Alpha-Mixing and Other Dependence Coefficients

Why Measure Dependence Between the Past and the Future?

For a time series, dependence is not captured completely by asking whether two observations are correlated. One can instead ask a broader question: how much information about the distant future is contained in the entire past? Mixing coefficients give precise answers by comparing events in those two collections of observations with what their probabilities would be under independence.

This matters because many large-sample results for time series require dependence to weaken as observations become farther apart. Mixing conditions express that idea in terms of probabilities, not just covariances. They are properties of the process's joint distribution; they do not describe the uncertainty of a particular estimate.

Past and Future Information

Let $(X_t)_{t\in\mathbb Z}$ be a stochastic process indexed by integer time. For a fixed time $t$ and a positive integer separation $k$, write $\mathcal F_{-\infty}^{t}$ for the sigma-field generated by all observations $X_s$ with $s\leq t$, and $\mathcal F_{t+k}^{\infty}$ for the sigma-field generated by all observations $X_s$ with $s\geq t+k$. A sigma-field represents the events that can be determined from the specified observations.

The separation $k$ leaves observations $X_{t+1},\ldots,X_{t+k-1}$ between the past and the future. For a strictly stationary process, shifting $t$ does not change the coefficient below, so it is enough to define it at $t=0$. Strict stationarity means that the joint distribution is unchanged by a common shift of the time indices.

Definition: The alpha-mixing coefficient at separation $k$ is $$ \alpha(k)=\sup_{t\in\mathbb Z}\ \sup_{\substack{A\in\mathcal F_{-\infty}^{t}\\B\in\mathcal F_{t+k}^{\infty}}} \left|P(A\cap B)-P(A)P(B)\right|. $$ For a strictly stationary process, the outer supremum over $t$ can be omitted. The process is alpha-mixing (or strongly mixing) if $\alpha(k)\to 0$ as $k\to\infty$.

The quantity inside the absolute value is the difference between the actual probability that both events occur and the product of their probabilities. That product is what the joint probability would be if the events were independent. The supremum selects the pair of past and future events with the greatest discrepancy. Thus $\alpha(k)$ is small when every such pair is nearly independent, and it is zero exactly when the two sigma-fields are independent.

Beta-Mixing: A Stronger Comparison

Alpha-mixing checks one pair of events at a time. Beta-mixing instead compares the joint distribution across two collections of events, using finite partitions. A finite partition is a finite list of disjoint events whose union is the whole sample space.

Definition: The beta-mixing coefficient (also called the absolute-regularity coefficient) at separation $k$ is $$ \beta(k)=\sup_{t\in\mathbb Z}\ \sup_{\substack{(A_i)\text{ finite partition in }\mathcal F_{-\infty}^{t}\$B_j)\text{ finite partition in }\mathcal F_{t+k}^{\infty}}} \frac{1}{2}\sum_i\sum_j \left|P(A_i\cap B_j)-P(A_i)P(B_j)\right|. $$ The process is beta-mixing if \(\beta(k)\to 0$ as $k\to\infty$.

The factor $1/2$ is part of the convention used here. For any fixed pair of partitions, the sum measures the total discrepancy between the joint cell probabilities and the probabilities under independence. In particular, beta-mixing controls the dependence across whole partitions, not only one chosen event from each side.

Why Beta-Mixing Implies Alpha-Mixing

The two definitions are related directly. Take any past event $A$ and future event $B$. Use the two partitions $(A,A^c)$ and $(B,B^c)$ in the definition of beta. Set $\delta=P(A\cap B)-P(A)P(B)$. The four signed discrepancies for the partition cells are $\delta,-\delta,-\delta,\delta$. Their absolute values sum to $4|\delta|$, so the partition expression is $2|\delta|$. Taking suprema gives the bound below.

2\alpha(k)\leq \beta(k).

Consequently, if $\beta(k)\to 0$, then $\alpha(k)\to 0$. The reverse implication does not follow from this inequality: controlling each single event discrepancy does not by itself establish that the total discrepancy across partitions is small. Beta-mixing is therefore the stronger condition under these conventions.

