I have the following equation:

$$x(n)=\frac{14}{24}x(n-1)+\frac{9}{24}x(n-2)-\frac{1}{24}x(n-3)+w(n)$$ where, $w(n)$ is a stationary white noise process with variance $\sigma^2_w$

Now, I want to determine the auto correlation sequence $\gamma_{xx}(m)$, for $m = 0,1,\ldots,5$

Now, I know that the autocorrelation function for AR process which is given below:

$$\gamma_{xx}(m)= \begin{cases}\displaystyle-\sum_{k=1}^p a_k\gamma_{xx}(m-k), & m > 0\\ \displaystyle -\sum_{k=1}^p a_k\gamma_{xx}(m-k) + \sigma_w^2, & m = 0\\ \gamma_{xx}^*(-m), & m < 0\\ \end{cases}$$

Here, I already found the coefficients of ${a_k}$ which are $a_1=-\frac{14}{24}, a_2=-\frac{9}{24}, a_3=\frac{1}{24}$.

Now, I don't get, how can I calculate $\gamma_{xx}(0)$ to $\gamma_{xx}(5)$ recursively?

Please someone tell me how can I solve $\gamma_{xx}(0)$ and $\gamma_{xx}(1)$? Then I can do the the rest.


1 Answer 1


Looking at your definition of $\gamma_{xx}(m)$, it seems that you have 4 equations with 4 unknown (for $m=0,\ldots,3$):

\begin{align} 24\gamma_{xx}(0) &= 14\gamma_{xx}(1) + 9\gamma_{xx}(2) - \gamma_{xx}(3) + 24\sigma_w^2 \\ 24\gamma_{xx}(1) &= 14\gamma_{xx}(0) + 9\gamma_{xx}(1) -\gamma_{xx}(2) \\ 24\gamma_{xx}(2) &= 14\gamma_{xx}(1) + 9\gamma_{xx}(0) - \gamma_{xx}(1)\\ 24\gamma_{xx}(3) &= 14\gamma_{xx}(2) + 9\gamma_{xx}(1) - \gamma_{xx}(0) \end{align}

Solving the above system of equations is straight forward, and after that you can easily obtain $\gamma_{xx}(m)$ for $m=4,5$

  • $\begingroup$ Hi: I'm not sure where you got the formula for $\gamma_{xx}(0)$ but the zero lag autocorrelation for any stochastic process is 1.0. $\endgroup$
    – mark leeds
    Jan 23, 2018 at 16:10
  • $\begingroup$ @mark leeds: I don't think that is right. Using the definition from en.wikipedia.org/wiki/Autocorrelation#Signal_processing, an autocorrelation can take any value, depending on the signal amplitude. $\endgroup$ Jul 22, 2018 at 21:57
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    $\begingroup$ @JonasSchwarz Hi Jonas: Sometimes the defintion of autocorrelation in DSP is what autocovariance is defined as in statistics. So, yes, the autocovariance ( in statistics ) can take on any value,. But, if one normalizes the autocovariance at lag zero, then by definition, it will be 1.0. The normalized expression ( at any lag ) is what statisticians refer to as the autocorrelation. . My background is statistics background so this issue has come up before in other threads. $\endgroup$
    – mark leeds
    Jul 24, 2018 at 4:54
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    $\begingroup$ Jonas: I only glanced but I think pages 19 and 20 of this link provide the method for solving the original question in case you are interested. All the best. euclid.ucc.ie/Eric/st4064/course_docs/ST4064-TSA_notes.pdf $\endgroup$
    – mark leeds
    Jul 24, 2018 at 5:24

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