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I can calculate a likelihood ratio statistic to measure how well my data fits either of two mutually exclusive models:

$\Lambda({\text{data}}) = \frac{ f( \theta_0 | {\text{data}} )}{f( \theta_1 | {\text{data}} )} $

Calculating this statistic is straightforward to me, but I'd like some measure of statistical confidence to be carried along with the likelihood ratio. Is there something analogous to a confidence interval for a likelihood ratio? Or is all of the information about the relative confidence of each model?

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Yes, in principle, the distribution function of the likelihood ratios statistic can be computed, since it is a function of the random variables "data" and the model parameters theta_0 and theta_1. Suppose, that cumulative distribution function is calculated, the p-value can be computed giving a confidence estimate. One example where the distribution of a likelihood ratio is known is when the distribution function "f" is Gaussian (For example in the case of Anova). In this case the likelihood ratio is F-distributed (it is a ratio of two chi-squared random variables) and the p-values can be explicitely computed.

In the case where "f" is not Gaussian , in general there is no explicit expression for the distribution function of the likelihood ratio, however, asymptotic distributions in the case of large samples can be obtained, see for example the following article by: S.S. Wilks, where a chi squared asymptotic distribution of the log-likelihood is obtained.

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  • $\begingroup$ I find the answer above confused. Yes one may find the probability distribution of the likelihood-ratio statistic give either of the two values of the parameters. But that's not about confidence intervals at all. A confidence interval is a subset of the parameter space, which in this case has only two points! And the F statistic is a "likelihood ratio test statistic" only when that term is construed to mean something more general than the simple-versus-simple hypothesis testing scenario described in this question. $\endgroup$
    – Michael Hardy
    Commented Jul 6, 2010 at 22:23
  • $\begingroup$ Can you clarify the aim the sentence: "In the case where "f" is not Gaussian , in general there is no explicit expression for the distribution function of the likelihood ratio"? My understanding is that: There are an infinite number of parametric pairs of distributions for which the likelihood ratio distribution can be computed, Gaussian is a well known example. $\endgroup$
    – robin girard
    Commented Jul 8, 2010 at 6:17
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The answer to the question as phrased is no. As David Bar Moshe said "the distribution function of the likelihood ratios statistic can be computed", but that is certainly not a confidence interval.

A confidence interval is a subset of the parameter space. The parameter space in this case contains only two points, $\theta_0$ and $\theta_1$, so the word "interval" is not appropriate. A what one would have would be a confidence interval FOR $\theta$, not FOR the likelihood ratio statistic.

Mathoverflow frequently sees these answers to statistical questions from people who've read things about the subject but whose way of using basic statistical terminology is that of confused freshmen.

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  • $\begingroup$ One other comment: $$ \Lambda(\text{data}) = \frac{f(\text{data} | \theta_0)}{f(\text{data} | \theta_1)}. $$ $\endgroup$
    – Michael Hardy
    Commented Jul 8, 2010 at 14:15

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