64c3814
==== TABLE OF CONTENTS ====
Threshold-parallel-repetition-for-finite-dimensional-entangled-games//main.tex
Threshold-parallel-repetition-for-finite-dimensional-entangled-games//sections/00-introduction.tex
Threshold-parallel-repetition-for-finite-dimensional-entangled-games//sections/00b-marginals.tex
Threshold-parallel-repetition-for-finite-dimensional-entangled-games//sections/01-operators.tex
Threshold-parallel-repetition-for-finite-dimensional-entangled-games//sections/02-exposure.tex
Threshold-parallel-repetition-for-finite-dimensional-entangled-games//sections/04-threshold.tex
Threshold-parallel-repetition-for-finite-dimensional-entangled-games//sections/05-root-support.tex
Threshold-parallel-repetition-for-finite-dimensional-entangled-games//sections/06-consequences.tex
Threshold-parallel-repetition-for-finite-dimensional-entangled-games//references.bib
===========================
==== BEGIN Threshold-parallel-repetition-for-finite-dimensional-entangled-games//main.tex ====
\documentclass[11pt]{article}
\usepackage[T1]{fontenc}
\usepackage{lmodern,microtype}
\usepackage{amsmath,amssymb,amsthm,mathtools,mathrsfs}
\usepackage[margin=1.08in]{geometry}
\usepackage{booktabs,array,needspace}
\usepackage[colorlinks=true,linkcolor=blue,citecolor=blue,urlcolor=blue]{hyperref}
\hypersetup{pdftitle={Threshold parallel repetition for finite-dimensional entangled games},pdfauthor={OpenAI}}
\pdfinfoomitdate=1
\pdftrailerid{}
\pdfsuppressptexinfo=15
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{hypothesis}[theorem]{Hypothesis}
\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}
\setlength{\emergencystretch}{2em}
\widowpenalty=10000
\clubpenalty=10000
\setcounter{tocdepth}{1}
\title{Threshold parallel repetition for finite-dimensional entangled games}
\author{OpenAI}
\date{September 25, 2026}
\begin{document}
\maketitle
\begin{abstract}
For every finite two-player game with entangled value $v<1$, we prove
exponential decay for the probability of winning at least a $v+\delta$
fraction of $k$ independent repetitions, uniformly over all
finite-dimensional joint strategies. The result allows arbitrary correlated
question distributions and holds for every $0<\delta<1-v$ and $k\ge1$.
Its universal rate is proportional to
$\delta^5/(1+\log(|\mathcal A||\mathcal B|))$, where
$\mathcal A,\mathcal B$ are the answer alphabets. For each fixed question
distribution, we also obtain an explicit cubic rate in $\delta$.
\end{abstract}
\tableofcontents
\clearpage
\input{sections/00-introduction}
\input{sections/04-threshold}
\input{sections/01-operators}
\input{sections/02-exposure}
\input{sections/05-root-support}
\input{sections/06-consequences}
\clearpage
\bibliographystyle{plain}
\bibliography{references}
\end{document}
==== END Threshold-parallel-repetition-for-finite-dimensional-entangled-games//main.tex ====
==== BEGIN Threshold-parallel-repetition-for-finite-dimensional-entangled-games//sections/00-introduction.tex ====
\section{Introduction}
A repeated game allows two players to coordinate their answers across
many independently sampled question pairs. Even when their success in a
single game is bounded away from one, the wins in a joint strategy need
not be independent. Threshold parallel repetition asks whether the
probability that the fraction of wins exceeds the single-game value by
any fixed positive amount is exponentially small. Such a bound permits
error amplification even when an intended strategy sometimes loses
individual games.
\subsection{Games and the main theorem}
A finite two-player one-round game consists of nonempty finite question
sets $\mathcal X,\mathcal Y$, nonempty finite answer sets
$\mathcal A,\mathcal B$, a probability distribution $\mu$ on
$\mathcal X\times\mathcal Y$, and a predicate
\[
V:\mathcal X\times\mathcal Y\times\mathcal A\times\mathcal B
\longrightarrow\{0,1\}.
