0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.11 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1 − F (ˆ ρ,ρ ) r 2 ≡ 2Tr ρ $^{2}$− 1 Infidelity: N=10 MLE HMLE ( β = 1 $_{2}$) 0.25 0.3 0.35 0.4 0.45 0.5 0.55 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 || ˆ ρ − ρ || 2 r 2 ≡ 2Tr ρ $^{2}$− 1 Euclidean Error: N=10 MLE HMLE ( β = 1 $_{2}$) 0 0.005 0.01 0.015 0.02 0.025 0.03 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1 − F (ˆ ρ,ρ ) r 2 ≡ 2Tr ρ $^{2}$− 1 Infidelity: N=100 MLE HMLE ( β = 1 $_{2}$) 0.09 0.1 0.11 0.12 0.13 0.14 0.15 0.16 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 || ˆ ρ − ρ || 2 r 2 ≡ 2Tr ρ $^{2}$− 1 Euclidean Error: N=100 MLE HMLE ( β = 1 $_{2}$) 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.004 0.0045 0.005 0.0055 0.006 0.75 0.8 0.85 0.9 0.95 1 1 − F (ˆ ρ,ρ ) r 2 ≡ 2Tr ρ $^{2}$− 1 Infidelity: N=1000 MLE HMLE ( β = 1 $_{2}$) 0.036 0.037 0.038 0.039 0.04 0.041 0.042 0.043 0.044 0.045 0.7 0.75 0.8 0.85 0.9 0.95 1 || ˆ ρ − ρ || 2 r 2 ≡ 2Tr ρ $^{2}$− 1 Euclidean Error: N=1000 MLE HMLE ( β = 1 $_{2}$) 4 p ′ k = β N and p ′ j = $^{(}$1 − β N ) n$_{j}$ N for j = k , then Likelihood ratios between e − 1 and e are “barely worth mentioning” [15], so if β < 1, then p$_{H}$ is essentially as plausible as p$_{MLE}$ . Actually, p$_{MLE}$ comprises K − 1 independent parameters, and in this case likelihood ra- tios between e − K and e K are insignificant. [Typically, L ( p$_{true}$ ) ≈ e − $^{K}$L ( p$_{MLE}$ ), so tighter significance criteria would reject the true state.] If p$_{MLE}$ assigns zero prob- ability to M < K different events, and p$_{H}$ hedges all M of them, then the argument leading to Eq. 11 gives a likelihood ratio of e − $^{Mβ}$, which is not significant. For quantum HMLE, it’s possible to show the same result for the HMLE estimate ˆ ρ$_{H}$ : L ( p$_{H}$ ) ( p$_{MLE}$ ) =$^{(}$1 − β N$^{)}$$^{N}$ ≈ e − $^{β}$. (11) L (ˆ ρ$_{H}$ ) (ˆ ρ$_{MLE}$ ) ≥ e − $^{dβ}$. (12) The proof is a bit long, and can be found in Appendix A of the arxiv.org version. Conclusions: Hedging is a simple, well-motivated so- lution to the zero eigenvalue problem. It is also easy to implement – unlike, for instance, Bayesian techniques (which tend to be an order of magnitude harder to cal- culate). HMLE can be implemented by a near-trivial change to any MLE routine. Because the hedged likeli- hood goes smoothly to zero near the boundary, no ex- plicit positivity constraint is needed. So in fact it may be easier than MLE, as simple gradient-crawling meth- ods should work (though care is necessary for small β , where the boundary roll-off becomes sharper). Because ˆ ρ$_{H}$ is always full-rank, it can safely be used for predictive tasks like gambling and data compression. HMLE works well for qubits, providing improved accu- racy by almost all metrics. Further studies will reveal how well it works for larger systems. Pure states are best estimated using very small values of β , and in gen- eral the optimal value of β is not clear. This contrast with the classical case, where β ≈ 1 2 is known to be asymptoti- cally optimal, suggests that alternative hedging functions (which do not correspond to “add β ”) may work better for quantum estimation. FIG. 2: Methodology : See Fig. 1. MLE and HMLE (with several β values) estimates were calculated, and for each ρ$_{true}$ , Euclidean distance and infidelity were averaged over all datasets. (Note that the trace and Euclidean distances are equivalent for qubits: || ˆ ρ − ρ ||$_{1}$ = √ 2 || ˆ ρ − ρ ||$_{2}$ .) Re- sults : The same three regimes are apparent as in Fig. 1. For highly mixed states, hedging provides a small but consistent improvement. Slightly mixed states see substantial improve- ment from hedging. For nearly-pure states, hedging decreases accuracy regardless of β . However, nearly-pure states are typically estimated with greater accuracy than more mixed states, so the best overall performance is achieved by hedg- ing. β = 0 . 25 . . . 1 seems to be optimal; for β > 1, the error in pure state estimation outweighs benefits for mixed states. if L (ˆ ρ$_{H}$ ) is almost as large as L (ˆ ρ$_{MLE}$ ), then ˆ ρ$_{H}$ is al- most as plausible as ˆ ρ$_{MLE}$ . If they have identical proper- ties, we may as well pick the more plausible one – but if ˆ ρ$_{H}$ is substantially different in some way, then it should be considered on its merits unless ˆ ρ$_{MLE}$ has significantly higher likelihood. We have already seen that the HMLE estimate has substantially different properties, so let’s confirm that it is not significantly less likely. Acknowledgments: This paper owes much to dis- cussions with Alexei Gilchrist, Daniel James, Jordan La- Pointe, and Rob Spekkens. I am supported by the Gov- ernment of Canada through Industry Canada and by the Province of Ontario through the Ministry of Research & Innovation. Consider the classical case. If n$_{k}$ = 0, then p$_{MLE}$ assigns ˆ p$_{k}$ = 0. But it’s equally plausible that p$_{k}$ > 0, since if p$_{k}$ < 1 N then “ k ” probably won’t appear in the first N samples. The likelihood function bears this out: the most likely state sets ˆ p$_{k}$ = 0, but nearby states with nonzero p$_{k}$ have almost the same likelihood. If p$_{H}$ assigns The point of this section is to demonstrate that the hedged maximum likelihood estimate ˆ ρ$_{H}$ is never signifi- Appendix A