<p>I think the interpretation here is one of correlation if we assume second-order stationarity. That is, the coefficient in your example is simply the correlation between a contemporaneous value of your dependent variable and its one-period lag. </p>

CC BY-SA 3.0
2017-06-30T20:14:13.477Z
Isnt correlation represented by$\rho=\frac{cov(x_t,x_{t-1})}{\sigma_{x_t} \sigma_{x_{t-1}}}$? coefficients in this context would be represented by $\beta_1=\frac{cov(x_t,x_{t-1})}{\sigma_{x_t}}$.
EconJohn
CC BY-SA 3.0
2017-07-01T11:33:42.523Z
in your example, $\beta_1 = COV(x_t,x_{t-1})/VAR(x_{t-1})$. Since $x$ is assumed to be stationary, $VAR(x_{t-1}) = SD(x_{t-1})SD(x_{t})$
Tobias
CC BY-SA 3.0
2017-07-02T02:43:03.173Z
@Tobias I didn't know that, can you link me a tutorial/pdf that discusses this?
EconJohn
CC BY-SA 3.0
2017-07-02T16:32:24.263Z
@EconJohn This is just the definition of stationarity, see the answer above.
Tobias
CC BY-SA 3.0
2017-06-30T19:34:16.483Z
2017-06-30T19:34:16.483Z
123
<p>You can think of the coefficient in the AR(1) model as telling you something about the dynamics of the process. For example, if the model is of wage growth, then a coefficient >0 suggests that higher wages yesterday are associated with higher wages today. If the coefficient is &lt;1 then there is not a complete pass-through from yesterdays wage growth (i.e. a stationary process) whereas if it were >1 then you have a non-stationary process where wage increases are accelerating. In the wage growth or inflation case this could be suggestive of hyperinflation.</p>

CC BY-SA 3.0
2017-07-01T22:56:47.433Z
2017-07-01T22:56:47.433Z
Andrew M
<p><strong>For a second-order stationary series it is the correlation coefficient between the dependent value and its lag.</strong>
Specify
$$y_{t+1} = a+ \beta y_t + u_{t+1}\qquad u_{t+1}= \text{white noise}$$</p>

<p>The correlation coefficient between $y_{t+1}$ and $y_{t}$ is defined as usual</p>

<p>$$\rho_{(1)} = \frac{\text{Cov}(y_{t+1},y_{t})}{\sigma(y_{t+1})\sigma(y_t)}$$</p>

<p>$$\text{Cov}(y_{t+1},y_{t}) = E(y_{t+1}y_{t}) - E(y_{t+1})E(y_{t})$$</p>

<p>$$ = E\Big((a+\beta y_t+u_{t+1})y_{t}\Big) - E(y_{t+1})E(y_{t}) = aE(y_t)+\beta E\Big(y_t^2+u_{t+1}y_{t}\Big) - E(y_{t+1})E(y_{t})$$</p>

<p>We have $E(u_{t+1}y_{t}) =0$. Also, under first-order stationarity we have $E(y_t)=E(y_{t+1}) = \frac{a}{1-\beta}$</p>

<p>Using these we get</p>

<p>$$\text{Cov}(y_{t+1},y_{t}) = \frac{a^2}{1-\beta}+\beta E(y_t^2) - \frac{a^2}{(1-\beta)^2}$$</p>

<p>By definition the variance is </p>

<p>$$\text{Var}(y_t) = E(y_t^2) - [E(y_t)]^2 = E(y_t^2) -\frac{a^2}{(1-\beta)^2}$$ </p>

<p>$$\implies E(y_t^2)  = \text{Var}(y_t) + \frac{a^2}{(1-\beta)^2}$$</p>

<p>Substituting,</p>

<p>$$\text{Cov}(y_{t+1},y_{t}) = \frac{a^2}{1-\beta}+\beta \text{Var}(y_t) + \beta \frac{a^2}{(1-\beta)^2}   - \frac{a^2}{(1-\beta)^2}$$</p>

<p>Things cancel out and we are left with</p>

<p>$$\text{Cov}(y_{t+1},y_{t}) = \beta\text{Var}(y_t) $$</p>

<p>Under the assumption of  2nd-order stationarity, $\text{Var}(y_t) = \text{Var}(y_{t+1}) = \text{Var}(y)$</p>

<p>Inserting all this back to the correlation coefficient</p>

<p>$$\rho_{(1)} = \frac{\beta\text{Var}(y)}{\sigma(y)\sigma(y)} = \frac{\beta\text{Var}(y)}{\text{Var}(y)} = \beta. $$</p>

<p>Note that the presence of the constant $a$ does not affect the correlation -it would be the same if $a=0$. This is because location parameters do not affect second-order statistics like the covariance and the variance.</p>

CC BY-SA 3.0
2017-07-04T21:18:42.000Z
So. This is exactly what I said without the math. Glad to know I was on the correct path. As always...good answer.
123
CC BY-SA 3.0
2017-07-04T21:47:11.317Z
@123 Thank you. Indeed it is the math behind the words.
Alecos Papadopoulos
CC BY-SA 3.0
2017-07-01T14:24:43.320Z
2017-07-01T14:57:12.450Z
2017-07-01T14:57:12.450Z
Alecos Papadopoulos
<p>An AR(1) process is given as:</p>

<p>$$x_t=\rho_0+\rho_{t-1}x_{t-1}+\epsilon_t$$</p>

<p>This regression tells us that $x_{t}$ is a function of its value at time $t-1$. </p>

<p>My question is, how do you interpret its coefficient $\rho_{t-1}$? by comparison in a labor economics example (where only cross sectional data is used) for a case where you regress education on wage.
$$y_{wage}=\beta_0+\beta_{1} x_{educ}+u$$
If I were to take partials of this regression with respect to $x_{educ}$ I can interpret $\beta_1$ as the marginal returns to wage from education.</p>

<p>In the AR(1) process, following the same steps as before, I am not sure how to interpret the coefficient $\rho_{t-1}$. </p>

<p>What is its meaning?</p>

<econometrics><time-series>
Interpetation of coefficent in AR(1) model
EconJohn
# Interpetation of coefficent in AR(1) model

(Asked by: username_0 on 2017-06-30)

An AR(1) process is given as:

$$x\_t=\rho\_0+\rho\_{t-1}x\_{t-1}+\epsilon\_t$$

This regression tells us that $x\_{t}$ is a function of its value at time $t-1$.

