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In their book "Robust Statistics", Maronna et al. consider the following model for robust regression: $y_i = \beta x_i + u_i$, where $u_i$ are independent of the $x_i$, and are i.i.d, with finite variance. They go on to provide a robust estimate $\hat{\beta}$ of $\beta$ which is asymptotically normal and give the covariance matrix for $\hat{\beta}$. My question is that without any knowledge about the distribution of $u_i$, is it possible to provide a prediction interval for $y$ (without using bootstrap)? I'm asking this because

library(MASS)
robustModel = rlm(formula = myFormula, data = myData, method = "MM")
predict.rlm(robustModel, newdata = myNewData, interval = "prediction") 

in R generates a prediction interval. For reference, this is the code for predict.rlm:

predict.rlm <- function (object, newdata = NULL, scale = NULL, ...)
{
## problems with using predict.lm are the scale and
## the QR decomp which has been done on down-weighted values.
object$qr <- qr(sqrt(object$weights) * object$x)
    predict.lm(object, newdata = newdata, scale = object$s, ...)
}

It seems to me that the prediction interval that is obtained this way is for normally distributed $u_i$. Is that correct? What am I missing here?

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It would be easier to answer the question if we had the actual formula for the estimator. But generally speaking the exact distribution of the estimator should depend on the error distribution. However the covariance matrix can be estimated from the data. An exact prediction interval would seem to depend on the distribution of the error term and hence cannot be determined without specifying the error distribution. But that doesn't mean that you could not get an approximate prediction interval (note that the problem is the same for confidence intervals). For the example where beta is one-dimensional the covariance matrix is just a single variance. In the case where beta is multidimensional then there would be covariance terms. I think the result in the book deals with a robust estimator in the more general multidimensional case.

Going back to the one-dimensional case the estimates standard deviation could be estimated from the data and the asymptotic normal distribution could be used to get approximate confidence and prediction intervals which the theory says would have the approximate confidence level for large n.

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  • $\begingroup$ thank you for response. The estimator is obtained using an iterative procedure, so there's really no closed form for it. I can understand your comment about the confidence interval: we're looking for a confidence interval for $\hat{\beta} x$ (assume 1-d) and $\hat{\beta}$ is asymptotically normal, and $\Var(\hat{\beta}x) = $x^2Var(\hat{\beta})$ and from here we can get the confidence interval for $\hat{\beta}x$. But without any idea about the distribution of $u$, I can't see how to get the a (approx) confidence interval for $\hat{\beta} x + u$ (i.e. a prediction interval). $\endgroup$ – user765195 Oct 2 '12 at 1:57
  • $\begingroup$ Thank you for response @michaelchernick. The estimator is obtained using an iterative procedure, so there's really no closed form for it. I can understand your comment about the confidence interval: we're looking for a confidence interval for $\hat{\beta} x$ (assume 1-d) and $\hat{\beta}$ is asymptotically normal, and $Var(\hat{\beta}x) = x^2Var(\hat{\beta})$ and from here we can get the confidence interval for $\hat{\beta}x$. But without any idea about the distribution of $u$, I can't see how to get the a (approx) confidence interval for $\hat{\beta} x + u$ (i.e. a prediction interval). $\endgroup$ – user765195 Oct 2 '12 at 2:03
  • $\begingroup$ my apologies. I've tried editing my comment to make it more readable, but it just doesn't seem to work. Thank you again for your answer. $\endgroup$ – user765195 Oct 2 '12 at 2:33

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