# p-values and measure theory

Self-reassurance that p-value properties donâ€™t depend on regularity assumptions on the test statistic.

Valerio Gherardi https://vgherard.github.io
2023-06-07

Let $$(\Omega, \mathcal E, \text{Pr})$$ be a probability space, where $$\Omega$$ is the space of random outcomes, $$\mathcal E$$ the $$\sigma$$-algebra of measurable events, and $$P$$ the probability measure.

Given a random variable $$T\colon \Omega \to \mathbb R$$, define $$p_T\colon \Omega \to \left[0,1\right]$$ as:

$p_T(\omega) = \text{Pr}(\{\omega'\in \Omega\,\vert\, T(\omega')\geq T(\omega)\})$

Theorem. $$p_T$$ is measurable and $$\text{Pr}(p_T\leq \alpha) \leq \alpha$$ for all $$\alpha \in \left[0,1\right]$$. Equality holds if and only if there exists a sequence $$\{\omega_n\}_{n\in \mathbb N}$$ such that $$p_T(\omega_n) \leq \alpha$$, and $$p_T(\omega _n)\to \alpha$$ as $$n \to \infty$$.

Proof. Let $$\alpha\in\left[0,1\right]$$, and denote: $E_T(\omega) = \{\omega'\in \Omega\,\vert\, T(\omega')\geq T(\omega)\},$ so that $$p_T(\omega) = \text{Pr}(E_T(\omega))$$.

Assume first that there exists $$\omega_\alpha \in p_T^{-1}(\alpha)$$, that is to say $$\text{Pr}(E_T(\omega)) = \alpha$$. We can show that:

$N_T(\omega_\alpha) = \{\omega \vert p_T(\omega) \leq \alpha\} \backslash E_T(\omega_\alpha)$ is a measurable, zero probability set, which proves the thesis for this particular case. As a matter of fact, for any $$\omega \in \Omega$$, if $$p_T(\omega)\leq \alpha$$ and $$T(\omega) <T(\omega_\alpha)$$, then we must have:

$\text{Pr}(\{\omega'\in \Omega\,\vert\, T(\omega_\alpha)>T(\omega')\geq T(\omega)\} ) = p_T(\omega) - \alpha=0.$ If $$t_* = \inf_{p_T(\omega)\leq \alpha}T(\omega)$$ and $$\{a _n\}_{n \in \mathbb N}$$ is a sequence in $$\Omega$$ such that $$T(a _n)\to t_*$$ as $$n\to \infty$$, then:

$N_T(\omega _\alpha) \subseteq \cup _n \{\omega'\in \Omega\,\vert\, T(\omega_\alpha)>T(\omega')\geq T(a_n)\},$ the right hand side being a probability zero set.

If $$p_T^{-1}(\alpha)$$ is empty, let $$\alpha^* = \sup _{p_T(\omega)\leq \alpha}p(\omega)$$, and let $$\{b _n\}_{n\in \mathbb N}$$ be a sequence in $$\Omega$$ such that $$p_T(b_n)\to \alpha^*$$ as $$n\to \infty$$. Then:

$\{\omega \vert p_T(\omega) \leq \alpha\}= \{\omega \vert p_T(\omega) \leq \alpha^*\}= \cup _n \{\omega \vert p_T(\omega) \leq p_T(b_n)\},$ so that, from the particular case proved earlier, we have:

$\text{Pr}(p_T \leq \alpha) = \lim _{n \to \infty} \text{Pr}(p_T \leq p_T(b_n)) \leq \lim _{n \to \infty} p_T(b_n) = \alpha ^* \leq \alpha,$ as was to be proved.

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### Reuse

Text and figures are licensed under Creative Commons Attribution CC BY-SA 4.0. Source code is available at https://github.com/vgherard/vgherard.github.io/, unless otherwise noted. The figures that have been reused from other sources don't fall under this license and can be recognized by a note in their caption: "Figure from ...".

### Citation

Gherardi (2023, June 7). vgherard: p-values and measure theory. Retrieved from https://vgherard.github.io/posts/2023-06-07-p-values-and-measure-theory/
@misc{gherardi2023p-values,
}