I know that the log likelihood statistic, i.e. $X^2=-2(LL_1 - LL_2)$ where $LL_1$ and $LL_2$ are the maximum log likelihoods of two models, one nested in the other, is asymptotically distributed as a chi-squared variate with degrees of freedom equal to the difference in the number of free parameters between the two models as $n \rightarrow \infty$; $n$ is the sample size of the data.

My question: is the difference between the sampling distribution of $X^2$ and the theoretical chi-squared distribution always decreasing as n increases? If we compare the difference given two different sample sizes, is the approximation of a theoretical chi-square distribution always better at the larger sample size?

I ask because, in simulations of a rather complicated model, the sampling distribution of $X^2$ does come closer to the theoretical distribution as $n$ increases up to a point and then, as $n$ increases further, the fit between the two gets worse. I did not expect this behaviour and cannot find an explanation.

  • $\begingroup$ For how many simulations (repeated samples) have you assessed the convergence to the Chi2? That is, you are evaluating the convergence at different n. Perfect. Then how many simulations/repeated samples for each n have you performed? $\endgroup$
    – Fr1
    Sep 4 '19 at 15:49
  • $\begingroup$ I have used 5000 independent simulations each time, and have varied the sample size (n) from 50, 100, 200, 500 and 1000 $\endgroup$ Sep 13 '19 at 14:22
  • $\begingroup$ Could you include some more details of the model and simulations? Maybe show some plots? $\endgroup$ Mar 13 '21 at 8:07

It's not clear to me if this is implied in the original question, but if the data follow the alternate hypothesis rather than the null one, the statistic will tend asymptotically toward a noncentral, rather than standard, $\chi^2$. Also, there are circumstances in which the asymptotic result doesn't hold, eg https://stats.stackexchange.com/a/52855/163989. More information on the problem, including some plots, could help.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.