-1
$\begingroup$

Can anyone explain to me why A white noise process (εt) and An IID process (εt) are martingales?

$\endgroup$
3
$\begingroup$

A stochastic process $\{X_t\}$ is called a martingale if

$$ \operatorname{E}[X_{t+1} \mid X_{t}, \ldots, X_1\} = X_t $$

That is, the expectation of the future conditional on the past is the present.

Fumio Hayashi's Econometrics defines a process $\{Z_t\}$ as white noise if $\operatorname{E}[Z_t] = 0$ and for any $j \neq 0$ $\operatorname{E}[Z_tZ_{t+j}] = 0$.

Let process $\{Y_t\}$ be a series of independent flips of a fair coin where $Y_t = 1$ if heads and $Y_t = -1$ if tails. Observe that:

  • $\{Y_t\}$ is white noise
  • $\{Y_t\}$ is NOT a martingale. If we flip a coin heads, we don't expect the next flip to be heads! The conditional expectation of $Y_t$ is always zero, not $Y_{t-1}$.

Perhaps what you're thinking? (or what your Prof is leading you to...)

A process $\Delta_t$ is called a martingale difference sequence if the conditional expectation of $\Delta_t$ given past information $\mathcal{F}_{t-1}$ is zero, that is, $\operatorname{E}[\Delta_t \mid \mathcal{F}_{t-1}] = 0$. Consequently a white noise process is a martingale difference sequence. Why is $\Delta_t$ called a martingale difference sequence? Define $X_t = X_{t-1} + \Delta_t$. Then $X_t$ is a martingale.

(Note also that a martingale difference sequence need not be white noise.)

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

if i recall correctly a martingale is a stochastic process for which the expectation of the next value in the sequence is equal to the present oberservation, even when we know all the earlier observations

Would a white noise not fall under that definition? having zero mean, and finite constant variance?

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Please clarify what process you are referring to when you mean "white noise". It seems to me that when a time series of (continuous) variables is independent, then the conditional mean of future values does not depend on the previous values. However, the previous values almost surely are not equal to their expectations. Thus, you seem to be saying that the values of random variables can alter the expectations of other random variables that are independent of them! What does form a martingale in this situation is the cumulative sum of the variables--not the variables themselves. $\endgroup$ – whuber Apr 25 '17 at 19:36

Not the answer you're looking for? Browse other questions tagged or ask your own question.