Worked Example: Independent Observations

Suppose $X_t$, for integer $t$, are mutually independent and identically distributed random variables. This is a strictly stationary process. For any $k\geq 1$, the observations generating the past sigma-field and those generating the future sigma-field are disjoint collections of independent variables. The two sigma-fields are therefore independent.

For every past event $A$ and future event $B$, this gives $P(A\cap B)=P(A)P(B)$. Hence the supremum defining alpha is zero: $\alpha(k)=0$ for every $k\geq1$. For any finite partitions, joint cell probabilities also factor, so every term in the beta sum is zero and $\beta(k)=0$ for every $k\geq1$. Independence is stronger than merely having coefficients that approach zero: the coefficients vanish immediately at all positive separations.

Coefficient Decay in Concrete Processes

The next examples distinguish immediate independence, dependence with a finite reach, and dependence that fades gradually. A coefficient is a property of the entire process distribution, so its behavior can differ substantially from what is suggested by looking at only a few observations.

Worked Example: A Process With Finite Dependence Range

Let $(\varepsilon_t)$ be independent Bernoulli random variables with $P(\varepsilon_t=1)=P(\varepsilon_t=0)=1/2$, and define the two-component process $X_t=(\varepsilon_t,\varepsilon_{t-1})$. This construction is strictly stationary because shifting all time indices leaves the joint distribution of the independent innovations unchanged.

The past through time 0 is generated by innovations with indices at most 0. The future from time $k$ is generated by innovations with indices at least $k-1$. When $k\geq2$, these index sets are disjoint. The corresponding sigma-fields are independent, so $$ \alpha(k)=\beta(k)=0,\qquad k\geq2. $$

At $k=1$, the past event $A=\{\text{first component of }X_0=1\}$ and future event $B=\{\text{second component of }X_1=1\}$ both mean $\{\varepsilon_0=1\}$. Therefore $P(A\cap B)=1/2$, while $P(A)P(B)=1/4$, and the discrepancy is $1/4$. Thus $\alpha(1)\geq1/4$: the adjacent past and future are not independent. The important feature is that this dependence has a finite reach, as the exact zero coefficients from separation 2 onward show.

Worked Example: A Two-State Markov Chain

Let $X_t$ take values 0 and 1. Suppose that, at each step, the chain stays in its current state with probability $0.8$ and switches with probability $0.2$. Its stationary distribution assigns probability $1/2$ to each state, and we consider the chain initialized in that distribution. Write $r=0.8-0.2=0.6$.

After $k$ steps, the difference between the probabilities of being in state 1 given initial states 1 and 0 is $r^k$. This follows because a one-step difference is multiplied by $r$ at each transition. In particular, $$ P(X_k=1\mid X_0=1)=\frac{1+r^k}{2}, \qquad P(X_k=1\mid X_0=0)=\frac{1-r^k}{2}. $$

For the events $A=\{X_0=1\}$ and $B=\{X_k=1\}$, the discrepancy is $$ \left|P(A\cap B)-P(A)P(B)\right| =\left|\frac{1+r^k}{4}-\frac14\right| =\frac{r^k}{4}. $$ This attains the alpha-mixing supremum. To see why it is also an upper bound, condition on $X_0$. The Markov property says that, given $X_0$, the earlier past and later future are conditionally separated. The difference between the conditional probabilities of any future event given $X_0=1$ and $X_0=0$ is at most the total variation distance between the two conditional future distributions, which is $r^k$. For any past event $A$, its association with the binary variable $X_0$ is at most $1/4$. Thus every event-pair discrepancy is at most $r^k/4$, establishing $$ \alpha(k)=\frac{0.6^k}{4}. $$

For beta-mixing, the conditional distribution of the future given the past depends on the past only through $X_0$. Given $X_0$, the future from $k$ onward starts with the distribution of $X_k$ and then follows the same transition rule. Applying that common transition rule to two starting distributions does not increase their total variation distance. The distance between the conditional distribution of $X_k$ given $X_0$ and its stationary distribution is $r^k/2$; the future path retains this distance because it includes $X_k$. Averaging over the two equally likely states gives $$ \beta(k)=\frac{0.6^k}{2}. $$ At $k=1$, the values are $\alpha(1)=0.15$ and $\beta(1)=0.30$; at $k=3$, they are $\alpha(3)=0.0540$ and $\beta(3)=0.108$. Both decay geometrically to zero, with beta twice alpha in this example.