My question is, how do you interpret its coefficient $\rho\_{t-1}$? by comparison in a labor economics example (where only cross sectional data is used) for a case where you regress education on wage.
$$y\_{wage}=\beta\_0+\beta\_{1} x\_{educ}+u$$
If I were to take partials of this regression with respect to $x\_{educ}$ I can interpret $\beta\_1$ as the marginal returns to wage from education.

In the AR(1) process, following the same steps as before, I am not sure how to interpret the coefficient $\rho\_{t-1}$.

What is its meaning?

**Tags:** econometrics, time-series


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## 3 answers

### Answer by username_4 on 2017-07-01. Score: 3 | Accepted answer

**For a second-order stationary series it is the correlation coefficient between the dependent value and its lag.**
Specify
$$y\_{t+1} = a+ \beta y\_t + u\_{t+1}\qquad u\_{t+1}= \text{white noise}$$

The correlation coefficient between $y\_{t+1}$ and $y\_{t}$ is defined as usual

$$\rho\_{(1)} = \frac{\text{Cov}(y\_{t+1},y\_{t})}{\sigma(y\_{t+1})\sigma(y\_t)}$$

$$\text{Cov}(y\_{t+1},y\_{t}) = E(y\_{t+1}y\_{t}) - E(y\_{t+1})E(y\_{t})$$

$$ = E\Big((a+\beta y\_t+u\_{t+1})y\_{t}\Big) - E(y\_{t+1})E(y\_{t}) = aE(y\_t)+\beta E\Big(y\_t^2+u\_{t+1}y\_{t}\Big) - E(y\_{t+1})E(y\_{t})$$

We have $E(u\_{t+1}y\_{t}) =0$. Also, under first-order stationarity we have $E(y\_t)=E(y\_{t+1}) = \frac{a}{1-\beta}$

Using these we get

$$\text{Cov}(y\_{t+1},y\_{t}) = \frac{a^2}{1-\beta}+\beta E(y\_t^2) - \frac{a^2}{(1-\beta)^2}$$

By definition the variance is

$$\text{Var}(y\_t) = E(y\_t^2) - [E(y\_t)]^2 = E(y\_t^2) -\frac{a^2}{(1-\beta)^2}$$

$$\implies E(y\_t^2) = \text{Var}(y\_t) + \frac{a^2}{(1-\beta)^2}$$

Substituting,

$$\text{Cov}(y\_{t+1},y\_{t}) = \frac{a^2}{1-\beta}+\beta \text{Var}(y\_t) + \beta \frac{a^2}{(1-\beta)^2} - \frac{a^2}{(1-\beta)^2}$$

Things cancel out and we are left with

$$\text{Cov}(y\_{t+1},y\_{t}) = \beta\text{Var}(y\_t) $$

Under the assumption of 2nd-order stationarity, $\text{Var}(y\_t) = \text{Var}(y\_{t+1}) = \text{Var}(y)$

Inserting all this back to the correlation coefficient

$$\rho\_{(1)} = \frac{\beta\text{Var}(y)}{\sigma(y)\sigma(y)} = \frac{\beta\text{Var}(y)}{\text{Var}(y)} = \beta. $$

Note that the presence of the constant $a$ does not affect the correlation -it would be the same if $a=0$. This is because location parameters do not affect second-order statistics like the covariance and the variance.

> **Comments:**
> **username_1**: So. This is exactly what I said without the math. Glad to know I was on the correct path. As always...good answer.  
> **username_4**: @username_1 Thank you. Indeed it is the math behind the words.  

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### Answer by username_1 on 2017-06-30. Score: 2

I think the interpretation here is one of correlation if we assume second-order stationarity. That is, the coefficient in your example is simply the correlation between a contemporaneous value of your dependent variable and its one-period lag.

> **Comments:**
> **username_0**: Isnt correlation represented by$\rho=\frac{cov(x_t,x_{t-1})}{\sigma_{x_t} \sigma_{x_{t-1}}}$? coefficients in this context would be represented by $\beta_1=\frac{cov(x_t,x_{t-1})}{\sigma_{x_t}}$.  
> **username_2**: in your example, $\beta_1 = COV(x_t,x_{t-1})/VAR(x_{t-1})$. Since $x$ is assumed to be stationary, $VAR(x_{t-1}) = SD(x_{t-1})SD(x_{t})$  
> **username_0**: @username_2 I didn't know that, can you link me a tutorial/pdf that discusses this?  
> **username_2**: @username_0 This is just the definition of stationarity, see the answer above.  

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### Answer by username_3 on 2017-07-01. Score: 1

You can think of the coefficient in the AR(1) model as telling you something about the dynamics of the process. For example, if the model is of wage growth, then a coefficient >0 suggests that higher wages yesterday are associated with higher wages today. If the coefficient is <1 then there is not a complete pass-through from yesterdays wage growth (i.e. a stationary process) whereas if it were >1 then you have a non-stationary process where wage increases are accelerating. In the wage growth or inflation case this could be suggestive of hyperinflation.


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