Equivalent MA(1) Representations And Invertibility

Why an MA(1) Process Can Have More Than One Representation

An MA(1) model describes a series using a current random shock and the shock from one period earlier. Its autocovariances reveal the dependence created by those shocks, but they do not always determine a unique coefficient and innovation variance. In fact, two distinct parameter pairs can produce the same mean, variance, and lag-one autocovariance. In applications, the invertibility condition supplies a standard way to choose between them.

The key is to compare pairs by the autocovariances they imply, rather than by their coefficients alone. This tutorial works through that comparison, including the special boundary case and an important feasibility check.

Definition: Consider the MA(1) model $$ X_t=\mu+\varepsilon_t+\theta\varepsilon_{t-1}, $$ where $\mu$ is constant and $\{\varepsilon_t\}$ is white noise with mean zero, variance $\sigma_\varepsilon^2>0$, and zero covariance between distinct time points. The parameter $\theta$ is the MA coefficient, and $\sigma_\varepsilon^2$ is the innovation variance. Two parameter pairs $(\theta,\sigma_\varepsilon^2)$ are second-order equivalent if they imply the same mean and all the same autocovariances. This is equivalence of second-order properties; without additional distributional assumptions, it does not necessarily imply identical full probability distributions.

Autocovariances and the Information They Contain

Write $\gamma_k=\operatorname{Cov}(X_t,X_{t-k})$ for the lag-$k$ autocovariance. Under the model above, the mean is $\mu$, and the autocovariances are

\gamma_0=(1+\theta^2)\sigma_\varepsilon^2,\qquad \gamma_1=\theta\sigma_\varepsilon^2,\qquad \gamma_k=0\quad\text{for }k\geq 2.

The variance includes contributions from both $\varepsilon_t$ and $\theta\varepsilon_{t-1}$. For lag one, the two consecutive observations share one shock: the coefficient of that shared shock is $1$ in $X_t$ and $\theta$ in $X_{t-1}$. At lags of two or more, the observations have no shocks in common, so their covariance is zero.

These formulas show why identifying $\theta$ and $\sigma_\varepsilon^2$ separately can be ambiguous. The observed ratio of lag-one autocovariance to variance is

\frac{\gamma_1}{\gamma_0}=\frac{\theta}{1+\theta^2}.

The right-hand side has the same value at $\theta$ and $1/\theta$, provided $\theta\neq 0$. Thus, reciprocal coefficients can yield the same autocovariance ratio. The innovation variance must also change for the full autocovariances to agree.

Deriving the Equivalent Parameter Pair

Suppose one representation has coefficient $\theta\neq 0$ and innovation variance $\sigma_\varepsilon^2$. Seek another representation with coefficient $\theta^*=1/\theta$ and variance $\sigma_{\varepsilon^*}^2$. To match the lag-one autocovariance, require

\theta^*\sigma_{\varepsilon^*}^2 =\theta\sigma_\varepsilon^2.

Substituting $\theta^*=1/\theta$ gives $\sigma_{\varepsilon^*}^2=\theta^2\sigma_\varepsilon^2$. This same variance adjustment matches the marginal variance:

\left(1+(\theta^*)^2\right)\sigma_{\varepsilon^*}^2 =\left(1+\frac{1}{\theta^2}\right)\theta^2\sigma_\varepsilon^2 =(1+\theta^2)\sigma_\varepsilon^2.

Both representations have zero autocovariances at lags $k\geq2$, so their entire autocovariance sequences agree. This proves the reciprocal relationship rather than merely suggesting it from the ratio.

Theorem: For $\theta\neq0$, the MA(1) parameter pair $(\theta,\sigma_\varepsilon^2)$ has the second-order equivalent pair $$ \left(\frac{1}{\theta},\ \theta^2\sigma_\varepsilon^2\right). $$ If $|\theta|\neq1$, these are two distinct pairs. If $\theta=1$ or $\theta=-1$, the reciprocal coefficient is unchanged, so there is only one pair, and it lies on the invertibility boundary.

Recovering Representations from Autocovariances

Given $\gamma_0>0$ and $\gamma_1$, let $r=\gamma_1/\gamma_0$. Any MA(1) representation must satisfy

r=\frac{\theta}{1+\theta^2}, \qquad r\theta^2-\theta+r=0.

