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I have a AR(1) process $X_t = 0.4X_{t-1} + Z_t$. I multiplied by $X_{t-k}$, took the expectation of both sides, divided by the variance (which I assume is stationary), and got the Yule-Walker equation:

$$\rho(k) = 0.4\rho(k-1)$$

I thought that this is true for every $k$, but I was told that this is true only for $k\ge 1$. Why is that? where did I assume that in my derivation?

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Because, for $k=0$, you'll have: $$\begin{align}\rho(0)&=E[X_t^2]=E[(0.4X_{t-1}+Z_t)^2]=(0.4)^2\underbrace{E[X_{t-1}^2]}_{\rho(0)}+0.8\underbrace{E[X_{t-1}Z_t]}_{0}+E[Z_t^2]\\&=0.16\rho(0)+\sigma_z^2\rightarrow \rho(0)=\frac{\sigma_z^2}{1-0.16}\end{align}$$ Typical assumptions are $E[Z_t]=0$ and $E[X_{t-1}Z_t]=0$.

But, let $k\geq1$: $$\begin{align}\rho(k)&=E[X_tX_{t-k}]=E[(0.4X_{t-1}+Z_t)X_{t-k}]=0.4E[X_{t-1}X_{t-k}]+\underbrace{E[X_{t-k}Z_t]}_{0}\\&=0.4\rho(k-1)\end{align}$$

You can't assume $E[X_{t-k}Z_t]=0$ when $k=0$ because $$E[X_tZ_t]=0.4\underbrace{E[X_{t-1}Z_t]}_{0}+E[Z_t^2]=\sigma_z^2$$ You can only assume the independence of current output and the future input; not current input and the current output.

Note: for all $k$, we have $\rho(k)=\rho(-k)$ so you don't need to consider $k\leq -1$ separately.

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  • $\begingroup$ isn't it (1-0.16) in the denominator? $\endgroup$
    – ihadanny
    Commented Aug 25, 2019 at 6:54
  • $\begingroup$ so you mean that it's true for every $k\neq0$ $\endgroup$
    – ihadanny
    Commented Aug 25, 2019 at 6:56
  • $\begingroup$ Oh right, I've corrected the denominator, not the exact relation, $\rho$ is symmetric around $0$. For example: $\rho(2)=0.4\rho(1)\rightarrow \rho(-2)=0.4\rho(-1)$ $\endgroup$
    – gunes
    Commented Aug 25, 2019 at 7:22
  • $\begingroup$ so it's not true for k < 0! there it's $\rho(k) = 0.4\rho(k+1)$ - why is that? where in your derivation did you assume that k>=1 and not simply $k\neq0$? $\endgroup$
    – ihadanny
    Commented Aug 25, 2019 at 10:14
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    $\begingroup$ Right, the nomenclature for auto-correlation is not stable across disciplines: see here: wikiwand.com/en/Autocorrelation#/Normalization ; in my answer, $\rho$ denotes the auto-covariance function, but simple scaling with $\lambda(0) $yields the autocorrelation function. So, the relation still holds. Mean of $X_t$ will be $0$ when mean of $Z_t$ is $0$; which typically is, since $Z$ is modelled as a noise process. $\endgroup$
    – gunes
    Commented Aug 29, 2019 at 20:27

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