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I have $\gamma_l = Cov(r_t, r_{t-l})$ as a definition in my notes and now I need to find $gamma_l$ for a series $r_t- m = p(r_{t-1} - m) + a_t$ where $r_t$ is a linear time series with expected value $m$ and $a_t$ is a white noise.

Here is my attempt: $Cov(r_t - m, r_{t-l} - m) = E[(r_t - m)(r_{t-l} -m )] = E[p^2(r_{t-1} - m)(r_{t-l-1} -m ) + pa_t(r_{t-l-1} -m ) + pa_{t-l}(r_{t-1} - m) + a_ta_{t-l}] = p^2E[(r_{t-1} - m)(r_{t-l-1} -m )] + 0 + pE[a_{t-l}(r_{t-1} - m)] + E[a_t]E[a_{t-l}]$

Now if $l=0, = p^2E[(r_{t-1} - m)(r_{t-l-1} -m )] + pE[a_{t-l}(r_{t-1} - m)] + s_a^2 $

Otherwise, $= p^2E[(r_{t-1} - m)(r_{t-l-1} -m )] + pE[a_{t-l}(r_{t-1} - m)]$.

In either case, I'm stuck here. I am supposed to get everything in terms of $\gamma_{l-1}$ (or $\gamma_1$ in the case of 0, but I am not sure how to proceed.

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Let $v_t=r_t-m$, then $v_t=pv_{t-1}+a_t$. Covariance is invariant under de-meaning: $\gamma_l=cov(v_t,v_{t-l})$. Let’s first solve $\gamma_0$:

$$\gamma_0=cov(pv_{t-1}+a_t, pv_{t-1}+a_t)=p^2\gamma_0+\sigma_a^2 \rightarrow \gamma_0=\frac{\sigma_a^2}{1-p^2}$$

Note that $cov(v_{t-1},a_t)=0$ because previous output is irrelevant of the current white noise.

For $l=1$: $$cov(v_t,v_{t-1})=cov(pv_{t-1}+a_t,v_{t-1})=pcov(v_{t-1},v_{t-1})=p\gamma_0$$

For $l=2$: $$cov(v_t,v_{t-2})=cov(pv_{t-1}+a_t,v_{t-2})=pcov(v_{t-1},v_{t-2})=p\gamma_1=p^2\gamma_0$$

If you go on like this, you’ll have $\gamma_l=p^l\gamma_0$. Since $\gamma_l=\gamma_{-l}$, by definiton, this formula generalizes into the following:

$$\gamma_l=p^{|l|}\frac{\sigma_a^2}{1-p^2}$$

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