<p>By recursive substitution you obtain <span class="math-container">$y_t = \sum_{j=0}^t e_{t-j}.$</span> Thus <span class="math-container">$\forall p \in \mathbb{N} \hspace{.2cm} y_{t-p} = \sum_{j=0}^{t-p} e_{t-p-j}.$</span> </p>

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

<p><span class="math-container">$Var(y_t) = E(\sum_{j=0}^{t} e_{t}^2) = (t+1)\sigma_e^2 \hspace{.2cm} \wedge Var(y_{t-p}) = E(\sum_{j=0}^{t-p} e_{t-p-j}^2)(t-p+1)\sigma_e^2.$</span></p>

<p><span class="math-container">$\Rightarrow \rho_t(p) = \frac{(t-p+1)\sigma_e^2}{\sqrt{(t+1)\sigma_e^2}\sqrt{(t-p+1)\sigma_e^2}} = \frac{\sqrt{(t-p+1)}}{\sqrt{(t+1)}}.$</span></p>

<p><span class="math-container">$\Rightarrow lim_{p\rightarrow 0} \rho_t(p) = 1 \hspace{.2cm} \wedge \Rightarrow lim_{p\rightarrow t} \rho_t(p) =  \frac{1}{\sqrt{(t+1)}}$</span>.</p>

<p><span class="math-container">$\Rightarrow$</span> for fixed <span class="math-container">$p$</span> <span class="math-container">$lim_{t\rightarrow \infty} \rho_t(p) = lim_{t\rightarrow \infty} \frac{\sqrt{(t-p+1)}}{\sqrt{(t+1)}}= 1$</span></p>

CC BY-SA 4.0
2020-02-20T22:24:05.827Z
Thank you, what is the intuition behind this result?
Bob Charles
CC BY-SA 4.0
2020-02-20T22:40:04.837Z
@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 $y_{t-p}$.
Grada Gukovic
CC BY-SA 4.0
2020-02-20T20:08:07.970Z
2020-02-20T20:56:20.270Z
2020-02-20T20:56:20.270Z
Grada Gukovic
<p>Here's a good discussion on random walks: <a href="https://machinelearningmastery.com/gentle-introduction-random-walk-times-series-forecasting-python/" rel="nofollow noreferrer">https://machinelearningmastery.com/gentle-introduction-random-walk-times-series-forecasting-python/</a></p>

<p>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 <em>not</em> run over <span class="math-container">$t$</span>, it's run over <span class="math-container">$k$</span> where <span class="math-container">$k$</span> indexes the lag order - in that case, as @1muflon1 notes, running <span class="math-container">$k \rightarrow \infty$</span> should see the correlation fall to zero.</p>

CC BY-SA 4.0
2020-02-20T22:26:28.090Z
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?
Bob Charles
CC BY-SA 4.0
2020-03-04T03:25:11.153Z
@BobCharles You are gravely mistaken. The expression tends to 1 as $t \to \infty$, as shown in your other post: https://economics.stackexchange.com/questions/34102/limit-of-random-walk-auto-correlation-function
Art
CC BY-SA 4.0
2020-02-20T20:11:15.183Z
2020-02-20T20:11:15.183Z
heh
<p>What is the intuition behind the result that the autocorrelation function of a random walk process <span class="math-container">$y_{t}=y_{t-1}+e_{t}$</span> tends to 1 as <span class="math-container">$t\rightarrow 0$</span>? Thank you. </p>

CC BY-SA 4.0
2020-02-20T19:04:23.453Z
Tends to 1? Can you provide source for that statement? It should be 0 for the process you describe
1muflon1
CC BY-SA 4.0
2020-02-20T20:05:38.680Z
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
CC BY-SA 4.0
2020-02-20T22:24:35.397Z
@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?
Bob Charles
CC BY-SA 4.0
2020-02-20T22:25:53.973Z
@heh You are gravely mistaken, and I refer you to my proof to this effect commented above.
Bob Charles
CC BY-SA 4.0
2020-02-21T06:29:50.380Z
@BobCharles So in the question you say $t \to \infty$ but in the comment you say it's $t \to 0$?
Art
CC BY-SA 4.0
2020-02-21T14:56:32.710Z
@Art no, you are mistaken. As a general rule, please read carefully the original post before making remarks of this nature. (this ought to be obvious).
Bob Charles
CC BY-SA 4.0
2020-03-03T16:01:34.147Z
@Art Given you have yet to express but an iota of penitence regarding the mistake you have made—let alone demonstrated a will to, in the future, conduct yourself otherwise—I have no option but to report you.
Bob Charles
CC BY-SA 4.0
2020-03-03T17:03:12.783Z
@BobCharles If you look at the "edit" history you could see very clearly that you edited from $t \to \infty$ to $t \to 0$ at 14:55. And you said I'm "mistaken" at 14:56. Not very nice.
Art
<econometrics><statistics><stochastic-processes>
Autocorrelation function of a random walk process
Bob Charles
# Autocorrelation function of a random walk process

(Asked by: username_0 on 2020-02-20)

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.

**Tags:** econometrics, statistics, stochastic-processes

> **Comments:**
> **username_1**: Tends to 1? Can you provide source for that statement? It should be 0 for the process you describe  
> **username_2**: 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.  
> **username_0**: @username_1 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?  
> **username_0**: @username_2 You are gravely mistaken, and I refer you to my proof to this effect commented above.  
> **username_3**: @BobCharles So in the question you say $t \to \infty$ but in the comment you say it's $t \to 0$?  
> **username_0**: @username_3 no, you are mistaken. As a general rule, please read carefully the original post before making remarks of this nature. (this ought to be obvious).  
> **username_0**: @username_3 Given you have yet to express but an iota of penitence regarding the mistake you have made—let alone demonstrated a will to, in the future, conduct yourself otherwise—I have no option but to report you.  
> **username_3**: @BobCharles If you look at the "edit" history you could see very clearly that you edited from $t \to \infty$ to $t \to 0$ at 14:55. And you said I'm "mistaken" at 14:56. Not very nice.  

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## 2 answers

### Answer by username_4 on 2020-02-20. Score: 1

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 $p$ $lim\_{t\rightarrow \infty} \rho\_t(p) = lim\_{t\rightarrow \infty} \frac{\sqrt{(t-p+1)}}{\sqrt{(t+1)}}= 1$

> **Comments:**
> **username_0**: Thank you, what is the intuition behind this result?  
> **username_4**: @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 $y_{t-p}$.  

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### Answer by username_2 on 2020-02-20. Score: 0

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 @username_1 notes, running $k \rightarrow \infty$ should see the correlation fall to zero.

> **Comments:**
> **username_0**: 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?  
> **username_3**: @BobCharles You are gravely mistaken. The expression tends to 1 as $t \to \infty$, as shown in your other post: https://economics.stackexchange.com/questions/34102/limit-of-random-walk-auto-correlation-function  

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