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I'm trying to find the autocorrelation function $\rho_x(h)$ for the stationary process

$X_t + 0.1X_{t-1} = Z_t$

So I have that

$\rho_x(h) = Corr[X_t, X_{t+h}] = \frac{\gamma_x(h)}{\gamma_x(0)}$ where $\gamma_x(h) = \frac{\phi^{|h|}\sigma_z^2}{1-\phi^2}$ and $\gamma_x(0) = \frac{\sigma_z^2}{1-\phi^2}$

This simplfies to

$\rho_x(h) = \phi^{|h|}$

but I'm not sure how to compute this.

I've been trying to use the recursive definition $X_t = \phi^hX_{t-h} = \sum_{j=0}^{h-1}\phi^jZ_{t-j}$, subbing in $0.1$ for $\phi$, but I have no idea how to figure this out.

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  • $\begingroup$ What you want to show is that $\rho_x(h) = \gamma_x(h)/\gamma_x(0)$? $\endgroup$ – caverac Sep 24 '17 at 19:58
  • $\begingroup$ What do you mean with "but I'm not sure how to compute this"? If you have the value for $\gamma_x(h)$ and how it relates to $\rho_x(h)$ then what are you missing? Do you want to show that $\gamma_x(h) = \phi^{\lvert h\rvert} \sigma_z^2/(1-\phi^2)$? $\endgroup$ – Therkel Sep 25 '17 at 12:33
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A few preliminaries

$X_t = \sum_{j = 0}^{+\infty}\phi^jZ_{t-j}$

You can prove this one this one just by writing

\begin{eqnarray} X_t &=& \sum_{j = 0}^{+\infty}\phi^jZ_{t-j} = \phi^0Z_{t-0} + \sum_{j = 1}^{+\infty}\phi^jZ_{t-j} \\ &=& Z_t + \phi \sum_{j = 1}^{+\infty}\phi^{j-1}Z_{t-j} \stackrel{k=j-1}{=} Z_t + \phi \sum_{k = 0}^{+\infty}\phi^{k}Z_{t- (k + 1)}\\ &=& Z_t + \phi\sum_{k=0}^{+\infty}\phi^kZ_{(t-1)-k} \\ &=& Z_t +\phi X_{t-1} \tag{1} \end{eqnarray}

where $Z_t$ is white noise with $\mathbb{E}[X_t] = 0$ and ${\rm Var}[X_t] = \sigma^2$

Roots of the characteristic equation

the characteristic equation for this process is

$$ 1 -\phi z =0 \tag{2} $$

which has only one root $\lambda = 1/\phi$. The process is thus covariance-stationary if $|\lambda| > 1$, or equivalently if $|\phi| < 1$

${\rm Var(X_t) = \sigma^2/(1-\phi^2)}$

If $|\phi|<1$ then $\{X_t\}$ is covariante stationary, and ${\rm Var}[X_t] = \sigma_X^2$ for all time $t$, therefore

\begin{eqnarray} {\rm Var}[X_t] &=& {\rm Var}[\phi X_{t-1}] + {\rm Var}[Z_t] \\ \sigma_X^2 &=& \phi^2 \sigma_X^2 + \sigma^2 \\ \sigma_X^2 &=& \frac{\sigma^2}{1 - \phi^2} \tag{3} \end{eqnarray}

${\rm Cov}(X_t, X_{t-1}) = \phi \sigma_X^2$

\begin{eqnarray} {\rm Cov}(X_t, X_{t- 1}) &=& {\rm Cov}(\phi X_{t- 1} + Z_t, X_{t-1}) = \phi {\rm Cov}(X_{t-1}, X_{t-1}) = \phi {\rm Var}[X_{t-1}] = \phi \sigma_X^2 \tag{4} \end{eqnarray}

We have then \begin{eqnarray} {\rm Cov}(X_t, X_{t- 1}) &=& \phi {\rm Cov}(X_{t-1}, X_{t-1}) = \phi \sigma_X^2\\ {\rm Cov}(X_t, X_{t- 2}) &=& {\rm Cov}(\phi X_{t - 1} + Z_t, X_{t-2}) = \phi{\rm Cov}( X_{t - 1}, X_{t-2}) \stackrel{(4)}{=} \phi^2\sigma_X^2 \\ &\vdots& \\ {\rm Cov}(X_t, X_{t- h}) &=& \phi^h \sigma_X^2 \tag{5} \end{eqnarray}

${\rm Corr}(X_t, X_{t-1}) = \phi^h$

If you call

$$ {\rm Cov}(X_t, X_{t- h}) =\gamma(h) \tag{6} $$

Then

$$ {\rm Corr}(X_t,X_{t-h}) = \frac{\gamma(h)}{\gamma(0)} = \phi^h \tag{7} $$

with

$$ \gamma(0) = {\rm Var}[X_t] = \sigma_X^2 $$

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  • $\begingroup$ I'm following all the preliminaries and theories, but when it comes to doing an actual example, I don't know what to compute. I see that $\phi = 0.1$ but I don't know what else to "plug in". How do I find $\sigma$ and $h$? $\endgroup$ – jon givony Sep 26 '17 at 1:48
  • $\begingroup$ @jongivony In your example $\phi = -0.1$. $\sigma$ is the variance of the "noise" term, in a practical application you can measure it from the residual of a fit of the form $X_{t} = c + \beta X_{t-1} + \epsilon$, and then $\sigma^2 = {\rm Var}[\epsilon]$. Finally $h$ is just the lag, i.e. how many observations in the past are you interested in, what Eq. (7) is telling you is that if you take an observation today in this process, it correlates with past observations all the way back to the beginning, but the earlier you go, the weaker the correlation $\endgroup$ – caverac Sep 26 '17 at 5:45

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