\]
The referee samples $(x,y)$ from $\mu$, sends $x$ to Alice and $y$ to
Bob, and accepts answers $(a,b)$ when $V(x,y,a,b)=1$. The players do
not communicate. A finite-dimensional entangled strategy consists of
a unit vector $\psi\in\mathcal H_A\otimes\mathcal H_B$ and, for each
local question, a positive operator-valued measurement (POVM):
positive operators $M_x^a,N_y^b$ satisfying
$\sum_aM_x^a=I$ and $\sum_bN_y^b=I$. Its success probability is
\[
\sum_{x,y,a,b}\mu(x,y)V(x,y,a,b)
\langle\psi,M_x^a\otimes N_y^b\psi\rangle.
\]
Using pure states causes no loss, since a finite-dimensional mixed state
can be purified by adding a finite local register. The entangled value
$\omega^*(G)$ is the supremum of this expression over all finite
dimensions and all such strategies; the supremum need not be attained.
In $k$ repetitions the referee samples
$(X_1,Y_1),\ldots,(X_k,Y_k)$ independently from $\mu$ and sends each
player the entire local question string. A strategy may use arbitrary
joint POVMs $M_{\mathbf x}^{\mathbf a}$ and
$N_{\mathbf y}^{\mathbf b}$, with answer strings in
$\mathcal A^k$ and $\mathcal B^k$. Write
\[
I_i=V(X_i,Y_i,A_i,B_i),\qquad W_k=\sum_{i=1}^k I_i,
\qquad d=|\mathcal A||\mathcal B|.
\]
All logarithms below are natural.
\begin{theorem}[Threshold parallel repetition]\label{thm:main}
There is a universal constant $\kappa_0>0$ such that the following holds.
Let $G$ be any finite two-player one-round game with Boolean acceptance
predicate and entangled value $v=\omega^*(G)<1$. For every
$0<\delta<1-v$, every integer $k\ge1$, and every finite-dimensional
joint strategy for $k$ repetitions,
\begin{equation}\label{eq:main}
\Pr\!\left[W_k\ge\lceil(v+\delta)k\rceil\right]
\le \exp\!\left(-\frac{\kappa_0\delta^5}
{1+\log d}\,k\right).
\end{equation}
The same inequality holds for the supremum over those strategies.
\end{theorem}
The distribution $\mu$ may be correlated and may have zero entries or
disconnected support. Neither the question alphabet sizes nor the
strategy dimension enter the displayed rate. We obtain this theorem by
combining a transfer from conditioning bounds to threshold bounds with
Song's conditioning estimate~\cite[Lemma~4.4]{song2026}. The transfer is
proved in Section~\ref{sec:threshold}; its tested-set selection uses
only the joint law of the binary win indicators.
A second bound records the geometry of a fixed question distribution.
Form the graph whose vertices
are the positive-probability pairs $H=\{(x,y):\mu(x,y)>0\}$, joining two
pairs when they share either question. For its connected components
$H_j$, let $\mu_j=\mu(\,\cdot\mid H_j)$ and
$\mu_{j,\min}=\min_{z\in H_j}\mu_j(z)$, and set
\[
\kappa_\mu=8\max_j\frac{|H_j|-1}{\mu_{j,\min}}.
\]
In particular, a singleton component contributes zero.
Corollary~\ref{cor:support-threshold} proves, under the same hypotheses and for
every $k\ge1$, the complementary estimate
\[
\Pr\!\left[W_k\ge\lceil(v+\delta)k\rceil\right]
\le\exp\!\left(-\frac{\delta^3}
{16384(1+\kappa_\mu)(1+\log d)}\,k\right).
\]
Both estimates apply to the same strategy, so one may use the larger
exponential rate. The cubic bound is proved in
Section~\ref{app:root-rounding} by constructing one reference state per
support component. Its constant depends on the fixed distribution through
$\kappa_\mu$, whereas the fifth-power constant is universal. Both bounds
hold at every repetition length, including when rounding the threshold
to an integer matters. When $v=0$, Lemma~\ref{lem:mean-bound} gives the
sharper probability zero.
\subsection{Related work}
The classical value of a game is its largest success probability using
only shared randomness and local response functions. Raz proved that,
when this value is less than one, the probability of winning every
coordinate decays exponentially under repetition~\cite{raz1998}.
Holenstein's information-theoretic proof introduced correlated sampling
into this argument and improved its quantitative dependence on the gap
from one~\cite{holenstein2009}. Rao then proved exponential concentration
above the classical value for general games~\cite[Theorem~7]{rao2011}.
