## p-values, Bayes factors, and sufficiency

Posted in Books, pictures, Statistics with tags , , , , , , , , , on April 15, 2019 by xi'an

Among the many papers published in this special issue of TAS on statistical significance or lack thereof, there is a paper I had already read before (besides ours!), namely the paper by Jonty Rougier (U of Bristol, hence the picture) on connecting p-values, likelihood ratio, and Bayes factors. Jonty starts from the notion that the p-value is induced by a transform, summary, statistic of the sample, t(x), the larger this t(x), the less likely the null hypothesis, with density f⁰(x), to create an embedding model by exponential tilting, namely the exponential family with dominating measure f⁰, and natural statistic, t(x), and a positive parameter θ. In this embedding model, a Bayes factor can be derived from any prior on θ and the p-value satisfies an interesting double inequality, namely that it is less than the likelihood ratio, itself lower than any (other) Bayes factor. One novel aspect from my perspective is that I had thought up to now that this inequality only holds for one-dimensional problems, but there is no constraint here on the dimension of the data x. A remark I presumably made to Jonty on the first version of the paper is that the p-value itself remains invariant under a bijective increasing transform of the summary t(.). This means that there exists an infinity of such embedding families and that the bound remains true over all such families, although the value of this minimum is beyond my reach (could it be the p-value itself?!). This point is also clear in the justification of the analysis thanks to the Pitman-Koopman lemma. Another remark is that the perspective can be inverted in a more realistic setting when a genuine alternative model M¹ is considered and a genuine likelihood ratio is available. In that case the Bayes factor remains smaller than the likelihood ratio, itself larger than the p-value induced by the likelihood ratio statistic. Or its log. The induced embedded exponential tilting is then a geometric mixture of the null and of the locally optimal member of the alternative. I wonder if there is a parameterisation of this likelihood ratio into a p-value that would turn it into a uniform variate (under the null). Presumably not. While the approach remains firmly entrenched within the realm of p-values and Bayes factors, this exploration of a natural embedding of the original p-value is definitely worth mentioning in a class on the topic! (One typo though, namely that the Bayes factor is mentioned to be lower than one, which is incorrect.)

## Darmois, Koopman, and Pitman

Posted in Books, Statistics with tags , , , , , , , , on November 15, 2017 by xi'an

When [X’ed] seeking a simple proof of the Pitman-Koopman-Darmois lemma [that exponential families are the only types of distributions with constant support allowing for a fixed dimension sufficient statistic], I came across a 1962 Stanford technical report by Don Fraser containing a short proof of the result. Proof that I do not fully understand as it relies on the notion that the likelihood function itself is a minimal sufficient statistic.

## that the median cannot be a sufficient statistic

Posted in Kids, Statistics, University life with tags , , , , , on November 14, 2014 by xi'an

When reading an entry on The Chemical Statistician that a sample median could often be a choice for a sufficient statistic, it attracted my attention as I had never thought a median could be sufficient. After thinking a wee bit more about it, and even posting a question on cross validated, but getting no immediate answer, I came to the conclusion that medians (and other quantiles) cannot be sufficient statistics for arbitrary (large enough) sample sizes (a condition that excludes the obvious cases of one & two observations where the sample median equals the sample mean).

In the case when the support of the distribution does not depend on the unknown parameter θ, we can invoke the Darmois-Pitman-Koopman theorem, namely that the density of the observations is necessarily of the exponential family form,

$\exp\{ \theta T(x) - \psi(\theta) \}h(x)$

to conclude that, if the natural sufficient statistic

$S=\sum_{i=1}^n T(x_i)$

is minimal sufficient, then the median is a function of S, which is impossible since modifying an extreme in the n>2 observations modifies S but not the median.

In the other case when the support does depend on the unknown parameter θ, we can consider the case when

$f(x|\theta) = h(x) \mathbb{I}_{A_\theta}(x) \tau(\theta)$

where the set indexed by θ is the support of f. In that case, the factorisation theorem implies that

$\prod_{i=1}^n \mathbb{I}_{A_\theta}(x_i)$

is a 0-1 function of the sample median. Adding a further observation y⁰ which does not modify the median then leads to a contradiction since it may be in or outside the support set.

Incidentally, if an aside, when looking for examples, I played with the distribution

$\dfrac{1}{2}\mathfrak{U}(0,\theta)+\dfrac{1}{2}\mathfrak{U}(\theta,1)$

which has θ as its theoretical median if not mean. In this example, not only the sample median is not sufficient (the only sufficient statistic is the order statistic and rightly so since the support is fixed and the distributions not in an exponential family), but the MLE is also different from the sample median. Here is an example with n=30 observations, the sienna bar being the sample median: