# Does a martingale difference sequence (mds) imply strong mixing?

I read this from an econometrics paper

" The typical hypothesis which is imposed in the time series literature is that the $$u_t$$'s are either independent and identically distributed (i.i.d.) or a martingale difference sequence (m.d.s.). In this work, we do not impose such strong assumptions..... We only assume that it satisfies a strong mixing condition. "

By strong mixing, the authors mean an $$\alpha$$-mixing condition.

Question: Does a m.d.s. imply strong mixing? Thanks

P.s. I did some googling for a few days and I cant seem to find satisfactory answers.

mds is yet another condition of weak dependence, but I do not think it necessarily implies strong-mixing. I have never seen this mentioned in the literature, even though it might take some time to elaborate a counterexample. Strong mixing is implied by several other mixings (beta, delta and uniform mixing for example). However, and this might be good news to you, if it helps, it is easy to check that a mds is necessarily a mixingale (you can definitely see it trivially assuming adaptedness and taking everything in the Lebesgue space of integrable random variables, just to simplify the mixingale definition to its bone).

• "...Strong mixing is implied by several other mixings (...uniform mixing for example)"---that is not quite true. Uniform mixing implies strong mixing, not the other way around. – Michael Oct 6 '20 at 4:10
• Isn t that what I wrote? Unless I m wrong about the english grammar passive construction (not a native speaker), writing that strong mixing is implied by uniform mixing is the same as writing that uniform mixing implies strong mixing, which is what you wrote... – GGG Oct 6 '20 at 17:43
• Right. Your statement was correct. Sorry about that. – Michael Oct 6 '20 at 20:30

Martingale conditions and mixing-type conditions are different. Neither one implies the other.

Here is one simple example of a MDS that is not even weak-mixing. Let $$X_1$$ take values $$1$$ or $$-1$$ with equal probability $$\frac12$$. Define $$X_t = - X_{t-1}$$ for $$t \geq 2$$. Then $$(X_t)$$ is a MDS. On the other hand, for any $$t$$ and $$h \geq 1$$, $$P (X_t = 1, X_{t+2h} = 1) - P (X_t = 1)P(X_{t+2h} = 1) = \frac14, \,$$ which does not approach zero as $$h \rightarrow \infty$$.

Mixing-type conditions also need not imply MDS. The fractional Gaussian noise, for example, is weak-mixing but not martingale. An MA(2) series $$X_t = \epsilon_t + \theta \epsilon_{t-1}, \;\; (\epsilon_t) \stackrel{i.i.d}{\sim}(0, \sigma^2)$$ would be an example of strong-mixing but not martingale.

(As previous answer points out, both martingales and strong-mixing processes are special cases of mixingales, first introduced by McLeish 1977 (Annals of Probability). Martingales are trivially mixingales. Strong-mixings are mixingales by mixing inequality.)