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What is the intuition behind the result that the autocorrelation function of a random walk process $y_{t}=y_{t-1}+e_{t}$ tends to 1 as $t\rightarrow 0$? Thank you.

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  • $\begingroup$ Tends to 1? Can you provide source for that statement? It should be 0 for the process you describe $\endgroup$ – 1muflon1 Feb 20 at 19:04
  • $\begingroup$ This isn't true - by inspection, the correlation between successive random walk outputs (for any $t$) should be very high, differing from 1 only to the extent that $var[e_t]$ is large. With that said, I don't think the problem is being posed correctly. $\endgroup$ – heh Feb 20 at 20:05
  • $\begingroup$ @1muflon1 Given the random walk process $y_{t}=y_{t-1}+e_{t}$, the auto correlation function is given by $corr(y_{t}, y_{t-h})=(\frac{t-h}{t})^{1/2}=(1-\frac{h}{t})^{1/2}$, which tends to 0 as t tends to infinity. What is the intuition behind this result? $\endgroup$ – Bob Charles Feb 20 at 22:24
  • $\begingroup$ @heh You are gravely mistaken, and I refer you to my proof to this effect commented above. $\endgroup$ – Bob Charles Feb 20 at 22:25
  • $\begingroup$ @BobCharles So in the question you say $t \to \infty$ but in the comment you say it's $t \to 0$? $\endgroup$ – Art Feb 21 at 6:29
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By recursive substitution you obtain $y_t = \sum_{j=0}^t e_{t-j}.$ Thus $\forall p \in \mathbb{N} \hspace{.2cm} y_{t-p} = \sum_{j=0}^{t-p} e_{t-p-j}.$

Under the usual white noise assumption for the errors $i \neq j \Rightarrow E(e_je_i) =0\Rightarrow Cov(y_t, y_{t-p}) = E(\sum_{j=0}^{t-p} e_{t-p-j}^2) = (t-p+1)\sigma_e^2.$

$Var(y_t) = E(\sum_{j=0}^{t} e_{t}^2) = (t+1)\sigma_e^2 \hspace{.2cm} \wedge Var(y_{t-p}) = E(\sum_{j=0}^{t-p} e_{t-p-j}^2)(t-p+1)\sigma_e^2.$

$\Rightarrow \rho_t(p) = \frac{(t-p+1)\sigma_e^2}{\sqrt{(t+1)\sigma_e^2}\sqrt{(t-p+1)\sigma_e^2}} = \frac{\sqrt{(t-p+1)}}{\sqrt{(t+1)}}.$

$\Rightarrow lim_{p\rightarrow 0} \rho_t(p) = 1 \hspace{.2cm} \wedge \Rightarrow lim_{p\rightarrow t} \rho_t(p) = \frac{1}{\sqrt{(t+1)}}$.

$\Rightarrow$ for fixed $p$ $lim_{t\rightarrow \infty} \rho_t(p) = lim_{t\rightarrow \infty} \frac{\sqrt{(t-p+1)}}{\sqrt{(t+1)}}= 1$

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  • $\begingroup$ Thank you, what is the intuition behind this result? $\endgroup$ – Bob Charles Feb 20 at 22:24
  • $\begingroup$ @BobCharles $y_{t-p}$ is entirely contained in $y_{t}$. The larger $t$ is the smaller is the relative contribution of the p values that are in $y_{t}$ but not in $y_{t-p}$. $\endgroup$ – Grada Gukovic Feb 20 at 22:40
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Here's a good discussion on random walks: https://machinelearningmastery.com/gentle-introduction-random-walk-times-series-forecasting-python/

Hopefully after reading it, you'll be able to see how your question is probably not well-posed, and correct it so that we can help you. For example, the autocorrelation function is not run over $t$, it's run over $k$ where $k$ indexes the lag order - in that case, as @1muflon1 notes, running $k \rightarrow \infty$ should see the correlation fall to zero.

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  • $\begingroup$ But, given the random walk process $y_{t}=y_{t-1}+e_{t}$, the auto correlation function is given by $corr(y_{t}, y_{t-h})=(\frac{t-h}{t})^{1/2}=(1-\frac{h}{t})^{1/2}$, which tends to 0 as t tends to infinity. What is the intuition behind this result? $\endgroup$ – Bob Charles Feb 20 at 22:26
  • $\begingroup$ @BobCharles You are gravely mistaken. The expression tends to 1 as $t \to \infty$, as shown in your other post: economics.stackexchange.com/questions/34102/… $\endgroup$ – Art Mar 4 at 3:25

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