_t\}$, and not the forecasts-process $\{X^e_{t+1|t}\}$.

Answer 2

Martingale is a very broad term, sometimes just basically meaning "the future is independent conditioned on today". Of course, any random walk has this property. So do Markov chains.

"Martingale" also usually refers to a real-valued random variable that changes over time, but whose expectation is always equal to its current value. An unbiased random walk also satisfies this because, given that the current location of the random walk is $x$, the expected location after $50$ steps is still $x$. It is hard to see how a Markov chain relates to this property in general because its state space might not be the real numbers.

The idea of the Doob martingale is that these two cases are almost the same: any time you have a Markov-chain type structure where the future is independent conditioned on the present, and you have a real-valued random variable $X_t$ that is changing over time, you can construct a new random variable $Y_t = \mathbb{E}[X_T | X_t]$, where $T$ is the time that the process stops. Now by conditional independence, $Y_t$ is a martingale.

Finally, a Bayesian agent's expectation of a random variable is a martingale, as they receive new information over time. To see this, consider a two-stage process: The agent begins with just a prior distribution on $Z$, then receives a signal $A$ and updates to a posterior on $Z$. Naturally, the posterior will be $\mathbb{E}[ Z \mid A]$. But the prior will just be $\mathbb{E} Z$, and the law of iterated expectation says that $$ \mathbb{E}_A\left[ \mathbb{E}_Z [Z \mid A] \right] = \mathbb{E} Z. $$ You can see this by working it out for yourself using the definition of expectation.