This question concerns the asymptotic variance of an $\text{ARMA}(p,q)$ process. Suppose that an $\text{ARMA}$ process can be rewritten as an $\text{MA}(\infty)$ process, and from this we can in principle derive the variance. First, let's start with a stationary $\text{AR}(1)$ process, whose variance is given by:

$$\mathbb{V}(X_t) = \frac{\sigma^2}{1-\phi^2}.$$

If we rewrite the model in lag notation we have $(1-\phi L) X_t = \epsilon_t$ which we can invert to get to:

$$X_t = \frac{\epsilon_t}{1-\phi L}.$$

Now if we take the variance of both sides, this should yield:

$$\mathbb{V}(X_t) = \frac{\mathbb{V}(\epsilon_t)}{(1-\phi L )^{2}} = \frac{\mathbb{V}(\epsilon_t)}{(1-\phi)^{2}} \neq \frac{\mathbb{V}(\epsilon_t)}{1-\phi^2}.$$

This already confuses me, since it gives a different expression than the proper expression for the variance. It is clear that if we re-write this as a geometric sum and take variances of the error terms (noticing that covariance of errors at different times is zero due to IID assumption) we receive the correct expression for the variance. How is it then possible that $(1-\phi L)^2 = 1-\phi^2$ when evaluating $(1-\phi L)$ at $L=1$?

Furthermore, in the general case of a stationary and invertible $\text{ARMA}(p,q)$ model, textbooks (like Hamilton) indicate that the variance can be derived by inverting the $\text{AR}$ polynomial, giving us a $\Psi$ polynomial (Wold decomposition) and then squaring the value of this polynomial evaluated at $L=1$ multiplied by the error variance --- i.e. $\mathbb{V}(X_t) = \sigma^2 \cdot \Psi(1)^2$. It seems that already in the simple case of an $\text{AR}(1)$ model this argument already doesn't work, so how could it possibly work in the much more complicated case of a general $\text{ARMA}(p,q)$?

  • $\begingroup$ Please edit your post to include citations to the sources of the assertions you find confusing. Most of them look wrong, so it would be useful to see where these assertions are coming from. $\endgroup$
    – Ben
    Jun 9, 2020 at 22:58

2 Answers 2


Preliminary matters involving lag operators: The lag operator is a function that operates on a sequence of values to push the index $t \in \mathbb{Z}$ backward by one point in time. Despite a notational convention that makes it look like it does, the lag operator does not operate on the actual values in the sequence, but the sequence itself. Thus, for a sequence $\mathbf{x} \equiv \{ x_t | t \in \mathbb{Z} \}$ we get the lagged sequence:

$$L \mathbf{x} = \{ x_{t-1} | t \in \mathbb{Z} \}.$$

When we refer to $L x_t = x_{t-1}$ this is actually a shorthand notation for $L x_t \equiv (L \mathbf{x})_t = x_{t-1}$ --- i.e., the lag operator is not actually operating on the value $x_t$ even though the notation makes it look like it does. Instead, it operates on the sequence, and then we substitute the time index afterwards.

This brings us to the issue of applying a lag operator to a constant. Strictly speaking, the operator only operates on a sequence indexed by $t \in \mathbb{Z}$, but we generally allow it to operate on a constant by using the convention of treating the constant as a constant sequence. That is, for any constant $\phi \in \mathbb{R}$ we suppose that there is a corresponding sequence $\boldsymbol{\phi} \equiv \{ \phi_t | t \in \mathbb{Z} \}$ with $\phi_t = \phi$ for all $t \in \mathbb{Z}$. We then have $L \boldsymbol{\phi} = \{ \phi_{t-1} | t \in \mathbb{Z} \}$, so we get $L \phi \equiv (L \boldsymbol{\phi})_t = \phi_{t-1} = \phi$. By treating the constant as if it were a sequence of constants, we can apply the lag operator and this moves the time index backward by one, which does not change the constant value that is the output ---i.e., we have the rule $L \phi = \phi$ for any constant $\phi \in \mathbb{R}$.

Your error using the lag polynomial: Since $L \phi = \phi$ you also have $(1-L\phi)^2 = (1-\phi)^2$, so your error does not lie in the step you have mentioned in your question. Rather, your error lies in the implicit use of the (erroneous) step where you take:

$$\mathbb{V} \bigg( \frac{\epsilon_t}{1-\phi L} \bigg) = \frac{\mathbb{V} (\epsilon_t)}{(1-\phi L)^2} \quad \quad \quad (\text{Erroneous equation})$$