This threshold conclusion is stronger than an all-wins bound: the event
that every coordinate is won is only a subset of the event that a
specified fraction is won.
For entangled strategies, early repetition results used structure in
the acceptance predicate or in the question distribution. Cleve,
Slofstra, Unger, and Upadhyay proved perfect repetition for XOR games,
whose answers are bits and whose acceptance depends on their parity:
the all-wins value of a product is the product of the individual
values~\cite{csuu2008}. Kempe, Regev, and Toner proved exponential
repetition for unique games, whose constraints are permutations between
the answer sets~\cite{krt2010}. Dinur, Steurer, and Vidick extended
exponential repetition to projection games, in which one player's
answer determines the other player's required answer, and developed
quantum correlated sampling~\cite{dsv2015}. Jain, Pereszl\'enyi, and Yao
treated free games, where the players' questions are independently
distributed~\cite{jpy2014}; Chailloux and Scarpa subsequently improved
the rate~\cite{cs2015}. These results require restrictions on either
the predicate or the question distribution.
For complete-support question distributions, which may be correlated,
Chailloux and Scarpa also proved exponential all-wins decay with a rate
depending on the distribution~\cite[Corollary~1, corrected version]{cs2014}.
That result does not cover arbitrary zero entries in the question law.
Yuen proved polynomial all-wins decay for arbitrary finite two-player
entangled games~\cite[Theorem~1]{yuen2016}. His proof combines
conditioning with classical and quantum correlated sampling; its
tested-coordinate selection already samples indices with
replacement~\cite[Proposition~9, full version]{yuen2016}.
Bavarian, Vidick, and Yuen obtained exponential amplification by
anchoring, a transformation that adds questions to the
game~\cite{bvy2022}. Their threshold theorem concerns these anchored
games~\cite[Section~5.4, Theorem~5.6]{bvy2017}. Rao's and their
concentration arguments are precedents for extracting a threshold bound
from control after conditioning on tested coordinates. Unlike the
anchoring transformation, the present bounds retain the given question
distribution throughout.
Chapter~6 of \emph{Ten Advances in Mathematics and Theoretical Computer
Science} obtains exponential all-wins repetition for arbitrary finite
entangled games~\cite{tenadvances2026}. Its operator entropy estimate
and conditioning construction are the analytic antecedents of the
argument below. We reproduce that chapter's resolvent purification and
its estimate from Definition~4.2 and Lemma~4.3, and adapt its
Section~4 weighted reveal argument to a fixed ordering of the untested
coordinates.
Song replaces the quantum sampler in that framework by a smooth
construction and improves the all-wins gap dependence to a cubic
power~\cite{song2026}. His generic conditioning lemma supplies our
fifth-power threshold theorem. The cubic all-wins result uses an
additional estimate for near-perfect conditional success; it does not
by itself imply a cubic threshold rate above an arbitrary value.
Our contribution is a quantitative transfer from tested-set
conditioning to threshold decay, together with its universal
fifth-power application and the explicit cubic bound for each fixed
question distribution. The universal application uses Song's sampler
independently of the Dinur--Steurer--Vidick sampling theorem, while
retaining the conditioning machinery of Chapter~6. Since $0<\delta<1$,
the weaker rate
$\kappa_0\delta^{13}/(1+\log d)$ already follows from
Theorem~\ref{thm:main}.
\subsection{The proof in outline}
Fix a repeated strategy. For a tested set $C\subsetneq[k]$, let $E_C$
be the event that all coordinates in $C$ are won, and suppose
$p=\Pr(E_C)>0$. Put $m=k-|C|$ and
\[
\tau=\frac{\log(1/p)+|C|\log d}{m}.
\]
The numerator records the rarity of the conditioning event and the
logarithm of the number of possible tested answer strings. Song's
estimate, stated in Proposition~\ref{prop:song-conditioning}, bounds
the average success on the remaining coordinates by
\begin{equation}\label{eq:outline-conditioning}
\frac1m\sum_{i\notin C}\Pr(I_i=1\mid E_C)
\le v+B_4\tau^{1/4}\qquad(0\le\tau\le1),
\end{equation}
where $B_4$ is universal. Thus a small conditioning budget cannot raise
the average win probability far above $v$.
