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If i understand correctly, the ARIMA function produces an estimate for the mean of the process instead of the intercept. It is possible to transform the mean into the intercept: mean= 1-Sum(AR-Coefficients). Is it also possible to transform the standard error of the mean into standard error of the intercept?

As an example:

arima(x = example, order = c(2, 0, 2), method = "ML")

Coefficients:
          ar1      ar2     ma1     ma2  intercept
      -0.7508  -0.0367  1.3156  0.5433    -0.0183
s.e.   0.2591   0.1772  0.2424  0.1958     0.0244

sigma^2 estimated as 0.03748:  log likelihood = 35.67,  aic = -59.33

# mean:
> 1+0.7508+0.0367 
[1] 1.7875

Thanks a lot

laterstat

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The mapping between the mean (denoted $\mu$) and the intercept (denoted $\alpha$) in an ARMA(1,q) process is given by:

$$ \mu = \frac{\alpha}{1 - \phi} \,. $$

It is easy to check for an AR(1) process:

\begin{eqnarray} \begin{array}{lcl} y_t &=& \alpha + \phi y_{t-1} + \epsilon_t \,, \quad \epsilon_t \sim NID(0, \sigma^2) \\ E(y_t) \equiv \mu &=& \alpha + \phi E(y_{t-1}) + E(\epsilon_{t-1}) \\ &=& \alpha + \phi E(\alpha + \phi y_{t-2} + \epsilon_{t-2}) + 0 \\ &=& \alpha + \phi \alpha + \phi^2 E(y_{t-2}) + 0 + 0 \\ &=& \alpha + \phi \alpha + \phi^2 \alpha + \phi^3 E(y_{t-3}) + 0 \\ &=& \alpha + \phi \alpha + \phi^2 \alpha + \phi^3 \alpha + \phi^4 E(y_{t-4}) + 0 \\ &=& \dots \\ E(y_t) \equiv \mu &=& \alpha \left(1 + \sum_{t=1}^n\phi^t \right) + y_0 \,. \end{array} \end{eqnarray}

For a stationary AR process, where $-1 < \phi < 1$, the term in the big parentheses converges to $1/(1 - \phi)$.

For a general AR(p) process the derivation following the above strategy is more tedious, but I think that the mapping can be generalized as $\mu = \alpha/(1 - \sum_{i=1}^p \phi_i)$.

Now, the standard error of the intercept -given the output from stats::arima where we get an estimate of the mean and its standard error- can be obtained following the result known as the Delta method. Basically, it says that given a function of the parameter estimate, $g(\hat{\mu})$, and the standard error of the estimate $\hat{\mu}$ (denoted $s_{\hat{\mu}}$), then the standard error of $\hat{\alpha}$ is given by

$$ s_{\hat{\alpha}} = g'(\hat{\mu}) s_{\hat{\mu}} \,, $$

where $g'(\hat{\mu})$ is the derivative of $g(\hat{\mu})$. In our case $g(\hat{\mu}) = \hat{\mu} (1 - \hat{\phi})$, which yields $\hat{\alpha}$, and $g'(\hat{\mu}) = (1 - \hat{\phi})$.

Let's see how it works in practice. Below I generate 5000 series from an AR(1) model and fit the same model. The estimate of the intercept is obtained as shown above. In order to use the standard error derived before, the $t$-statistics for the mean and the intercept are obtained as well. Rejections at the 5% significance level are counted.

set.seed(123)
niter <- 5000
sum.alpha <- sum.mu <- 0
rejections.alpha <- rejections.mu <- rep(NA, niter)
for(i in seq(niter))
{
  x <- arima.sim(200, model = list(ar = 0.6))
  fit <- arima(x, order = c(1,0,0), include.mean = TRUE)
  b <- coef(fit)
  gprime <- 1 - b["ar1"]
  alpha <- as.vector(b["intercept"] * gprime)
  sum.alpha <- sum.alpha + alpha
  sum.mu <- sum.mu + b["intercept"]
  alpha.se <- sqrt(gprime^2 * fit$var.coef[2,2])
  tstat.alpha <- alpha / alpha.se
  tstat.mu <- b[2] / sqrt(fit$var.coef[2,2])
  rejections.alpha[i] <- abs(tstat.alpha) > 1.96
  rejections.mu[i] <- abs(tstat.mu) > 1.96
}
# average estimate of "mu"
sum.mu / niter
# -0.002008448 
# average estimate of "alpha"
sum.alpha / niter
# [1] -0.0009473526
# number of rejections alpha=0 over number of series
sum(rejections.alpha) / niter
# [1] 0.0634
# number of rejections mu=0 over number of series
sum(rejections.mu) / niter
# [1] 0.0634
identical(rejections.alpha, rejections.mu)
# [1] TRUE

The same empirical levels of the tests are obtained for the intercept and the mean, $0.0634$ and are close to the nominal level $0.05$. In fact, the null that these parameters are zero is rejected for the same simulated series, suggesting that the calculations for the standard error of the intercept are valid.

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