This is part of your more general error of assuming that $\mathbb{V}(\Psi(L) \epsilon_t) = \Psi(L)^2 \mathbb{V}(\epsilon_t)$, which is not a valid equation. To see the correct equation in this case, let's expand out $(1-\phi L)^{-1}$ into its geometric expansion to give:

$$\begin{aligned} \mathbb{V} \bigg( \frac{\epsilon_t}{1-\phi L} \bigg) &= \mathbb{V} \bigg( \sum_{k=0}^\infty \phi^k L^k \epsilon_t \bigg) \\[6pt] &= \mathbb{V} \bigg( \sum_{k=0}^\infty \phi^k \epsilon_{t-k} \bigg) \\[6pt] &= \sum_{k=0}^\infty \phi^{2k} \mathbb{V}(\epsilon_{t-k}) \\[6pt] &= \sigma^2 \sum_{k=0}^\infty \phi^{2k} \\[6pt] &= \frac{\sigma^2}{1-\phi^2} \\[6pt] &\neq \frac{\sigma^2}{(1-\phi L)^2}. \\[6pt] \end{aligned}$$

As you can see, the quadratic effect of the variance operation means that the $\phi$ values end up getting squared but the lag operations do not. This leads to a result that is different from the erroneous equation shown above, which you have implicitly used in your working.

Application to invertible ARMA models: You should now be able to see what happens when you apply the lag operator to the objects in your question. However, even with this done correctly, several of the equations in your question do indeed appear to be incorrect. (Since you have not shown us the sources of those equations, I have no idea whether you actually got these from textbooks or just misinterpreted other results.) If we present an $\text{ARMA}(p,q)$ model in its $\text{MA}(\infty)$ form then we have:

$$X_t = \Psi(L) \varepsilon_t = \sum_{k=0}^\infty \psi_k \varepsilon_{t-k}.$$

Assuming that the error terms are IID with zero mean and variance $\mathbb{V}(\varepsilon_t) = \sigma^2$, we then have:

$$\begin{aligned} \mathbb{V}(X_t) &= \mathbb{V} \bigg( \sum_{k=0}^\infty \psi_k \varepsilon_{t-k} \bigg) \\[6pt] &= \sum_{k=0}^\infty \mathbb{V}( \psi_k \varepsilon_{t-k}) \\[6pt] &= \sum_{k=0}^\infty \psi_k^2 \mathbb{V} ( \varepsilon_{t-k} ) \\[6pt] &= \sigma^2 \sum_{k=0}^\infty \psi_k^2. \\[6pt] \end{aligned}$$

Since $\Psi(1)^2 = (\sum_{k=0}^\infty \psi_k)^2$ it is not generally true that $\Psi(1)^2 = \sum_{k=0}^\infty \psi_k^2$. Thus, if you have seen assertions that $\mathbb{V}(X_t) = \sigma^2 \Psi(1)^2$ then that is wrong. However, it can be shown that $|\Psi(1)|<\infty$ is a sufficient condition for $\sum_{k=0}^\infty \psi_k^2 < \infty$ so that the variance of the process exists.

  • $\begingroup$ Thank you, that makes a lot of sense. I now found that I confused two formulas: One for the variance (or 0th covariance) of the ARMA process, and one for the variance of the mean of the ARMA process. As I said, clearly $(1-\phi)^2\neq1-\phi^2$, which confused me. But since the formulas are for different things, they don't need to equal each other (and in most cases, can't) $\endgroup$
    – Orandronor
    Jun 10, 2020 at 9:24

Hi: It's not a big deal but, in the fourth line, you left out the lag operator $L$.

As far as your argument, I'm not a math person but I don't think squaring the denominator is legal because there's a lag operator in there. The evaluation at $L = 1$ has to be done AFTER the summation is carried out. Maybe someone on math.stackexchange could explain why that is ?

See page 78 of below for an example of where the evaluation at L = 1 is done at the end. The link uses $B$ instead of $L$ but the meaning is the same.


  • $\begingroup$ I think Ben just answered my question but I have to read it more carefully. Thanks Ben for detailed explanation. $\endgroup$
    – mlofton
    Jun 9, 2020 at 23:04
  • $\begingroup$ You can certainly square an equation with the lag operator in it. Indeed, we commonly use lag polynomials in ARMA models. $\endgroup$
    – Ben
    Jun 10, 2020 at 4:57
  • $\begingroup$ Hi Ben: But if you can square an expression that contains the lag operator, then how come the variance of the AR(1) in Orandronor's example, namely, $var(x_t) = \frac{\sigma^2}{(1- \phi)^2}$, doesn't result in the correct expression for the long tern variance of the AR(1). The expression is immediately before where he writes: "This already confuses me". Thanks. $\endgroup$
    – mlofton
    Jun 10, 2020 at 21:25
  • $\begingroup$ I have added an additional section to my answer to expand on this. $\endgroup$
    – Ben
    Jun 11, 2020 at 0:48
  • 1
    $\begingroup$ Thanks Ben. I'll check it out tomorrow. Brain was switched off for the rest of night. Trying to turn back on doesn't work. $\endgroup$
    – mlofton
    Jun 11, 2020 at 3:16

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