The statement that you are claiming you have read is incorrect. The chi-square and inverse chi-square densities evaluated at x are not 'similar' up to a constant (in fact there is no such concept as far as I am aware. If are equal up to a constant then the constant is 1. Also, in your R code which generates a co-plot of two densities, you plot the chi-squared in black and a scaled chi-squared in (color # 2). There is no inverse chi-squared in your code. The inverse chi-squared is the distribution of 1/X when X is chi-squared-- therefor the variable 'x' appears in two places-- raised to a negative exponent and to the power {-1} inside the exponential function. So while you can play all sorts of games with scales and shapes obtaining various identities, the most important thing to keep in mind is that it is the

Inverse gamma density of rate 'r' and shape 's-1' and the chi-squared of n degrees of freedom that are conjugate -- the former usually used as the prior for the latter. Stated more generally,

the inverse gamma density of rate 'r1' and shape 's1-1' and the gamma of rate 'r2' and shape 's2' are conjugate.

It appears that you are trying to work with Bayes estimators. If that is the case, you should be aware that Bayesians often parameterize the Normal family in terms of the mean and precision, where the precision parameter is the inverse of the variance.

When parameterized in this way, the conjugate prior of the precision is the Gamma distribution. When parameterized in terms of the variance, the conjugate prior of the variance is the inverse Gamma distribution.

However, in Bayesian analysis, it is also true that conjugate priors are becoming less important. Markov Chain Monte Carlo methods are replacing the mathematical convenience of conjugate priors.

Thanks for your answers. Actually, I am not trying to work with Bayes estimators. I simply tried to understand the t-test or, more general, linear models. I found that the t-test actually is a likelihood ratio test, so I had to learn about likelihoods. This is where RA Fisher stayed. Bayesians use the likelihoods to get the posterior prbability distribution; for a flat (uniform) prior, this will just be the normalized likelihood function.

So I'm not interested in the property of "conjugacy" or prior/posterior things (at least not in this context). I am really only looking at likelihoods: the "information" about parameters, given the data (can one say so? Likelihood will only turn into probability when multiplied with a probability distribution... there we get to Bayes).

When our expectation about the residuals is described by the normal distribution we have to deal with this nuisance parameter (sigma) to get the precision of our etimate of mu. I read that "integrating out" such parameters will give the marginal likelihood as the solution to this problem. I further read that this marginal likelihood here has the shape of the t-distribution. I tried to follow this all with numerical examples and I found that I end up with a t-distribution with n d.f. - not with n-1 d.f. So there must be some error. I thought this error is already in my calculation of the likelihood for sigma(²).

@Grant: There is the inverse, because dscinvchisq uses exp(-x). So I thought. I got this formula from Wikipedia. Was it not correct or is my implementation wrong?

Grant seems to be right. I think your definition of the scaled inverse chi^2 distribution is not correct. You can use the implementation of the scaled inverse chi^2 distribution from the geoR package (http://rss.acs.unt.edu/Rdoc/library/geoR/html/InvChisquare.html) for crosschecking.

The chi-square/squared tests forms an enormous families of statistical tests.

We may speak also about the theory of chi-squared tests.

It is very interesting and very nice theory of statistical tests, used for testing very different hypotheses. There are many papers published from 1900, when the first paper of Karl Pearson on chi-square test was published, see also

P.Greenwood, M.Nikulin "A Guide to chi-squared testing" (1996), J.Wiley ans sons,

where were exposed many results on parametric models for complete data.

From 1960/1970 , in survival analysis and reliability people to use the right censored and left truncated data. There are many papers on construction of the chi-squared tests for censored data, can see, for example, V.Bagdonavicius, J.Kruopis, M.Nikulin « Nonparametric Tests for Censored Data », WILEY-ISTE( 2011). More about different applications of chi-squared tests, see in different papers and books via Google, for example.

Which hypothsis you want to test? What you know about the possible alternative for your null hypothesis?

Any chi-squared tests as all other tests has some advantages and desadvantages.

Suppose that you have no any alternative the null hypothesis. What you shall do if your test reject the null hypothesis? Any chy-squared tests likes that a statisticain knows the alternative.

His behavior is very natural in such a case, and even more, he can chose the interval for data groupping by special way to have good power.

Ok, I hope I found a point to clarify my problem:

The formulas are in http://www-clmc.usc.edu/~adsouza/notes/t_distribution.pdf.

Instead of the variance, its reciprocal, the precision, is used (as stated before by Dennis). This is said to be gamma-distributed. Why? How is this distribution derived (it may well be via some other distribution where is was shown that the gamma is a generalization that is useful as a prior)? How is it related to the likelihood of the precision (or the variance)?

(Just an example: for a binomial variable the likelihood of the proportion is a beta-distribution. This can be relatively easily derived from the binomial pdf. Another one: for a given variance s², the likelihood of the mean is again a normal, but with variance s²/n. This also follows from the formulas).

You do know that the chi-squared distribution on n degrees of freedom is so named because its the distribution of the sum of squares of n independent variables having mean zero and unit variance. You also are aware that the chi-square distribution on n-degrees of freedom is a gamma of shape n/2 and rate 1/2. In F tests we have a ratio of sum of squares--of residuals about the alternative hypothesis means in the numerator and of residuals about the null hypothesis means in the denominator. A good example of the need for a shrinkage variance estimate is in the analysis of microarrays in which we are conducting per gene F tests with tens of thousands of genes, but only tens of microarrays. If the number of arrays is smaller than 20 it may serve us well to stabilize the per gene estimate of the denominator sums of squares by borrowing strength from the whole ensemble and using what is called a stabilized F-statistic.

A stabilized F-statistic is of the form

SS_top / ( SS_bottom + k )

The idea of 'borrowing' from the ensemble is kind of a random variance model-- the observed per gene sum of squares about the null hypothesis means are assumed to be distributed inverse gamma shape 's' and rate 'r' . Now we insert your question here-- why inverse gamme ? Let V be the random variance. Divide top and bottom by V

SS_top/V / ( SS_bottom/V + k / V )

now, after appropriate normalizing by degrees of freedom, we have in the numerator a Non-Central Chi-Square and in the denominator we have the sum of a Central Chi-square and ... and ... bingo ... a gamma (since the inverse gamma is so named because its reciprocal has a gamma distribution). Since the sum of two gammas is gamma then the denominator is still gamma and the entire ratio has an F distribution.

Thats why

Technically, Snedecor's F is the ratio of independent Chi-square random variables divided by their degrees of freedom, rather than the ratio of the Chi-squares, directly. You acknowledge this in passing by mentioning normalization by degrees of freedom.

What exactly are you taking k and V to be in this? (Distributional assumptions and relationships to SS_Top and SS_Bottom, I mean.)

Without that, I can't figure out if I believe your claim. It looks to me like you end up a ratio of correlated Snedecor's F's or a doubly noncentral F. I can't see assumptions that make a central F pop out.