I am reading Taleb's note on p-value hacking (https://arxiv.org/pdf/1603.07532.pdf) and am pretty confused with Proposition 1.
Let $Z$ be a random normalized variable with realizations $\zeta$, from a vector $v$ of $n$ realizations, with sample mean $mv$, and sample standard deviation $sv$, $\zeta = (mv−mh)/(sv/\sqrt{n})$ (where $mh$ is the level it is tested against), hence assumed to be distributed Student T with $n$ degrees of freedom.
We have g($\zeta$) = P(Z > $\zeta$) which is the one-tailed survival function of the Student T distribution with zero mean and $n$ degrees of freedom. We also have f($\zeta; \bar{\zeta}$) which is the pdf of $\zeta$ given the sample mean $\bar{\zeta}$. In the proof of Proposition 1, the following transformation is made:
$\phi(p,\bar{\zeta}) = \frac{f(g^(-1)(p)}{|g'(g^(-1)(p))|}$
Let $p$ represent the sample-derived one-tailed p-value from the paired T-test statistic (unknown variance) with median value $p_M$ derived from a sample of n size. The distribution across the ensemble of statistically identical copies of the sample has for PDF $\phi(p,p_M)$ (In other words, this is the p-curve!). But I don't see how this is the corresponding p-curve, why this transformation makes sense or what it intuitively means.
Any help would be hugely appreciated.
Hope this helps.
If a random $\zeta$ is inserted into a function $g$ to produce a related random variable $p=g(\zeta)$ then their respective pdfs (denoted $\phi$ and $f$ respectively) satisfy the equality of infinitesimal probability $d {\Bbb P} = \phi(p) dp= f(\zeta) d\zeta$. Next by calculus $dp= g'(\zeta) d\zeta$. Thus $\phi(p)= \frac{f(\zeta)}{g'(\zeta)}$ and since $\zeta=g^{-1}(p)$ this becomes $ \phi(p)= \frac{f( g^{-1}(p))}{ g'( g^{-1}(p))}$