Complexity Limitations on One-turn Quantum Refereed Games

Abstract

This paper studies complexity theoretic aspects of quantum refereed games, which are abstract games between two competing players that send quantum states to a referee, who performs an efficiently implementable joint measurement on the two states to determine which of the player wins. The complexity class QRG(1) contains those decision problems for which one of the players can always win with high probability on yes-instances and the other player can always win with high probability on no-instances, regardless of the opposing player’s strategy. This class trivially contains QMA ∪co-QMA and is known to be contained in PSPACE. We prove stronger containments on two restricted variants of this class. Specifically, if one of the players is limited to sending a classical (probabilistic) state rather than a quantum state, the resulting complexity class CQRG(1) is contained in ∃⋅PP (the nondeterministic polynomial-time operator applied to PP); while if both players send quantum states but the referee is forced to measure one of the states first, and incorporates the classical outcome of this measurement into a measurement of the second state, the resulting class MQRG(1) is contained in P ⋅PP (the unbounded-error probabilistic polynomial-time operator applied to PP).

Appendix A: Hoeffding’s inequality for dependent random variables with bounded conditional expectation

In the proof of Theorem 17 we used a slight variant of Hoeffding’s inequality, where the assumption of independence is replaced by a bound on conditional expectation. We expect that a bound along these lines has been observed before, but we have not found a suitable reference. (A similar bound is proved in [4] for Bernoulli random variables, but we require the bound to hold more generally for discrete random variables.)

It is, however, straightforward to adapt the most typical proof of Hoeffding’s inequality to obtain this bound, as we now explain. We begin with Hoeffding’s lemma, which is the essential ingredient in the proof, and which we state without proof. (A proof may be found in [7], among many other references).

Lemma 19

Let X be a random variable taking values in [α, β], for real numbers α < β, and assume E(X) ≤ 0. For every λ > 0 it is the case that

$$ \text{E}\bigl(\exp(\lambda X)\bigr) \leq \exp\biggl(\frac{\lambda^{2}}{8(\beta - \alpha)^{2}}\biggr). $$

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Remark 20

The more typical assumption for this lemma is that E(X) = 0, but (as is not surprising) it is true assuming instead that E(X) ≤ 0. This follows immediately from the observation that if E(X) ≤ 0, then

$$ \text{E}(\exp(\lambda X)) \leq \text{E}(\exp(\lambda(X - \text{E}(X)))). $$

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The next lemma provides the inequality in the proof of Hoeffding’s inequality that would ordinarily follow from the assumption of independence. For simplicity we prove this lemma for discrete random variables, which suffices for our needs.

Lemma 21

Let X and Y be discrete random variables taking values in [α, β] for real numbers α < β, and assume that E(Y |X) ≤ 0. For every λ > 0 it is the case that

$$ \text{E}\left( \exp(\lambda (X + Y))\right. \leq \exp\biggl(\frac{\lambda^{2}}{8(\beta - \alpha)^{2}}\biggr) \text{E}(\exp(\lambda X)). $$

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Proof

We may write

$$ \text{E}\left( \exp(\lambda (X + Y))\right. = \sum\limits_{x} \exp(\lambda x) \text{E}(\exp(\lambda Y) | X = x) \text{Pr}(X = x), $$

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where the sum ranges over all possible values of X. By the assumption E(Y |X) ≤ 0, Hoeffding’s lemma implies

$$ \begin{array}{@{}rcl@{}} &&\sum\limits_{x} \exp(\lambda x) \text{E}(\exp(\lambda Y) | X = x) \text{Pr}(X = x) \\ && \leq \exp\biggl(\frac{\lambda^{2}}{8(\beta - \alpha)^{2}}\biggr) \sum\limits_{x} \exp(\lambda x) \text{Pr}(X = x)\\ &&= \exp\biggl(\frac{\lambda^{2}}{8(\beta - \alpha)^{2}}\biggr) \text{E}(\exp(\lambda X)), \end{array} $$

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as required. □

Finally, we state and prove the variant of Hoeffding’s inequality we have used (again for discrete random variables).

Theorem 22

Let X₁,…,X_n be discrete random variables taking values in [0, 1], let γ ∈ [0, 1], and assume that

$$ \text{E}(X_{k} | X_{1},\ldots,X_{k-1}) \leq \gamma $$

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for all k ∈{1,…,n}. For all ε > 0 it is the case that

$$ \text{Pr}\bigl(X_{1}+\cdots+X_{n} \geq (\gamma + \varepsilon)n\bigr) \leq \exp(-2n\varepsilon^{2}). $$

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Proof

For every λ > 0 we have that

$$ \begin{array}{@{}rcl@{}} &&\text{Pr}\bigl(X_{1}+\cdots+X_{n} \geq (\gamma + \varepsilon)n\bigr) \\ &&= \text{Pr}\bigl(\exp\bigl(\lambda (X_{1} + {\cdots} + X_{n} - \gamma n)\bigr) \geq \exp(\lambda\varepsilon n)\bigr) \\ &&\leq \frac{\text{E}\bigl(\exp\bigl(\lambda (X_{1} + {\cdots} + X_{n} - \gamma n)\bigr)\bigr)}{ \exp(\lambda\varepsilon n)} \end{array} $$

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by Markov’s inequality. Applying Lemma 21 iteratively yields

$$ \text{E}\bigl(\exp\bigl(\lambda (X_{1} + {\cdots} + X_{n} - \gamma n)\bigr)\bigr) \leq \exp\biggl(\frac{n\lambda^{2}}{8}\biggr). $$

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Choosing λ = 4ε yields the claimed bound. □