Two sophisms on products of independent p-values
It is usual for frequentists to say that multiplying independent p-values together is not permitted. I will show that the frequentists multiply independent p-values together all the time. It is usual for Bayesians to say that frequentist p-values are not Bayesian probabilities. I will show that many frequentist p-values in common use are indeed Bayesian probabilities. In neither sophism am I speaking of inequalities or approximations.
First sophism
The frequentists tell us to begin by taking the natural logarithms of the p-values, and then to add those logarithms together. However, the sum of logarithms is the logarithm of the product.
Solution of the first sophism
The product of the p-values is not yet the answer, and neither is the logarithm of that product. One must multiply by minus 2 and then look up the result in the table of chi-square.
Second sophism
Let us use Laplace’s model where exp( -(x-mu)^2/2 )/sqrt( 2*PI ) is the conditional probability density for x given mu, where x and mu lie between minus infinity and plus infinity, and where the a priori density of mu is uniform on its domain. If we take a sample of n numbers from this model, then the a posteriori probability density of mu is sqrt(n)*exp( -n*(mu-xBar)^2/2 )/sqrt(2*PI), where xBar is the sample mean.
Let the null hypothesis say that mu is uniformly distributed from minus infinity to zero, and let the alternative hypothesis say that mu is uniformly distributed from zero to plus infinity. Then the a posteriori probability that the null hypothesis is true is just the integral of the a posteriori probability density from minus infinity to zero. This is numerically equal to the integral with respect to w of
sqrt(n)*exp( -n*w^2/2 )/sqrt(2*PI)
from xBar to infinity. So it is equal to
the p-value of the corresponding frequentist inference. The p-values of this problem are continuously distributed between zero and unity. Conversely, any p-values that are continuously distributed between zero and unity can be mapped into Laplace’s model. (I am not doing two-sided inferences or discrete domains.)
Now suppose that we have two independent p-values of this kind. Each is mapped into a Laplace model, so each is equal to a Bayes a posteriori probability. These probabilities are independent, so we may multiply them together and get a Bayes probability.
Solution of the second sophism
Yes, it is a Bayes probability, but it does not belong to a Laplace model of the kind we are considering, so there is not any obvious way to change it back into a p-value.
Non-originality, date, e-mail address, links
I cannot claim to have invented the above sophisms. I have merely assembled them from pieces known to everybody. I put the sophisms on this web page to amuse students of statistical inference and their teachers. The date of this page is 25 February 2008. Comments on all this, both constructive and destructive, come to me, Harold Kaplan,
at dot
smtw2gh toadmail com
Harold Kaplan’s statistics.htm
John C. Pezzullo’s page