Section~\ref{sec:threshold} first proves the generic transfer. If the
upper tail has too much probability,
sampling a short list of coordinate indices with replacement identifies
a fixed tested set $C$ whose success probability is large and whose
conditional win rate outside $C$ exceeds $v$ by a fixed fraction of
$\delta$. Its budget is small enough to contradict
\eqref{eq:outline-conditioning}. This binary sampling argument gives
the exponential bound for sufficiently large repetition lengths. Taking
independent tensor copies of any shorter strategy then gives the same
rate for every length. The generic transfer turns a conditioning error
of order $\tau^{1/4}$ into the fifth-power rate in the main theorem.
The fixed-distribution cubic bound uses a different construction of a
single-game strategy. Sections~\ref{sec:operators} and
\ref{sec:exposure} represent the conditional answers by measurements on
vectors in common finite-dimensional spaces. Resolvent purifications
preserve the answer probabilities. To compare the vectors, order the
untested coordinates and expose Alice's questions before a chosen
coordinate and Bob's questions after it. Moving this dividing point
gives two entropy telescopes, with total cost at most
$p(\log(1/p)+|C|\log d)$ on each side. The resulting vector has two
nearby approximations, each determined by one player's question and a
transcript of exposed questions and tested answers.
Section~\ref{app:root-rounding} compares these vectors along paths in
each connected component of the question support. The resulting
strategies for the component games can be combined in one fixed
tensor-product state, since each player identifies the component from
the local question. The loss is $O_\mu(\sqrt\tau+\tau)$, and the same
transfer therefore gives cubic dependence on $\delta$.
Corollary~\ref{cor:error-amplification} applies the threshold theorem
to amplify both completeness and soundness.
==== END Threshold-parallel-repetition-for-finite-dimensional-entangled-games//sections/00-introduction.tex ====
==== BEGIN Threshold-parallel-repetition-for-finite-dimensional-entangled-games//sections/00b-marginals.tex ====
\Needspace{15\baselineskip}
\subsection{Marginal success probabilities}
\label{subsec:marginal-success}
We first record a bound that also settles the zero-value case.
\begin{lemma}\label{lem:mean-bound}
Let $G$ be a finite two-prover game with entangled value $v$, and let
$I_1,\ldots,I_k$ be the coordinate win indicators of any finite-dimensional
strategy for its $k$-fold repetition. Then
\[
\mathbb E I_i\le v\quad(1\le i\le k),
\qquad \mathbb E\sum_{i=1}^k I_i\le kv.
\]
In particular, if $v=0$, then every coordinate is lost almost surely.
\end{lemma}
\begin{proof}
Fix $i$ and fix the question pairs at all other coordinates. On input
$(x,y)$ to one copy of $G$, insert $x$ and $y$ at coordinate $i$, use the
fixed questions elsewhere, run the repeated strategy, and retain only the
answers at $i$. These are local measurements on the original finite shared
state, so their success probability is at most $v$. Averaging over the
fixed question pairs with distribution $\mu^{\otimes(k-1)}$ proves the first
claim. Summing proves the second; when $v=0$, each nonnegative indicator
has expectation zero.
\end{proof}
==== END Threshold-parallel-repetition-for-finite-dimensional-entangled-games//sections/00b-marginals.tex ====
==== BEGIN Threshold-parallel-repetition-for-finite-dimensional-entangled-games//sections/01-operators.tex ====
\section{A common purification of effects}
\label{sec:operators}
Conditioning on answers replaces the shared state by local positive
operators applied to that state. We need purifications of these operators
that are close whenever an operator entropy gap is small. For positive
definite operators, this gap is an average of Petz's operator Bregman
divergence~\cite{petz2007}; related resolvent estimates appear in
Kim~\cite{kim2014}. The following construction provides all purifications
in a common space and will let us retain the original answer measurements.
The purification formulation and its Jensen estimate are given in
\cite[Chapter~6, Definition~4.2, Lemma~4.3, and Appendix~A.4]{tenadvances2026};
we include the proof, including its finite-dimensional reduction.
For a finite-dimensional complex Hilbert space $\mathcal H$, an
\emph{effect} is a self-adjoint operator $T$ with $0\preceq T\preceq I$.