For $r\neq0$, solving the quadratic gives the candidate coefficients. The product of its two roots is $1$, so when there are two distinct real roots, they are reciprocals. For each candidate, calculate the innovation variance from $\gamma_0=(1+\theta^2)\sigma_\varepsilon^2$:

\sigma_\varepsilon^2=\frac{\gamma_0}{1+\theta^2}.

A useful feasibility condition follows from the coefficient equation. The quadratic has real roots only if its discriminant is nonnegative:

1-4r^2\geq0, \qquad\text{equivalently}\qquad \left|\frac{\gamma_1}{\gamma_0}\right|\leq\frac12.

If $\gamma_1=0$, the equation reduces to $-\theta=0$, so the only finite-coefficient MA(1) representation has $\theta=0$ and $\sigma_\varepsilon^2=\gamma_0$. At the boundary $|r|=1/2$, the quadratic has a repeated root, $\theta=1$ or $\theta=-1$, depending on the sign of $r$. That representation is not strictly invertible.

Worked Example: Transforming a Noninvertible Pair

A centered hypothetical process has an MA(1) representation with $\theta=2$ and $\sigma_\varepsilon^2=3$. Find its equivalent reciprocal-coefficient representation, verify the autocovariances, and identify the invertible pair. Use the method of matching the autocovariances; the common mistake to avoid is changing the coefficient while leaving the innovation variance unchanged.

Step 1: Calculate the original autocovariances. The mean is zero, and

\gamma_0=(1+2^2)(3)=5(3)=15,\qquad \gamma_1=(2)(3)=6,\qquad \gamma_k=0\quad(k\geq2).

Step 2: Find the reciprocal pair. The reciprocal coefficient is $\theta^*=1/2$. Its innovation variance must be $\theta^2\sigma_\varepsilon^2$, so

\sigma_{\varepsilon^*}^2=2^2(3)=12.

Step 3: Verify both nonzero autocovariances. For the new pair,

\gamma_0^*=\left(1+\left(\frac12\right)^2\right)(12) =\frac54(12)=15,\qquad \gamma_1^*=\left(\frac12\right)(12)=6.

Its autocovariances at lags $k\geq2$ are also zero. The pair $(1/2,12)$ is strictly invertible because $|1/2|<1$; the original pair $(2,3)$ is not.

Check the common mistake. If the reciprocal coefficient $1/2$ were paired incorrectly with variance $3$, it would imply $\gamma_0=(5/4)(3)=3.75$ and $\gamma_1=(1/2)(3)=1.5$, not 15 and 6.

Worked Example: Finding Both Pairs from Autocovariances

A centered hypothetical series is known to have an MA(1) representation, with $\gamma_0=13$ and $\gamma_1=-6$. Find every coefficient and innovation-variance pair consistent with these autocovariances, then select the invertible representation. Use the quadratic method and check the resulting moments; do not treat the ratio alone as a complete answer.

Step 1: Form the ratio and equation.

r=\frac{-6}{13},\qquad -\frac{6}{13}\theta^2-\theta-\frac{6}{13}=0.

Multiplying by $-13$ gives $6\theta^2+13\theta+6=0$. Factoring,

(3\theta+2)(2\theta+3)=0, \qquad \theta=-\frac23\quad\text{or}\quad\theta=-\frac32.

Step 2: Calculate each innovation variance. For $\theta=-2/3$,

\sigma_\varepsilon^2 =\frac{13}{1+4/9} =\frac{13}{13/9}=9.

For $\theta=-3/2$,

\sigma_\varepsilon^2 =\frac{13}{1+9/4} =\frac{13}{13/4}=4.

Step 3: Check both pairs. The pair $(-2/3,9)$ gives $\gamma_0=(13/9)(9)=13$ and $\gamma_1=(-2/3)(9)=-6$. The pair $(-3/2,4)$ gives $\gamma_0=(13/4)(4)=13$ and $\gamma_1=(-3/2)(4)=-6$. Both give zero autocovariance at every lag of two or more.

Conclusion. The equivalent pairs are $(-2/3,9)$ and $(-3/2,4)$. Only $(-2/3,9)$ is strictly invertible, since $2/3<1$.

Worked Example: Checking Feasibility and the Boundary

A centered hypothetical process is proposed to be MA(1), with $\gamma_0=20$ and $\gamma_1=8$. Find the possible parameter pairs and the invertible one. The trap is to assume that every pair of stated autocovariances can come from an MA(1) model without checking the ratio.

