I'm using robust linear regression in R (the rlm function from MASS package that uses the Huber M estimator by default). I'm wondering if I can conduct a likelihood-ratio test after running the nested and full model. I wanted to test if c1+c2+c3 combined improves the fit of the model. Can anyone answer from a statistical perspective? Here are the example code:

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    $\begingroup$ I am not sure what form of robust regression the rlm function uses. The likelihood ratio test assumes a form of a model which is a function of certain parameters. Robust regression does not make such assumptions. It attempts to fit a model that avoids the normality assumption. $\endgroup$ Apr 26, 2017 at 20:46
  • $\begingroup$ rlm uses the M-estimation by default, does this violate the assumption of the likelihood ratio test? Thank you~ $\endgroup$
    – gaol
    Apr 26, 2017 at 22:47
  • $\begingroup$ Given that some function names occur in more than one package, can you explicitly say which package the function rlm you're using is in? Do you mean the one in MASS? Further, your code isn't reproducible (since we don't have your x and y) -- a small reproducible example might be useful for potential answerers to illustrate their answers. $\endgroup$
    – Glen_b
    Apr 27, 2017 at 0:16
  • $\begingroup$ Yes! It's the one in MASS. $\endgroup$
    – gaol
    Apr 27, 2017 at 1:11
  • $\begingroup$ That information would be a good thing to edit into your question. If you're stuck for a small reproducible example you might be able to organize one from one of the built in data sets in R or in MASS $\endgroup$
    – Glen_b
    Apr 27, 2017 at 1:14

1 Answer 1


You can use a likelihood ratio test when the model is fitted by maximum likelihood (indeed, you need the likelihood under the null and alternative).

In the cases where the estimator you use does correspond to ML for some distributional assumption, under that same assumption you could generally get a likelihood ratio test.

In the case of M-estimators, it depends on the M-estimator. For example, a Huber M-estimator corresponds to ML for a particular distribution (one with a normal center and exponential tails) but the Hampel 3-part redescending M-estimator doesn't.

Note that this would assume either a known scale or a consistent estimator of it (which I think by applying Slutsky's theorem would then leave us still with the usual asymptotics for the LR test).

Even in cases where you use an estimator which wouldn't correspond to an ML estimator there are often tests that should have fairly good properties. In many cases, one could choose a permutation test based on some robust estimator, for example. In cases where you're testing all the regression coefficients at once, that should yield a distribution-free test. In the cases where you're only testing some of them, I think you would have an approximate test (you don't get exchangeability of residuals under the null -- except perhaps asymptotically, if your estimator is consistent). Alternatively, in many cases a parameter estimate might be shown to be asymptotically normal and so give a large sample test.

There's also the possibility that you might use a bootstrap test in large samples; I think that should work pretty well with the Huber psi function.

Note that even when a likelihood ratio test is possible, I'm not at all sure that rlm will return a (log-)likelihood (or something from which you can easily compute it) -- you might have to evaluate the (log-)likelihood yourself. I haven't checked that it doesn't though.

  • $\begingroup$ +1 for a good answer. It seems the OP is unaware of which psi function he is using. Your other points are well taken but doesn't fit in with the OPs constraints. Still it is worth including. $\endgroup$ Apr 27, 2017 at 0:16

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