Can a Process Be Markov Without Being a Martingale?

A martingale has its current value as the conditional mean of its future value. A Markov process has a conditional future distribution determined by its current state and time. These conditions answer different questions. Brownian motion satisfies both. Brownian motion with nonzero constant drift is Markov and fails the martingale condition.

Same mean or same distribution?

For an adapted integrable process $X$ and information $\mathcal F_s$ , the martingale identity is

\mathbb E[X_t\mid\mathcal F_s]=X_s,\qquad s\le t.

A Markov identity for the natural filtration is

\mathbb E[g(X_t)\mid\mathcal F_s]=\mathbb E[g(X_t)\mid X_s]

for every bounded measurable function

g

. Testing all bounded functions captures the full conditional distribution. A single linear function captures only the mean. In a Markov model with integrable values, one can check the martingale condition by computing the conditional mean from the transition law.

For

X_t=W_t+\mu t

, we have

X_t=X_s+\mu(t-s)+(W_t-W_s)

. The increment of

W

is independent of the past, so the conditional distribution is normal with mean

X_s+\mu(t-s)

and variance

t-s

. Its dependence on

X_s

and elapsed time establishes the Markov property. The mean establishes the martingale property exactly when

\mu=0

A martingale can retain predictive information in its history. Let $H$ take values $1$ and $2$ with equal probability. Set $M_0=H$ . At time 1, toss an independent fair coin and set $M_1=0$ or $2H$ according to its result. At time 2, set $M_2=M_1+H\varepsilon$ , with an independent fair sign $\varepsilon$ .

The conditional mean of $M_1$ given $M_0$ is $H$ . The conditional mean of $M_2$ given the full past is $M_1$ . Hence the process is a martingale in its natural filtration. However, when $M_1=0$ , the next value could be $\pm1$ or $\pm2$ depending on $M_0$ . The past changes the conditional distribution, so $M$ is not Markov. The enlarged state $(M_t,H)$ restores the missing information.

The martingale counterexample is easiest to test on the event

M_1=0

. Both values of

H

can get there. If the earlier value

M_0

was 1, the next move is plus or minus 1. If it was 2, the next move is plus or minus 2. For instance, the conditional probability of

|M_2|=2

is then either zero or one depending on

M_0

. The conditional mean is zero in both cases. This is why checking only the mean cannot establish the Markov property for the proposed state variable.

Can a Process Be Markov Without Being a Martingale?

Same mean or same distribution?

Turn a fair random walk into a biased one

References and next steps