Put $f(t)=t\log t$ for $t>0$ and $f(0)=0$, and define
\[
\mathcal F_T:\mathcal H\longrightarrow
L^2((0,\infty),ds;\mathcal H),\qquad
(\mathcal F_T h)(s)=T(T+sI)^{-1}h.
\]
The inverse in this expression exists for every $s>0$, including when
$T$ has a kernel.
\begin{lemma}[Resolvent purification]
\label{lem:resolvent}
For every effect $T$ on $\mathcal H$,
$\mathcal F_T^*\mathcal F_T=T$. If $A_1,\ldots,A_N$ are effects,
$q_1,\ldots,q_N>0$ sum to $1$, and $B=\sum_jq_jA_j$, then
\begin{equation}
\sum_j q_j(\mathcal F_{A_j}-\mathcal F_B)^*
(\mathcal F_{A_j}-\mathcal F_B)
\preceq \sum_jq_j f(A_j)-f(B).
\label{eq:resolvent-jensen}
\end{equation}
For any finite collection $\mathscr T$ of effects, all maps
$\mathcal F_T$, $T\in\mathscr T$, can instead take values in one
finite-dimensional space, preserving every product
$\mathcal F_S^*\mathcal F_T$, $S,T\in\mathscr T$.
\end{lemma}
\begin{proof}
The scalar identity
\[
\int_0^\infty\frac{t^2}{(t+s)^2}\,ds=t\qquad(t\ge0)
\]
and the spectral theorem give the first assertion.
To prove \eqref{eq:resolvent-jensen}, write $D_j=A_j-B$.
If $h\in\ker B$, positivity and $q_j>0$ imply $A_jh=0$ for
all $j$. This common kernel reduces every operator in question,
and both sides of \eqref{eq:resolvent-jensen} vanish there.
We may therefore restrict to $\operatorname{supp}B$, where $B>0$.
The scalar integral
\[
f(t)=\int_0^\infty
\left(\frac{t}{1+s}-\frac{t}{t+s}\right)ds
\qquad(t\ge0)
\]
is convergent also at $t=0$, with the integrand then zero. Functional
calculus and $\sum_jq_jD_j=0$ give
\begin{align}
\sum_jq_j f(A_j)-f(B)
&=\int_0^\infty s\sum_jq_j
\big((A_j+sI)^{-1}-(B+sI)^{-1}\big)\,ds \notag\\
&=\int_0^\infty s\sum_jq_j(B+sI)^{-1}D_j
(A_j+sI)^{-1}D_j(B+sI)^{-1}\,ds.
\label{eq:resolvent-gap}
\end{align}
For the second equality use the exact identity
\begin{align*}
(A_j+sI)^{-1}
={}&(B+sI)^{-1}-(B+sI)^{-1}D_j(B+sI)^{-1}\\
&+(B+sI)^{-1}D_j(A_j+sI)^{-1}D_j(B+sI)^{-1},
\end{align*}
then average over $j$. The final integrand in
\eqref{eq:resolvent-gap} is bounded near $s=0$, since $B>0$ and
$s(A_j+sI)^{-1}\preceq I$, and is $O(s^{-2})$ as
$s\to\infty$. Thus these are norm-convergent operator integrals.
The resolvent identity also gives
\[
A_j(A_j+sI)^{-1}-B(B+sI)^{-1}
=s(A_j+sI)^{-1}D_j(B+sI)^{-1}.
\]
Consequently the left side of \eqref{eq:resolvent-jensen} has integrand
\[
s^2\sum_jq_j(B+sI)^{-1}D_j
(A_j+sI)^{-2}D_j(B+sI)^{-1}.
\]
Since
$s(A_j+sI)^{-2}\preceq(A_j+sI)^{-1}$, congruence by
$D_j(B+sI)^{-1}$ bounds this integrand by the final integrand in
\eqref{eq:resolvent-gap}. Integration proves the inequality.
Finally, the span
\[
\mathcal K=\operatorname{span}\{\mathcal F_T h:
T\in\mathscr T,\ h\in\mathcal H\}
\]
has dimension at most $|\mathscr T|\dim\mathcal H$.
Viewing every map as taking values in this same subspace preserves all
inner products of their images, hence all the stated products.
An unused one-dimensional summand may be added if a nonzero codomain
is needed.
\end{proof}
The common codomain is essential: compressing different operators
separately would not in general preserve their distances. In our
application the number of effects is finite, although their number and
the resulting dimension need not have a uniform bound.