Worked Example: Dependence That Does Not Decay

Let $Z$ be a single Bernoulli random variable with $P(Z=1)=P(Z=0)=1/2$, and set $X_t=Z$ for every $t$. This process is strictly stationary: every finite collection of observations consists of repetitions of the same random variable, regardless of the time indices.

For any $k$, choose $A=\{X_0=1\}$ and $B=\{X_k=1\}$. Both events are $\{Z=1\}$, so $$ \left|P(A\cap B)-P(A)P(B)\right| =\left|\frac12-\frac14\right|=\frac14. $$ Thus $\alpha(k)\geq1/4$ for every $k$. In fact, since each event discrepancy for binary events is at most $1/4$, $\alpha(k)=1/4$. For beta-mixing, the partitions $(\{Z=1\},\{Z=0\})$ on each side give a sum of four absolute discrepancies equal to 1, hence $\beta(k)\geq1/2$. The stated inequality $2\alpha(k)\leq\beta(k)$ then implies $\beta(k)\geq1/2$; the two binary partitions attain this value, so $\beta(k)=1/2$. Neither coefficient tends to zero. Observing the past reveals the future completely, regardless of the separation.

Interpreting a Mixing Condition

A statement that a process is alpha-mixing or beta-mixing is a statement about the limit of the corresponding coefficients as $k$ grows. A numerical coefficient at one separation measures the largest permitted departure from independence at that separation; it is not the probability of a particular event, nor a correlation. The rate of decay also matters in applications: different asymptotic results impose different restrictions on how quickly the coefficients must approach zero.

Mixing is useful because it gives a disciplined way to formalize diminishing dependence between distant parts of a time series. It is not a claim that observations become literally independent at some finite lag, unless the coefficients are exactly zero there. Nor does stationarity alone ensure mixing: the permanently dependent process above is stationary but its coefficients do not decay.

Common Mistake

Wrong reasoning: “The autocorrelation becomes small at long lags, so the process must be alpha-mixing.” A small or zero correlation concerns particular numerical summaries of pairs of observations; alpha-mixing takes a supremum over all events in the past and future.
Correction: To establish alpha-mixing, show that the event-probability discrepancy is uniformly small over every such pair of events as the separation grows. The coefficient definition requires more than decay of a single dependence summary.
Wrong reasoning: “Alpha-mixing and beta-mixing are interchangeable names for the same condition.”
Correction: They are distinct conditions. With the convention used here, $2\alpha(k)\leq\beta(k)$, so beta-mixing implies alpha-mixing, but the inequality alone does not establish the reverse implication.

Key takeaway: Alpha-mixing measures the maximum discrepancy from independence for one past-event/future-event pair; beta-mixing measures a stronger total discrepancy across finite partitions. Decay to zero means dependence between the distant past and future weakens in the corresponding sense.

Check Your Understanding

Answer each question using the coefficient definitions and the examples above.

In the definition of $\alpha(k)$, what two probabilities are compared, and what does the supremum range over?
Why does the beta-mixing definition include a factor of $1/2$, and what convention for $\beta(k)$ has been used here?
For the process $X_t=(\varepsilon_t,\varepsilon_{t-1})$, why are the past through time 0 and the future from time 2 independent?
For the two-state Markov chain, calculate $\alpha(2)$ and $\beta(2)$ from the formulas in the worked example.
Why does strict stationarity not, by itself, imply alpha-mixing?