Is the ar1 coefficient from an AR(1) model the "first order correlation of the noise" of a time series?

I'm using R's aws package and one of the arguments of the aws function allows you to "specify a first order correlation of the noise". Instinctively, I did

arima (foo, c(1, 0, 0))

and used the ar1 coefficient as this correlation value. Plugging it into aws makes the results look more like what I'd expect, so I was quite happy…

Until it struck me that AR coefficients can fall outside of the range [-1, 1], while the correlation cannot. So maybe I just got lucky. Then it struck me that if the ar1 coefficient of an AR(1) model falls outside of [-1, 1], the model's not stationary, so perhaps a well-fitting AR(1) model's ar1 coefficient is essentially restricted to [-1, 1].

But then I thought: just because they share the same range doesn't mean they are the same. I did some searching and didn't find anything saying this is a correlation. I did some simulations and it appears that it is close to a correlation.

Anyone have an authoritative statement on the matter?

I've also looked at arima (foo, c(1, 0, 0), xreg=theDate) to account for a trend. I've thought that if the AR coefficient is a correlation: what if an AR(2) model is a better fit than an AR(1) model -- is the ar1 coefficient still what I'm looking for or do I have to combine it with the ar2 coefficient, or do I still go with the AR(1) model?

  • 2
    $\begingroup$ If you check here under "Examples for some Low-order AR(p) processes", then for an AR(1) it's the case that $\rho_1 = \phi_1$. This is when the series given at that first place converges, which is when $|\phi_1|<1$ (the part about stationarity under 'Definition' at the top is needed; when that's not the case, it's really an ARI model) $\endgroup$
    – Glen_b
    Aug 24, 2013 at 16:01

1 Answer 1


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

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

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

Now $$\text{Cov}(y_{t+1},y_{t}) = E(y_{t+1}y_{t}) - E(y_{t+1})E(y_{t})$$ $$ = E\Big((\beta y_t+u_{t+1})y_{t}\Big) - E(y_{t+1})E(y_{t}) = E\Big(\beta y_t^2+u_{t+1}y_{t}\Big) - E(y_{t+1})E(y_{t})$$

Now $u_{t+1}, y_{t}$ are independent. Also, the expected value of the $y$-series is zero, given the specification. Using these facts we end up with $$\text{Cov}(y_{t+1},y_{t}) =\beta E(y_t^2) = \beta\text{Var}(y_t) $$

Since we assume 2nd-order stationarity, $\text{Var}(y_t) = \text{Var}(y_{t+1}) = \text{Var}(y)$

Inserting all this back to the correlation coefficient

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


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