Step 1: Check feasibility. Here $r=8/20=2/5$, and $|r|=0.4<1/2$, so two real coefficient candidates are possible.

Step 2: Solve for the coefficients.

\frac25\theta^2-\theta+\frac25=0 \quad\Longrightarrow\quad 2\theta^2-5\theta+2=0 \quad\Longrightarrow\quad (2\theta-1)(\theta-2)=0.

Thus $\theta=1/2$ or $\theta=2$. For $\theta=1/2$, the innovation variance is $20/(5/4)=16$. For $\theta=2$, it is $20/5=4$.

Step 3: Verify and select. The pair $(1/2,16)$ gives $\gamma_0=(5/4)(16)=20$ and $\gamma_1=(1/2)(16)=8$. The pair $(2,4)$ gives $\gamma_0=5(4)=20$ and $\gamma_1=(2)(4)=8$. Both have zero autocovariance for lags of two or more, but only $(1/2,16)$ is invertible.

Invertibility Selects a Standard Representation

For an MA(1), the MA polynomial is $1+\theta z$, with root $-1/\theta$ when $\theta\neq0$. The usual strict invertibility condition requires that this root lie outside the unit circle, which is equivalent to $|\theta|<1$. When two distinct reciprocal coefficients are possible, exactly one has absolute value below one. This is why the invertible representation is the standard choice for expressing the observed series in terms of its shocks.

The equivalence can also be understood through the inverse filter. In the centered model, write $B X_t=X_{t-1}$ for the backshift operation. If $|\theta|<1$, the geometric series converges in the usual mean-square sense, giving

\varepsilon_t =\frac{1}{1+\theta B}(X_t-\mu) =\sum_{j=0}^{\infty}(-\theta)^j(X_{t-j}-\mu).

The condition $|\theta|<1$ makes the coefficients of this infinite representation decay in absolute value. If $|\theta|=1$, the strict condition fails; if $|\theta|>1$, this inverse series does not have decaying coefficients. These facts distinguish selecting an invertible representation from merely finding a pair that matches the autocovariances.

Common Exam Traps

Reciprocating the coefficient but not adjusting the variance. The matching variance is $\theta^2\sigma_\varepsilon^2$, not the original variance in general.
Calling both representations invertible. For distinct reciprocal coefficients, one has absolute value below one and the other above one.
Confusing a strict condition with a weak one. A coefficient with $|\theta|=1$ is on the boundary, not strictly invertible.
Ignoring feasibility. If $|\gamma_1/\gamma_0|>1/2$, no real-coefficient MA(1) with positive innovation variance has those autocovariances.
Assuming second-order equivalence always means identical distributions. Matching the mean and autocovariances establishes second-order equivalence; stronger distributional equivalence needs additional assumptions.

For an exam calculation, a reliable sequence is to check the ratio, solve for the coefficient candidates, calculate the innovation variance for each candidate, verify both autocovariances, and then apply $|\theta|<1$. This keeps the equivalence calculation separate from the invertibility decision.

Check Your Understanding

For each problem, assume the stated process has a constant mean and is an MA(1) driven by zero-mean white noise with positive variance.

A centered process has $\theta=-3$ and innovation variance $2$. Find its reciprocal equivalent pair and state which pair is invertible. (Answer: original autocovariances $\gamma_0=20,\gamma_1=-6$; equivalent pair $(-1/3,18)$; $(-1/3,18)$ is invertible.)
A centered process has $\gamma_0=10$ and $\gamma_1=4$. Find all compatible pairs and identify the invertible one. (Answer: $(1/2,8)$ and $(2,2)$; the invertible pair is $(1/2,8)$.)
A centered process is proposed to have $\gamma_0=8$ and $\gamma_1=4$. Determine the coefficient and innovation variance, and state whether the model is strictly invertible. (Answer: $\theta=1$, $\sigma_\varepsilon^2=4$; it is not strictly invertible.)
A centered process is proposed to have $\gamma_0=12$ and $\gamma_1=-7$. Determine whether any real-coefficient MA(1) with positive innovation variance can have these autocovariances. (Answer: No; $|\gamma_1/\gamma_0|=7/12>1/2$.)
A centered process has $\theta=-1/4$ and innovation variance $8$. Find the equivalent reciprocal pair, verify its autocovariances, and identify the invertible representation. (Answer: original autocovariances $\gamma_0=17/2,\gamma_1=-2$; equivalent pair $(-4,1/2)$; $(-1/4,8)$ is invertible.)

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