Interpetation of coefficent in AR(1) model

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?

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.

• So. This is exactly what I said without the math. Glad to know I was on the correct path. As always...good answer. – 123 Jul 4 '17 at 21:18
• @123 Thank you. Indeed it is the math behind the words. – Alecos Papadopoulos Jul 4 '17 at 21:47

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.

• 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 Jun 30 '17 at 20:14
• 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 Jul 1 '17 at 11:33
• @Tobias I didn't know that, can you link me a tutorial/pdf that discusses this? – EconJohn Jul 2 '17 at 2:43
• @EconJohn This is just the definition of stationarity, see the answer above. – Tobias Jul 2 '17 at 16:32

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.