quick question regarding your "residual" command for your lmer command - please help me
Again, I will defer to the R-Sig-Mixed-Models mailing list.
On Wed, Aug 29, 2012 at 6:52 AM, Yugo Nakamura <yugonakamura25 at gmail.com> wrote:
To Professor Bates, Thank you for your reply and for adding me on to the list. I will check your commands again. To shorten my question, I essentially get confused when researches study the level 1 residuals pooled across the groups and not within groups separately. I just do not see why and what this analysis entail or suggest for the multilevel analysis and assumptions that do not take the grouping into consideration. Your thoughts and expertise will be deeply appreciated. Above On Tue, Aug 28, 2012 at 11:25 PM, Douglas Bates <bates at stat.wisc.edu> wrote:
I have taken the liberty of cc:'ing the R-SIG-Mixed-Models at R-project.org mailing list on this reply. Many of those who read that list will be able to help you, often more quickly than I am able to do. Your question is a bit confusing in that you say you are using lmer and then quote the documentation for residuals.lme. The nlme package, containing lme and supporting methods, and the lme4 package, containing lmer, are different. You should not expect the documentation for one to apply to the other. On Tue, Aug 28, 2012 at 8:20 AM, Yugo Nakamura <yugonakamura25 at gmail.com> wrote:
To Professor Bates,
I am terribly sorry for this impromptu email. I have been using your
lmer
command for my dissertation (at the University of Washington) but I am
having some difficulty to fully understand the outputs. Your expertise
will
be deeply appreciated. I would like to thank you in advance for your
time.
My question pertains to the Residual command of the lmer and the
variance
estimates of the level 1 residuals in the lmer output. I am not
certain as
to how these figures are calculated and what these estimates really
imply.
The definition of the residual are as follow.
residuals.lme {nlme}
The residuals at level i are obtained by subtracting the fitted levels
at
that level from the response vector (and dividing by the estimated
within-group standard error, if type="pearson"). The fitted values at
level
i are obtained by adding together the population fitted values (based
only
on the fixed effects estimates) and the estimated contributions of the
random effects to the fitted values at grouping levels less or equal to
i
http://stat.ethz.ch/R-manual/R-patched/library/nlme/html/residuals.lme.html
Is this definition saying that, the residuals are defined by subtracting
the
fitted values from the fixed effects and the empirical Bayes estimate of
the
level 2 residuals?
I calculated the variance of these residuals and it gave me the same
estimate of the variance estimate of the "Residuals" (level 1) in the
lmer
output.
But my question then is, is this variance estimate the "within group
variance" explained in different textbooks of multilevel analysis? But
if
so I find it slightly too big.. I have learned that within group
variance
are normally distributed with mean zero and constant variance for each
and
every group. That is, the variance is calculated within/for each and
every
group and this variance is constant across all groups (quite strong
assumption).
But the variance estimate and the residual command above seems to give
us
the pooled residuals regardless of the groups. That is, the variance is
the
estimate of the variance across all the residuals regardless/ignoring of
the
group. Is this so? If this is the case, isn't this variance the sum
of
all the within group variances (simply because the sum of normal
distribution is also a normal distribution with the mean and variance
also
summed)? Thus, to get a within group variance (for each group) I should
divide your Residual variance estimate by the number of groups?
I hope I was able to make sense. It will be great if you could share me
your expertise. If there is a link that describes all the details of
your
command, that will be very helpful as well.
Thank you so much in advance for your time and cooperation!
yours,
Yugo Nakamura