# Re: residual variance

From: <arthur.gilmour_at_DPI.NSW.GOV.AU>
Date: Tue, 29 May 2007 10:16:37 +1000

Dear Felicity,
re:

variance from the output below as I need it to calculate heritability? I
don't suppose it's as simple as an arithmetic mean, harmonic mean or
quadratic mean, or is it a bit more complicated than that?

Source Model terms Gamma Component Comp/SE %
C
Residual 1260 1252
at(Trial,2).Rep 4 4 0.842880E-02 0.842880E-02 0.51 0
P
Female 281 281 0.211808 0.211808 5.99 0
P
Trial.Female 1124 1124 0.103475 0.103475 4.87 0
P
Variance[ 1] 492 0 0.700046 0.700046 12.54 0
P
Residual AR=AutoR 12 0.118128 0.118128 2.06 0
U
Residual AR=AutoR 41 0.167493 0.167493 3.18 0
U
Variance[ 2] 492 0 0.472797 0.472797 12.62 0
P
Residual AR=AutoR 12 0.195265E-01 0.195265E-01 0.32 0
U
Residual AR=AutoR 41 0.237822 0.237822 4.58 0
U
Variance[ 3] 120 0 0.520579 0.520579 6.11 0
P
Residual AR=AutoR 12 -0.345290E-01 -0.345290E-01 -0.25 0
U
Residual AR=AutoR 10 0.193473 0.193473 1.50 0
U
Variance[ 4] 156 0 0.662102 0.662102 6.74 0
P
Residual AR=AutoR 12 0.121112E-01 0.121112E-01 0.11 0
U
Residual AR=AutoR 13 0.300947 0.300947 2.98 0
U

Since I have not noticed anyone venturing a response, here is my attempt.

The usual way to pool variances is weighted by their individual degrees of
freedom.
That is a bit difficult in this case as the degrees of freedom are not
easily obtained, being affected
by the correlated model as well as by the fixed terms in the model.

So, it may be possible to extract the residual stratum variance degrees of
freedom from
separate analyses of each trial, fitting 'female' as random, and use that.

Otherwise, weighting just be number of plots (less fixed effects), would
not be too bad in this instance
given the two largest trials have the most diverse variances (.70 and .47)
which are not extremely different.

You have constrained the genetic variances to be common across trials, so
you could also
constrain the residual variance to be common and see what the 'common'
value is.

This of course highlights the general problem that 'heritability' was
inverted in the context of homogeneous variances
but now we want to use it with heterogeneous variances.

So, an alternative would be to fit

Trial.Female
with

Trial 0 CORUH 0.7
.31 .31 .31 .31
Female

and then calculate 4 heritabilites (each trial) since this would be more
informative.

It seems to me that heritability as a concept in plot trials is a bit
vague in any case because
it will be affected by plot size. I think Brian has an alternative way to
do the calculation for plot trials.

May you know Jesus Christ and His blessing in 2007,

Arthur Gilmour, His servant .

Mixed model regression mapping for QTL detection in experimental crosses.
Computational Statistics and Data Analysis 51:3749-3764 now available at
http://dx.doi.org/10.1016/j.csda.2006.12.031

Personal website: http://www.cargovale.com.au/

mailto:Arthur.Gilmour_at_dpi.nsw.gov.au, arthur_at_cargovale.com.au
Principal Research Scientist (Biometrics)
NSW Department of Primary Industries
Orange Agricultural Institute, Forest Rd, ORANGE, 2800, AUSTRALIA

fax: 02 6391 3899; 02 6391 3922 Australia +61
telephone work: 02 6391 3815; home: 02 6364 3288; mobile: 0438 251 426

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Received on Sat May 29 2007 - 10:16:37 EST

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