# Autocorrelation function of a random walk process

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.

• Tends to 1? Can you provide source for that statement? It should be 0 for the process you describe
– 1muflon1
Feb 20, 2020 at 19:04
• 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.
– heh
Feb 20, 2020 at 20:05
• @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? Feb 20, 2020 at 22:24
• @heh You are gravely mistaken, and I refer you to my proof to this effect commented above. Feb 20, 2020 at 22:25
• @BobCharles So in the question you say $t \to \infty$ but in the comment you say it's $t \to 0$?
– Art
Feb 21, 2020 at 6:29

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

• Thank you, what is the intuition behind this result? Feb 20, 2020 at 22:24
• @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 $y_{t-p}$. Feb 20, 2020 at 22:40

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.

• 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? Feb 20, 2020 at 22:26
• @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/…
– Art
Mar 4, 2020 at 3:25