Let $(P,Q)$ and $(P',Q')$ be pairs of random variables and let $\phi, \phi'$ be the most powerful level $\alpha$ tests between $P$ and $Q$ and between $P'$ and $Q'$, respectively. Suppose that $\phi$ and $\phi'$ have the same power $1-\beta$ at all levels $\alpha$ for both tests. Can we show that this is equivalent to $P=P'$ and $Q=Q'$ in distribution?
It's easy to show that, if $P=P'$ and $Q=Q'$ i.d., the ROC curves are pointwise equal by the construction of the ROC curve, by which one direction follows.
However, I'm not yet able to show the other direction. Any ideas? Can we perhaps show something even stronger, such as $\mathrm{TV}(P,P') = \mathrm{TV}(Q,Q') = 0$?
As a simple counterexample, suppose $P'$ is the law of $X +1,$ where $X \sim P,$ and similarly $Q'$ is the law of $X+1$, where $X \sim Q$. Then for any test for $P$ v/s $Q,$ we can generate a test for $P'$ v/s $Q'$ by first subtracting $1$ from each data point, and vice versa. This means that the ROC curves for the most powerful tests are equal. But of course $P' \neq P$ and $Q' \neq Q$. Same idea works with $X \mapsto X+1$ replaced by any measurable bijection.