[R-sig-ME] lme and prediction intervals

D Chaws cat.dev.urandom at gmail.com
Wed Apr 7 06:17:58 CEST 2010

Very interesting.  Is it really the state of the field that there is uncertainty
about the how to integrate random and fixed effects variance components?
That seems like quite a hole.  Does that mean the best approach is probably
an empirical one then, through something like MCMC?

-- DC

On Tue, Apr 6, 2010 at 5:00 PM, David Hsu <dhsu2 at uw.edu> wrote:
> D Chaws wrote, Tue Apr 6 04:48:33 CEST 2010:
>> As a professional statistician, what is the appropriate method to
>> compute the SE of these predictions which accounts for the variance of
>> the fixed and random effects?
> I asked the same question myself about prediction last month.  I think
> Doug's response to me was, in a nutshell, how can you statistically
> *define* the variance contribution of the random effects?  (Doug,
> please correct me if I am wrong or over-simplifying).  After thinking
> about it for awhile, I indeed can *not* see what the variance of the
> random effect means from a classical perspective.
> Nobody seems to have conceptual difficulties with the errors from the
> fixed effects, so I'll sketch out a toy example to show how I've
> thought about it in terms of the theoretical statistics that I dimly
> remember (again, corrections from anyone are most welcome).
> Writing it out, a simple linear model is Y ~ N( A + BX, s^2).  Our
> vector of parameters in the first distribution is (theta_1) = (A, B,
> s).  From a classical perspective, A and B are fixed parameters, so
> that when we estimate A, B, we are maximizing the likelihood L:
> L ( theta_1 | X ) = P ( Y=y | theta_1 ) = 1/sqrt (2 pi sigma^2) exp -
> (Y - A - BX )^2 / sigma^2
> where (theta_1) is set by definite numbers, (theta_1) = (A, B, s).
> Now, thinking about the random effects model, we're saying that A is a
> random effect, so it's not just a set of numbers.  Therefore A is a
> random variable, distributed by A ~ N(mu, sigma^2).  Our second vector
> of parameters for this distribution is (theta_2) = (mu, sigma).  On
> the face of it, this doesn't necessarily seem catastrophic, since we
> just want the marginal probability distribution function for Y based
> on the joint probability of Y, A,
> P (Y=y | theta_1) = \int^\infty_infty P ( Y=y, A=a | theta_1, theta_2) da
> However, this is where I get stuck in the classical perspective.  We
> need to state *something* about the quantity P ( Y=y, A=a | theta_1,
> theta_2) inside the integral, but it's not clear to me what we can say
> about the relationship of A and Y, and presumably theta_2 is derived
> from the observations of Y.  I think Doug's point is that if we have
> the probability P ( Y=y, A=a | theta_1, theta_2), you can't just use
> conditioning to break it down such as:
> P ( Y=y, A=a | theta_1, theta_2) *not* equal to P ( Y=y | theta_1)
> times P ( A=a | theta_2)
> because somehow you got (or would like to get) theta_2 based on the Y
> (data) as well.  So they A and Y are not independent.
> In the Bayesian perspective, I guess you'd say that the posterior =
> prior x likelihood, so that:
> P ( theta_1, theta_2 | A, Y)  is proportional to P ( theta_1) times P
> ( theta_2 | A, Y, theta_1)
> and then insert some belief about the the prior distribution of
> theta_1, and do this iteratively for each of the parameter values
> until you get convergence.
> Anyway, this is how I've thought about it.  Corrections welcome.
> David
> --
> David Hsu
> PhD Candidate, Urban Planning
> University of Washington
> (e-mail) dhsu2 at uw.edu
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