Testing for sigma different than zero from the natural log of sigma

[Alfonso Sánchez-Peñalver]

Hello everyone,

I apologize in advance if this has been asked already in the list, since I figure it would, but I have not been able to find anything on it online. Many estimation commands estimate the variance or the standard deviation in logarithmic form when using maximum likelihood so that the estimate doesn't break any boundary issue of the parameter being estimated. The command then presents a test of the natural log coefficient being different than zero, and an estimate of the variance or standard deviation with a confidence interval, 95% being the default. My question is if there is any information anywhere that shows how we can test that the actual estimate of the variance, or the standard deviation for that matter, is different than zero. Accessing the natural log estimate is perfectly explained by Buis (2011), and working with correlations is perfectly explained by Cox (2008). However, I can't find anything on this topic and it is killing me, because the test that the natural log of the standard deviation is different from zero is a test that the standard deviation is different from one, but that doesn't mean that the standard deviation is different from zero. Also, the natural logarithm of zero is indeterminate, so one cannot test on the actual logarithm provided. Finally, I'm weary of using -nlcom- using the exponential of the actual value provided because this is a test on the boundary of the variable, and the test presented in -nlcom- assumes normality.

I'd appreciate any help provided in this, and I apologize again if this was already discussed and I was unable to find it.

Alfonso Sanchez-Penalver, PhD

References: Buis, Maarten L., “Stata Tip 97: Getting at ρs and σs,” The Stata Journal, 2011, 11 (2), 1–3.
Cox, Nicholas J., “Speaking Stata: Correlation with Confidence, or Fisher’s z Revisited,” The Stata Journal, 2008, 8 (3), 413–439.

[Steve Samuels]

The standard test that \(\sigma=0\) when there are inequality constraints (as here \(\sigma \ge 0\)) is based on the chi-bar-square distribution. See page 14 of the Multi-level Mixed Effects manual, which states:

When there is only one variance being set to 0 in the reduced model, the asymptotic distribution of the LR test statistic is a 50:50 mixture of a \(\chi_p^2\) and a \(\chi_{p+1}^2\) distribution, where p is the number of other restricted parameters in the reduced model that are unaffected by boundary conditions. Stata labels such test statistics as chibar and adjusts the significance levels accordingly.

When \(p = 0\), the chi-bar-square distribution puts probability 0.5 on zero and 0.5 on a \(\chi_1^2\) distribution. So these tests are not based on confidence intervals for \(\operatorname{log}\sigma\) or \(\sigma^2\).

Steve

[Alfonso Sánchez-Peñalver]

Hi Steve, thank you for your answer. I'm using Roodman's cmp (SSC) to model a random effects multinomial probit, so the test is to see if there is significant variance at the clustering (higher) level. cmp doesn't provide a test for the significance of random effects, like mixed does, for example. I, therefore, have to do the test myself. My question to your explanation then is about the p, and what it represents, because I don't really understand what is meant by

where p is the number of other restricted parameters in the reduced model that are unaffected by boundary conditions

.

Also, how do I calculate the test statistic from the natural logarithm values? In particular, how do I get the right value for the standard errors of \(\sigma\)?

Thanks again,

Alfonso Sanchez-Penalver, PhD

[Steve Samuels]

Sorry, I'm unfamiliar with cmp and the multinomial probit models. You'd need to see the formula for the log-likelihoods with and without the assumption that \(\sigma >0\). In your case, \(p = 0\), as the only restricted parameter is \(\sigma\).

[Alfonso Sánchez-Peñalver]

Thanks again, Steve. So basically the log-likelihoods of the model without random effects and the model with random effects, right? The test statistic would be the likelihood ratio (log-likelihood difference), and the \(p\) is the number of variances tested to be jointly insignificant. Am I understanding this right?

Thanks,

Alfonso.

[Alfonso Sánchez-Peñalver]

Sorry, I have finally understood. The \(p\) is any other parameters that we are restricting but not in the boundary. Reading help j_mixedlr made me realize it. When you have more than one random effect variance to test jointly, there is still no exact method of testing, but it has been proved that the \(\chi^2\) distribution will provide a conservative estimate. Thanks again for your help.

Alfonso.

[Steve Samuels]

You are very welcome, Alfonso.
As you say, you compare 2log-likelihood for the models with and without random effects. The absolute difference is the log-likelihood ratio, say, c. Include any covariates that were generated by the group-level units. So, if the group-level was "school" and the individual level was "student", include whatever "school" covariates were in the full model.. If \(\sigma\) is the only SD for a random effect, then \(p =0\). Suppose c is 2.1, then the chibar2(01) p-value will given by

. di chi2tail(1,2.1)/2
.07364957

Steve

[Alfonso Sánchez-Peñalver]

Hi again Steve,

so let's say now that \(p = 2\), so using your example, would the following be correct?

. di (chi2tail(2,2.1) + chi2tail(3,2.1))/2
.45092526

[Steve Samuels]

If \(p=2\), that is correct.