Having that $\ v(n) = [x(n),x(n-1),x(n-2))]^T $, and being $\ x(n) $ an autoregressive process AR(1) with known variance $\ \sigma_v^2 $ and transfer function $\ H(z) ={ 1 \over {1-0.7z^-1}}$, how could I obtain the autocorrelation matrix $\ R_{vv} = E[v(n), v^T(n)] $ ? Any help is appreciated!


1 Answer 1


Hi: You're expression will be a 3 by 3 matrix with the 1's along the diagonal and the correlations in the respective places.

$\begin{matrix} 1 & \sigma_{12} & \sigma_{13} \\ \sigma_{21} & 1 & \sigma_{23} \\ \sigma_{31} & \sigma_{32} & 1 \end{matrix}$

The 12 and 21 elements are $\rho$ because the lag one autocorrelation of an AR(1) with parameter $\rho$ is $\rho$.

The 23 and 32 elements are also $\rho$ for the same reason as above: The lag one autocorrelation of an AR(1) with parameter $\rho$ is $\rho$.

The 13 and 31 elements are $\rho^2 $ because the lag 2 autocorrelation of an AR(1) with parameter $\rho$ is $\rho^2$.

The transfer function implies that $\rho = 0.7$. This is because

$ y_t = \rho \times y_{t-1} + \epsilon_t$ has transfer function $H(z) = \frac{1}{(1-\rho z^{-1})}$.

  • $\begingroup$ So that means that the autocorrelation function of $x(n)$, i.e. $r_{xx}(n)$ is equal to 1? Why does it happen? On the other side, I do not understand why in the diagonal we will have ones. Could you help me with these little questions? Thanks for the answer! @mark-leeds $\endgroup$
    – Ragnar
    Oct 10, 2017 at 12:21
  • $\begingroup$ Hi: the diagonals of any correlation matrix represent the scaled variances since they are the autocorrelations at lag zero. so, the diagonals of any correlation matrix are always one. this is the same as saying that diagonals of a covariance matrix are the variances because the correlation matrix is just the scaled covariance matrix. autocorrelation is always associated with a lag. the ones on the diagonal have lag zero so they are the correlations of x at lag zero which is 1.0. $\endgroup$
    – mark leeds
    Oct 10, 2017 at 17:25
  • $\begingroup$ not sure what you mean by $r_{xx}(n)$. If $x$ represents the 3 element vector then $r_{xx}$ is the correlation matrix. but $r_{xx}(n)$ denotes the correlation of the $n$ th element with itself which is 1.0. the correlation of an individual time series observation, $x_{t}$ at lag zero is always 1.0. $\endgroup$
    – mark leeds
    Oct 10, 2017 at 17:30
  • $\begingroup$ But $x$ doesn't represent the 3 element vector, $x$ is just an AR(1) with a given $\sigma_v^2$ and $H(z)$. So how could I get it's autocorrelation function $r_{xx}(m)$ or its power spectral density $S_{xx}(\omega)$? If you mean that it is 1, I still can't understand why :( Btw, thanks for the answer @mark-leeds $\endgroup$
    – Ragnar
    Oct 10, 2017 at 18:54
  • $\begingroup$ Hi: Can you tell me what $r_{xx}(m)$ is ? I'm from statistics so maybe you're notation is different ? If it's the mth lag autocorrelation of an AR(1), then it equals $\rho^m$. Right, the 3 element vector is $v$ and the matrix I wrote at the beginning is the covariance matrix of $v$. I'll have to look up the power spectral density but let me know if my understnad is correct. $\endgroup$
    – mark leeds
    Oct 11, 2017 at 15:03

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