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Suppose I have the following $AR(p)$ model.

$$X_t = \sum_{i=1}^{p} \phi_i X_{t-i} + \epsilon_t\,, $$

where $\epsilon_t$ has mean 0 variance $\sigma^2$. I am not interested in fitting this model, but instead interested in results for the following quantities.

It is my understanding from here that the spectral density at 0 (which is also $\lim_{n\to \infty} n \text{Var}(\bar{X}_n)$) for this process is, $$f(0) = \dfrac{\sigma^2}{(1 - \sum_{i=1}^{p} \phi_i)^2} \,.$$

I also understand that the lag $k$ autocovariance $\gamma(k) = \text{E}[X_{t} X_{t-k}]$ is estimated by solving the Yule-Walker equations, and as such closed form expressions do not exist.

But since $f(0) = \sum_{k=-\infty}^{\infty} \gamma(k)$ by definition, can someone direct me to two proofs

  1. A direct proof that yields $$f(0) = \sum_{k=-\infty}^{\infty} \gamma(k) = \dfrac{\sigma^2}{(1 - \sum_{i=1}^{p} \phi_i)^2}\,,$$ without going through the theory on spectral densities.

  2. Is there a closed form expression for $\sum_{k=0}^{\infty} k\, \gamma(k)$?. Since Part 1 above has a closed form expression, my guess is there is a closed form expression here as well. Could someone lead me to it?


AR(1) Model: A closed for expression is available for the AR(1) model. $$Y_t = \phi Y_{t-1} + \epsilon_{t} \,. $$

The variance in the stationary process is known to be $\sigma^2/(1- \phi^2)$. The spectral density at 0 is $$f(0) = \dfrac{\sigma^2}{(1 - \phi)^2}\,. $$

Next, we consider $\sum_{k=0}^{\infty}k \gamma(k)$. \begin{align*} \sum_{k=0}^{\infty}k \gamma(k) &:= \sum_{k = 1}^{\infty} k \text{Cov}(Y_1, Y_{1+k})\\ & = \sum_{k = 1}^{\infty} \dfrac{\sigma^2}{1-\phi^2}k \phi^k\\ & = \dfrac{\phi\sigma^2}{1-\phi^2} \sum_{k=1}^{\infty} k \phi^{k-1}\\ & = \dfrac{\phi\sigma^2}{(1-\phi^2)(1-\phi)^2}\,. \end{align*} Thus, we have a closed form expression.


VAR(1) Model: We can do the same thing as above for the VAR(1) model. Suppose now $\Phi$ is a $p \times p$ matrix, $Y_t \in \mathbb{R}^p$ and $\epsilon_t \in \mathbb{R}^p$ with mean vector 0 ad variance covariance matrix $\Sigma$. The VAR(1) model is $$Y_t = \Phi Y_{t-1} + \epsilon_t\,. $$

Using the reference here, it is then known that the stationary process has variance matrix $V$ such that $vec(V) = (I_{p^2} - \Phi \otimes \Phi)^{-1} vec(\Sigma)$, where $\otimes$ is the Kronecker product.

Also, in same reference, we find that $\gamma(s) = \Phi^s V$, and $\gamma(-s) = \gamma(s)^T$. So, \begin{align*} f(0) & = \sum_{s=-\infty}^{\infty} \gamma(s)\\ & = \sum_{s=0}^{\infty}\gamma(s) + \sum_{s=0}^{\infty}\gamma(s)^T - \gamma(0)\\ & = \sum_{s=0}^{\infty} \Phi^s V + \sum_{s=0}^{\infty} V(\Phi^{T})^s - V \\ & = (I -\Phi)^{-1}V + V(I - \Phi^T)^{-1} - V. \end{align*}

Similarly, \begin{align*} \sum_{s=0}^{\infty} s\gamma(s) & = \sum_{s=0}^{\infty} s\Phi^s V \\ & = \left(\sum_{s=0}^{\infty} s\Phi^{s-1} \right) \Phi V \\ & = (I -\Phi)^{-2}\Phi V. \end{align*} Which again, gives a closed form solution. The same proof technique doesn't follow for the AR(p), because we don't have the nice geometric series expression. But considering there should be a relation between VAR(1) and an AR(p), can we use the results here to get an answer to my question (2)?

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  • $\begingroup$ You can view your AR(p) as a VAR(1) if you put it on companion form. Have you checked Hamilton (1994)? Basically the $\Phi$ matrix contains your AR parameters in the first row, an identity matrix ($p-1$) in the bottom left block and the final column (below the first row) is 0. The covariance matrix has a single non-zero element in the (1,1) element. Maybe you can simplify the general VAR(1) expression given this structure. $\endgroup$ – hejseb Oct 17 '18 at 4:12
  • $\begingroup$ @hejseb That seems like a viable option. However, looks like Taylor's answer is sufficient for me. In addition, working with the VAR structure is burdensome due to the large inverses and Kronecker products involved. $\endgroup$ – Greenparker Oct 17 '18 at 12:46
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Regarding 1, the first inequality is by definition of the spectral density. That's the only theory on spectral densities you need. The rest is just bilinearity of covariance, and definitions. Write your (causal) AR(p) process as $$ \phi(B) X_t = \epsilon_t $$ where $\phi(z) = 1 - \phi_1z - \cdots - \phi_p z^p$. This is the same as as $$ X_t = \phi^{-1}(B)\epsilon_t = \psi(B)\epsilon_t = \sum_{i=0}^{\infty}\psi_i \epsilon_{t-i} $$ where $\psi(z) = 1 + \psi_1 z + \psi_2 z^2 + \cdots$ is the reciprocal of $\phi(z)$.

