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  • LR test of nested models yields chi-sq=0. Is this cause for concern?

    I am using the Likelihood-Ratio test to test for the presence of unobserved individual-level heterogeneity (frailty) in a duration model. The duration/survival literature refers to this unobserved heterogeneity as frailty, but the frailty is simply a random effect (intercept) included at the individual level. As a result, my concern of the 0 valued chi-sq statistic should be the same as the concern one would have when dealing with random-effect models.

    I am using mestreg to estimate the duration model. I assume that survival times are distributed exponentially. The LR Test returned by mestreg compares the model containing the frailty/random effect component, with the same model that doesn't contain the random effect. mestreg reports a chi-sq value of 0 (log likelihoods from both models are identical). I interpret this as telling me that the duration model with frailty is identical to the model without frailty, which would mean that there is no unobserved heterogeneity and I can simply use the no-frailty model. However, I am skeptical of this result because it seems highly unlikely that two models (one with random effects and the other without) would yield identical log likelihoods. I provide the command I run below. Can you please let me know if I am making a syntax error that might be causing me to get this result? If not, would you accept the result, or would you look elsewhere/where would you look?

    Here is the command that returns the 0 chi-sq (note the frailty/random effect is introduced at the X3 level which is an identifier for each individual):
    Code:
     mestreg X1 X1 || X3:, dist(exponential)
    I also fit a piece-wise exponential model using the following syntax, and I still get a chi-sq of 0:
    Code:
     mestreg X1 X1 i._t || X3:, dist(exponential)
    Lastly, I fit a Weibull model and this gives me a non-zero chi-sq
    Code:
     mestreg X1 X1 || X3:, dist(weibull)
    Although the weibull model yields a non-zero chi-sq, the value is small and the p-value is 0.74. It allows me to make the same qualitative conclusion as the LR tests for the exponential models i.e., that there is no unobserved heterogeneity at the individual level.

    I am not sure what is causing the chi-sq to be 0 in the exponential models, and I don't even know whether this is a problem. The fact that both exponential models yielded the 0 chi-sq, while the weibull yielded a non-zero value gave me pause and made me wonder if Stata's mestreg didn't like fitting random effect models with an assumed exponential distribution. I re-fit both exponential models by switching the level of the random effect from X3 to some other variable X4. This yielded non-zero chi-sq values. This leads me to believe that it is not an issue of mestreg not liking random effects in an exponential model. Instead, it just appears that there isn't much unobserved heterogeneity at the individual level (this is corroborated by the LR test for the weibull model). However, I am still uneasy with the 0 chi-sq. Should I be concerned?
    Last edited by Sharad Kumar; 30 Sep 2017, 17:13.

  • #2
    Originally posted by Sharad Kumar View Post
    I re-fit both exponential models by switching the level of the random effect from X3 to some other variable X4.
    Huh?

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    • #3
      Joseph, I mean I reran the exponential model as the following

      Code:
      mestreg X1 X2 || X4:, dist(exponential)
      Where the X3-level random effect in the OP has been replaced by X4. I 're ran the piecewise exponential with a similar substitution.

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      • #4
        Does anyone have an answer to this?

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