Fisher's method for combining p-values

Question

You have k independent p-values P1, \dots, Pk, each testing the same hypothesis on disjoint data (say different sample periods), and you want to combine them into a single test. Prove that -2\ln Pi \sim \chi^22 under the null, and hence that -2\sumi \ln Pi \sim \chi^2{2k}. Explain how to use this to pool the evidence.

Accepted Answer

One p-value: -2\ln P \sim \chi^22. Under the null P \sim 	ext{Uniform}(0,1). Let Y = -2\ln P. For y \ge 0, \mathbb{P}(Y \le y) = \mathbb{P}(-2\ln P \le y) = \mathbb{P}\!\left(\ln P \ge -	frac{y}{2}ight) = \mathbb{P}\!\left(P \ge e^{-y/2}ight) = 1 - e^{-y/2}. That is the CDF of an Exponential with mean 2, which is exactly the \chi^22 distribution (a chi-squared with 2 degrees of freedom is Exponential with mean 2). So -2\ln P \sim \chi^22. Summing independent tests. Chi-squared random variables add their degrees of freedom when independent: if Yi \sim \chi^22 independently, then X = -2\s…

Fisher's method for combining p-values

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