The transport of answer effects below follows the ideal-state extraction
in \cite[Chapter~6, Lemma~3.2 and Appendix~A.1]{tenadvances2026}.
\begin{lemma}[Transporting answer measurements]
\label{lem:measurement-transport}
Let $\mathcal H,\mathcal K$ be finite-dimensional complex Hilbert spaces,
let $T$ be an effect on $\mathcal H$, and let
$F:\mathcal H\to\mathcal K$ satisfy $F^*F=T$.
Suppose that $\mathcal A$ is a nonempty finite set and that positive
operators $(G_a)_{a\in\mathcal A}$ satisfy $\sum_aG_a=T$.
There is a POVM $(\widetilde G_a)_{a\in\mathcal A}$ on $\mathcal K$
with
\begin{equation}
F^*\widetilde G_aF=G_a\qquad(a\in\mathcal A).
\label{eq:measurement-pullback}
\end{equation}
The POVM depends only on $F$ and the operators $G_a$.
\end{lemma}
\begin{proof}
All inverses below act on the support of $T$ and are zero on its
kernel. Define $V_T=FT^{-1/2}$. Then
$V_T^*V_T=\Pi_T$, the orthogonal projection onto
$\operatorname{supp}T$, so $V_TV_T^*$ is an orthogonal projection
on $\mathcal K$. Each $G_a$ is supported on
$\operatorname{supp}T$, because $0\preceq G_a\preceq T$.
Fix $a_*\in\mathcal A$ and put
\[
\widetilde G_a
=V_TT^{-1/2}G_aT^{-1/2}V_T^*
+\mathbf1_{\{a=a_*\}}(I_{\mathcal K}-V_TV_T^*).
\]
These operators are positive and sum to $I_{\mathcal K}$.
Since $F=V_TT^{1/2}$, their pullbacks are $G_a$, as claimed.
The same formula applies when $T=0$: all $G_a$ and $V_T$ then vanish,
and the POVM always outputs $a_*$.
\end{proof}
In particular, if $\Psi=(\mathcal F_T\otimes\mathcal F_S)\psi$
and $\sum_aG_a=T$, $\sum_bH_b=S$, the transported measurements satisfy
\begin{equation}
\langle\Psi|\widetilde G_a\otimes\widetilde H_b|\Psi\rangle
=\langle\psi|G_a\otimes H_b|\psi\rangle.
\label{eq:joint-measurement-pullback}
\end{equation}
Thus the common purification preserves the exact answer probabilities,
not just the norms of the conditioned vectors.
==== END Threshold-parallel-repetition-for-finite-dimensional-entangled-games//sections/01-operators.tex ====
==== BEGIN Threshold-parallel-repetition-for-finite-dimensional-entangled-games//sections/02-exposure.tex ====
\section{Conditioning on a tested set}\label{sec:exposure}
Fix a finite-dimensional strategy for the $k$-fold game, with state
$\psi$ and POVMs $\{M_{\mathbf x}^{\mathbf a}\}$ and
$\{N_{\mathbf y}^{\mathbf b}\}$. Let $C\subseteq[k]$ be a set of tested
coordinates and let
\[
E_C=\{V(X_j,Y_j,A_j,B_j)=1\text{ for every }j\in C\},
\qquad p=\Pr(E_C)>0.
\]
Assume that $m=k-|C|\ge1$, and put
\begin{equation}\label{eq:conditioning-budget}
d_C=(|\mathcal A||\mathcal B|)^{|C|},\qquad
L=\log(d_C/p),\qquad \tau=L/m.
\end{equation}
We construct a vector and local measurements that reproduce the
conditional answers at an untested coordinate. Two approximating vectors,
one determined by each player's question and a transcript, will have
total squared error at most $2p\tau$ on average. These comparisons let
Section~\ref{app:root-rounding} replace the conditional vector by a fixed
state on each support component. Throughout this
section, conditional expectations of operators use the original,
unconditioned question distribution $\mu^{\otimes k}$. We adapt the
weighted reveal argument of \cite[Chapter~6, Section~4]{tenadvances2026},
using the fixed ordering below.