Then you just plug everything in:

\begin{align*} \sum_{k=-\infty}^{\infty} \gamma(k) &= \sum_{k=-\infty}^{\infty}\text{cov}\left(\sum_{i=0}^{\infty}\psi_i \epsilon_{t-i}, \sum_{j=0}^{\infty}\psi_j \epsilon_{t+k-j} \right)\\ &= \sum_{k=-\infty}^{\infty} \sum_{i=0}^{\infty}\sum_{j=0}^{\infty} \psi_i \psi_j \text{cov}(\epsilon_{t-i}, \epsilon_{t+k-j} ) \\ &= \sigma^2 \sum_{k=-\infty}^{\infty} \sum_{i=0}^{\infty}\sum_{j=0}^{\infty} \psi_i \psi_j \mathbb{1}(t-i = t+k-j)\\ &= \sigma^2 \sum_{k=-\infty}^{\infty} \sum_{i=0}^{\infty}\psi_i \psi_{k+i} \\ &= \sigma^2 \sum_{i=0}^{\infty}\sum_{k=-\infty}^{\infty} \psi_i \psi_{k+i} \\ &= \sigma^2 \sum_{i=0}^{\infty}\sum_{k=-i}^{\infty} \psi_i \psi_{k+i} \\ &= \sigma^2\left(\sum_{k=0}^{\infty}\psi_k \right)^2\\ &= \sigma^2 \left(\frac{1}{1 - \sum_{i=1}^p\phi_i } \right)^2. \end{align*}

Regarding the second question: not sure, let me think about it. Can you say a little more about why you're interested in it?

Edit: I see you're using the formula for a low-order polylogarithm. You can use that again, but it seems like it gets a little messier.

Recall the Yule-Walker equations: $$ \gamma(k) = \sum_{i=1}^p \phi_i \gamma(k-i) $$ for $k=0,\ldots,p$. Then \begin{align*} \sum_{k=1}^{\infty} k \gamma(k) &= \sum_{k=1}^{\infty} k \sum_{i=1}^p \phi_i \gamma(k-i) \\ &= \sum_{i=1}^p \sum_{k=1}^{\infty} k \phi_i \gamma(k-i) \\ &= \sum_{i=1}^p \phi_i \left( \sum_{k=1}^{\infty} k \gamma(k-i) \right) \\ &= \left(\sum_{i=1}^p \phi_i \sum_{k=1}^{i} k \gamma(k-i) \right) + \left( \sum_{i=1}^p \phi_i \sum_{k=i+1}^{\infty} k \gamma(k-i) \right) \\ &= \left(\sum_{i=1}^p \phi_i \sum_{k=1}^{i} k \gamma(k-i) \right) + \left( \sum_{i=1}^p \phi_i \sum_{s=1}^{\infty} (s+i) \gamma(s) \right) \\ &= \left(\sum_{i=1}^p \phi_i \sum_{k=1}^{i} k \gamma(k-i) \right) + \left( \sum_{i=1}^p \phi_i \left[ \sum_{s=1}^{\infty} s \gamma(s) + i \sum_{s=1}^{\infty} \gamma(s) \right] \right) \\ &= \left(\sum_{i=1}^p \phi_i \sum_{k=1}^{i} k \gamma(k-i) \right) + \left( \sum_{i=1}^p \phi_i \sum_{s=1}^{\infty} s \gamma(s) \right)+ \left( \sum_{i=1}^p \phi_i i \sum_{s=1}^{\infty} \gamma(s) \right) . \end{align*} All of these bits have closed-form expressions.

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  • $\begingroup$ Thanks for the details here; this is very useful. $B$ here is the backward operator, correct? Could you explain how you got the last step? How is $\sum_{k=0}^{\infty} \psi_k = 1/(1 - \sum \phi_i)$? Regarding (2), that term show up as a bias in a variance estimator. To estimate the bias, I need to know what that sum corresponds to. $\endgroup$ – Greenparker Oct 15 '18 at 19:53
  • $\begingroup$ @Greenparker yep, that's right. And $\psi(z)$ is defined to be the reciprocal of the $\phi(z)$ (complex) polynomial. In other words $(\psi_0 + \psi_1 z^1 + \psi_2 z^2 + \cdots)(1 - \phi_1 z - \phi_2 z^2 - \cdots - \phi_p z^p) = 1$. As long as the AR(p) model is causal, we won't divide by zero if we move the $\phi(B)$ to the other side in the second equation. $\endgroup$ – Taylor Oct 15 '18 at 23:09
  • $\begingroup$ Thanks, that clears things up. Regarding (2), the closed form expression is available for AR(1). I am intuitively certain there is a closed form expression for (2) for the general AR(p). I know the closed form expression for (2) for a VAR(1) model, and there must be a connection between an AR(p) and a $p$-variate VAR(1). $\endgroup$ – Greenparker Oct 16 '18 at 13:32
  • $\begingroup$ @Greenparker can I see a reference? I might be able to change a few pieces to make it work. $\endgroup$ – Taylor Oct 16 '18 at 13:47
  • $\begingroup$ Taylor, I don't have a reference since I a lot of the math on my own. I edited the question, to indicate the closed form expressions for AR(1) and VAR(1). Hopefully you can find this useful. $\endgroup$ – Greenparker Oct 16 '18 at 14:17

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