This is still my old problem...
Obviously, the likelihood and the "sampling distribution" are related. The shape of the likelihood converges against the shape of a normal distribution with mean = xbar and variance = s²/n (difference is just a scaling factor to make the integral equal unity).
Consider normal distributed data, the variance s² being a nuisance parameter. Ignoring the uncertainty of s², the likelihood again is similar to the normal sampling distribution. The confidence interval can be obtained from the likelihood as the central range covering 100(1-a)% of the area under the likelihood curve. This is explained for example in the attached link, p. 31, last paragraph.
The author shows that the limits of the intervals are finally given by sqrt(chi²(a))*se (equation 13) and states that sqrt(chi²(a)) is the same as the a/2-quantile of the normal distribution. Then he writes "The test uses the quantile of a normal distribution, rather than a Student t distribution, because we have assumed the variance is known.". Ok so far.
How is this done in the case when the variance is unknown? I suppose one would somehow come to sqrt(F)*se, so that sqrt(F) is the quantile of a t-distribution...
Other question: How is the likelihood (joint of mu and sigma²) related to the t-distribution? It there a way to express this in terms of conditional and/or marginal likelihoods? This could possibly help me to understand the principle when there are nuisance parameters in general.
http://www.math.mcmaster.ca/~bolker/emdbook/chap6A.pdf
If you want to use likelihood you need to use profile likelihood. This does not correspond exactly to the t-distribution.
Like in this lecture:
http://ocw.jhsph.edu/courses/MethodsInBiostatisticsI/PDFs/lecture9.pdf
(slides 22-24)?
It's clear that the philosophical background of likelihood and t-distribution is different, but are the shapes of these functions different? I cant show this analytically, but numerically they seem to have identical shapes (is there an analytical proof of proportinality of the profile likelihood and the t?).
I also understood that the limits of the profile likelihood confidence intervals are obtained as the intersections of the profile likelihood function with 1/8 or 1/16 of its maximum. It is clear that these values do not correspond exactly to the 0.025 and 0.975-quantiles of the t. But (essentially the same question as before): if one would normalize the profile likelihood to get an unity integal and take this as a probability density, would the 0.025 and 0.975-quantiles then be identical to the limits of the 95% CI as given by the t-distribution?
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