I have obtained different values of intercept from autoregressive AR(1) model calculated with lm and arima, but coefficients before lagged variable are almost exactly the same, intercept for arima is 0.1153 and 0.03014 for lm why does this difference occurs ?
> set.seed(123) > ar1 <- arima.sim(n = 1000, list(ar = 0.8, sd = 1)) #simulated dataset > arima(ar1, order=c(1,0,0)) #AR(1) model Call: arima(x = ar1, order = c(1, 0, 0)) Coefficients: ar1 intercept 0.7772 0.1153 s.e. 0.0200 0.1418 sigma^2 estimated as 1.006: log likelihood = -1422.16, aic = 2850.33 > summary(lm(ar1[-1]~ar1[-1000])) Call: lm(formula = ar1[-1] ~ ar1[-1000]) Residuals: Min 1Q Median 3Q Max -2.8835 -0.6563 -0.0029 0.6587 3.1793 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 0.03014 0.03181 0.947 0.344 ar1[-1000] 0.77612 0.02000 38.802 <2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 1.003 on 997 degrees of freedom Multiple R-squared: 0.6016, Adjusted R-squared: 0.6012 F-statistic: 1506 on 1 and 997 DF, p-value: < 2.2e-16 Best Answer
Stationary time series estimation first starts by centering. So the intercept estimation in arima is just mean(ar1).
Turning to lm. Did you not see that the intercept is not statistically different from 0, as ar1[-1] and ar1[-1000] have the same mean? That's what stationarity implies.
In summary, arima here is doing real estimation for ar1, while lm is only estimating its auto-correlation. You are doing different things!
If this is still not that clear for you, let's just focus on the lm usage in general.
# true model: y = 0.7 x + 0.5 x <- runif(1000) y <- 0.5 + 0.7 * x + rnorm(1000, 0, 0.05) # a proper estimation lm(y ~ x) # now center x and y so they have the same mean: 0 # this only estimates slope x1 <- x - mean(x) y1 <- y - mean(y) lm(y1 ~ x1) Similar Posts:
- Solved – Auto.Arima using xreg giving xreg as insignificant
- Solved – How to calculate the R-squared of a regression with arima errors using R
- Solved – How to avoid log(0) term in regression
- Solved – For lm() coefficient in R, why not give slope directly? Focus “slope”,not a parameter estimation method
- Solved – Estimation of AR($p$) model by `lm` versus `arima` in R: different results