Non-iid hypothesis testing: from classical to quantum

De Palma, Giacomo, Fanizza, Marco, Mowry, Connor, O'Donnell, Ryan

arXiv.org Artificial Intelligence 

We study hypothesis testing (aka state certification) in the non-identically distributed setting. A recent work (Garg et al. 2023) considered the classical case, in which one is given (independent) samples from $T$ unknown probability distributions $p_1, \dots, p_T$ on $[d] = \{1, 2, \dots, d\}$, and one wishes to accept/reject the hypothesis that their average $p_{\mathrm{avg}}$ equals a known hypothesis distribution $q$. Garg et al. showed that if one has just $c = 2$ samples from each $p_i$, and provided $T \gg \frac{\sqrt{d}}{ε^2} + \frac{1}{ε^4}$, one can (whp) distinguish $p_{\mathrm{avg}} = q$ from $d_{\mathrm{TV}}(p_{\mathrm{avg}},q) > ε$. This nearly matches the optimal result for the classical iid setting (namely, $T \gg \frac{\sqrt{d}}{ε^2}$). Besides optimally improving this result (and generalizing to tolerant testing with more stringent distance measures), we study the analogous problem of hypothesis testing for non-identical quantum states. Here we uncover an unexpected phenomenon: for any $d$-dimensional hypothesis state $σ$, and given just a single copy ($c = 1$) of each state $ρ_1, \dots, ρ_T$, one can distinguish $ρ_{\mathrm{avg}} = σ$ from $D_{\mathrm{tr}}(ρ_{\mathrm{avg}},σ) > ε$ provided $T \gg d/ε^2$. (Again, we generalize to tolerant testing with more stringent distance measures.) This matches the optimal result for the iid case, which is surprising because doing this with $c = 1$ is provably impossible in the classical case. We also show that the analogous phenomenon happens for the non-iid extension of identity testing between unknown states. A technical tool we introduce may be of independent interest: an Efron-Stein inequality, and more generally an Efron-Stein decomposition, in the quantum setting.