5
$\begingroup$

I'm new to R, so please be gentle.

I was under the impression that rstandard(model) returns the z-scores of the residuals in model. However, when I standardized the residuals myself in z, the result was different. In fact, rstandard(model) had a mean different from 0 and had a standard deviation different from 1. The differences seem to be nonnegligible. What exactly does rstandard(model) do? Or am I doing something wrong here?

> x = rnorm(30, mean = 100, sd = 15)
> y = rnorm(30, mean = 100, sd = 15)
> model = lm(y ~ x)
> z.res = residuals(model)/sd(residuals(model)) #standardizing it myself
> rstandard(model) - z.res #difference between rstandard and what i did
            1             2             3             4             5 
-4.422354e-04  1.556269e-04 -4.576832e-03 -1.274350e-03  1.048068e-01 
            6             7             8             9            10 
-2.333922e-02  1.820134e-02 -3.307542e-03  3.368978e-02 -1.804108e-04 
           11            12            13            14            15 
-1.100621e-01 -1.343715e-03 -1.300427e-03  1.509862e-03  3.246602e-03 
           16            17            18            19            20 
 3.734255e-03 -1.821539e-06 -1.153190e-02 -1.713254e-06 -2.185101e-02 
           21            22            23            24            25 
-2.681935e-02  2.562472e-03 -4.721114e-02 -1.084481e-04 -3.430827e-03 
           26            27            28            29            30 
 4.149684e-04  7.705807e-04  2.166815e-03  2.537837e-02  4.182761e-04 
> mean(z.res) 
[1] -9.428041e-18 
#as expected of z-scores, mean is 0
> sd(z.res)
[1] 1 
#as expected of z-scores standard deviation is 1
> mean(rstandard(model)) 
[1] -0.001990908
#not really 0
> sd(rstandard(model)) 
[1] 1.019699
#not really 1

Also, the way I understood Standardized residuals in R's lm output, rstandard is actually studentized residuals. But isn't there already rstudent?

I'm using R version 2.14.1 in Xubuntu 12.04.

Thank you.

$\endgroup$
  • $\begingroup$ The residuals don't have constant variance, even though the error term does. Points with more extreme 'x'-values have smaller variance. If you want to make them comparable, you don't divide by their average standard deviation, but by their individual standard deviation. This is what is going on. On the other hand, rstudent computes externally studentized residuals, not internally studentized residuals. $\endgroup$ – Glen_b Jul 7 '13 at 1:30
7
$\begingroup$

rstandard() produces standardised residuals via normalisation to unit variance using the overall error variance of the residuals/model.

rstudent() produces Studentized residuals in the same way, but it uses a leave-one-out estimate of the error variance.

The key line in rstandard() is

res <- infl$wt.res/(sd * sqrt(1 - infl$hat))

where sd is defined as

sqrt(deviance(model)/df.residual(model))

where model is the object returned by lm. But note this is not the same as sd(resid(model))

Note that the sd is also scaled by the hat values $1 - h_{ii}$ which together explain the discrepancy with your values.

The key line in rstudent() is

res <- res/(infl$sigma * sqrt(1 - infl$hat))

which is almost the same but sd is replaced via the leave-one-out estimate of the error variance (sd) infl$sigma

$\endgroup$
3
$\begingroup$

The two functions do different things, as I understand it.

residuals(model) gives the response minus the fitted values. (from help for lm)

rstandard(model) is part of leave one out influence diagnostics (from help for rstandard), which says:

This suite of functions can be used to compute some of the regression (leave-one-out deletion) diagnostics for linear and generalized linear models discussed in Belsley, Kuh and Welsch (1980), Cook and Weisberg (1982), etc.

You can examine exactly what rstandard does by looking at the code. For lm, you can do this with

> methods(rstandard) 
> getAnywhere(rstandard.lm)

.

$\endgroup$
  • $\begingroup$ As rstandard.lm is exported from the stats namespace, you can just type its name at the prompt without needing getAnywhere(). $\endgroup$ – Gavin Simpson Jul 6 '13 at 15:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.