using label="mean"

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wanlin's picture
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Joined: 10/28/2009

Dear all
When set
mxMatrix(type="Full", nrow=1, ncol=2, free=TRUE, values=c(0,0), label= "mean", name="expMean")
in twin ACE model, i found if i set label="mean", the
twinACEFit@output$estimate return 4 values with label (mean, a,c,e)
if the label item not set
mxMatrix(type="Full", nrow=1, ncol=2, free=TRUE, values=c(0,0), name="expMean")

twinACEFit@output$estimate return 5 values with label (NA,NA, a,c,e) and the first two mean values are not exactly the same.

Which one is correct?

neale's picture
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Joined: 07/31/2009
Wanlin Which is correct

Wanlin

Which is correct depends on which model you wanted to fit. The first constrains the means to be equal. Only one label was provided ("mean") which was automatically re-used by R, so both elements of expMean were set to the same free parameter. This is the usual way of specifying ACE models (constrain means of twin1 and twin2 to be equal). The second approach is unconventional, allowing different means for twin1 and twin2. Sometimes - typically when using raw data - this approach is used as part of a saturated model so that the fit of the ACE model can be compared to it.

wanlin's picture
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Joined: 10/28/2009
1. Thanks for clarification

1. Thanks for clarification this. for my situation, I am using univariate ACE model to perform the heritability analysis for brain structure data. As you said, if i provide the label "mean", which indicates means of twin1 and twin2 to be equal for both MZ and DZ. therefore 4 parameters (A,C,E and mean) were estimated. To compare the good fit of ACE model with saturated model, the degree of freedom is 10-4 = 6.

2. if the variance of twin1 and twin2 is not equal, could i standardize both twin1 and twin2 to make them have equal mean=0 and equal variance = 1?

3. As i don't know how to combine the regression into the ACE model, all my data were adjusted age and sex using glm before input to SEM, is it appropriate?

neale's picture
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Joined: 07/31/2009
Ah, brain structure data,

Ah, brain structure data, some of my favorite!

Unequal Twin 1 and Twin 2 variances should only happen by chance, if their ordering is random. If it is not (birth order perhaps) then it may be worth seeing if they cause loss of fit. Any loss of fit above expectations (if you run 20 ROI's then you would expect 1 test to be signficant at .05 after all) may indicate birth order effects which seem unlikely to me. Much more likely is failure of the assumption of multivariate normality. It may be best to hunt for outliers.

I would not standardize twin 1 and twin 2 just to make the model fit. Model failure is worth knowing about and investigating. It should not be swept under the carpet.

wanlin's picture
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Joined: 10/28/2009
Thanks neale. If the

Thanks neale.
If the means and variances of Twin1 and Twin2 for both MZ and DZ are not significantly different. Can i standardize them.

Some times, i found if the data were standardized, the ACE model will fit the saturated model very well with a fixed initial value.

Another question, sometimes i find the -2 log likelihood is minus, but results from examples or papers always report positive likelihood, should the value always positive?

neale's picture
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Joined: 07/31/2009
No, it is not a good idea to

No, it is not a good idea to standardize twin 1 and twin 2 separately. Doing so artificially makes the agreement between say var(T1) and var(T2) perfect. Sampling error should make them different. So if you did standardize separately you should really adjust the degrees of freedom downwards. Thus model fits look better than they really are if you have not adjusted the df.

-2lnL can be either a positive or a negative value. Usually, most of the data points have a likelihood less than 1, so the log-likelihoods are negative, and the -2lnL's are positive. However, with normal theory (FIML) analysis with variables with small variances, the likelihoods can often exceed 1.0, so the log-likelihoods are positive, and the -2lnL's are negative.

While we're on the topic, it's an idea to avoid variables with very small variances, simply because during estimation trial values for a variance of zero necessarily cannot be evaluated (likelihood is zero, -2lnL is negative infinity).