\subsection{Exposing one question at each coordinate}
Fix any ordering of the untested coordinates, and abbreviate their
questions as $X_1,\ldots,X_m$ and $Y_1,\ldots,Y_m$, with answers
$A_1,\ldots,A_m$ and $B_1,\ldots,B_m$. The tested questions
retain the notation $X_C,Y_C$. Position-indexed win indicators refer to
these same untested coordinates; sums over $i\notin C$ in the result
statements use the original coordinate labels. For $0\le j\le m$
define the flag
\[
\Omega_j=(X_C,Y_C,X_{\le j},Y_{>j}).
\]
Conditioned on a flag of positive probability, the full question vectors
$\mathbf X$ and $\mathbf Y$ are independent. Indeed, at a tested
coordinate both questions are fixed. At each untested coordinate one
question is fixed and the other has its conditional law under $\mu$;
these remaining choices are independent across coordinates. This
argument uses no independence between the two questions within a
coordinate.
For tested answer strings $\alpha\in\mathcal A^C$ and
$\gamma\in\mathcal B^C$, define the coarse-grained effects
\[
M_{\mathbf x}^{\alpha}
=\sum_{\mathbf a:\,a_C=\alpha}M_{\mathbf x}^{\mathbf a},
\qquad
N_{\mathbf y}^{\gamma}
=\sum_{\mathbf b:\,b_C=\gamma}N_{\mathbf y}^{\mathbf b},
\]
and the testing mask
\[
\theta(x_C,y_C,\alpha,\gamma)
=\prod_{j\in C}V(x_j,y_j,\alpha_j,\gamma_j).
\]
For $C=\varnothing$ there is one pair of empty answer strings and the
empty product is $1$. Write
\[
F_j^\alpha=\mathbb E[M_{\mathbf X}^\alpha\mid\Omega_j],\qquad
H_j^\gamma=\mathbb E[N_{\mathbf Y}^\gamma\mid\Omega_j],\qquad
\chi(A,B)=\langle\psi|A\otimes B|\psi\rangle.
\]
The answer strings here are fixed indices of the effects, not additional
conditions on the question distribution. These are positive contractions.
Conditional independence and linearity
give, for every $j$,
\begin{equation}\label{eq:exposure-mass}
\mathbb E_{\Omega_j}\sum_{\alpha,\gamma}
\theta\,\chi(F_j^\alpha,H_j^\gamma)=p.
\end{equation}
A mode is an index $i\in[m]$. Delete its question from the flag and set
\[
r_0=(X_C,Y_C,X_{<i},Y_{>i}),\qquad \nu_i=\operatorname{law}(r_0).
\]
Figure~\ref{fig:exposure} records which questions are retained.
\Needspace{14\baselineskip}
\begin{figure}[!h]
\centering
\begin{tabular}{c@{\qquad}ccc}
\toprule
& Before $i$ & Coordinate $i$ & After $i$\\
\midrule
Alice & $X_{<i}$ revealed & $X_i$ fresh & $X_{>i}$ unrevealed\\
Bob & $Y_{<i}$ unrevealed & $Y_i$ fresh & $Y_{>i}$ revealed\\
\bottomrule
\end{tabular}
\caption{The questions exposed in $r_0$, in addition to $X_C,Y_C$;
$i$ indexes the fixed ordering of the untested coordinates. Under the
original product question law, before conditioning on $E_C$, fixing
$r_0$ and $(X_i,Y_i)$ makes the remaining Alice and Bob question strings
independent. Replacing $Y_i$ by $X_i$ in successive flags produces the
two entropy telescopes.}
\label{fig:exposure}
\end{figure}
The variable $r_0$ is independent of $(X_i,Y_i)\sim\mu$. For each value
of $r_0$ of positive probability, put
\begin{align*}
F_x^\alpha&=\mathbb E[M_{\mathbf X}^\alpha\mid r_0,X_i=x],&
H_y^\gamma&=\mathbb E[N_{\mathbf Y}^\gamma\mid r_0,Y_i=y],\\
F_y^\alpha&=\sum_x\mu(x\mid y)F_x^\alpha,&
H_x^\gamma&=\sum_y\mu(y\mid x)H_y^\gamma.
\end{align*}
We use these formulas only on positive marginal probabilities. In each
conditional ensemble, zero weights are omitted before applying
Lemma~\ref{lem:resolvent}. Questions of zero marginal probability may be
assigned arbitrary complete local POVMs with answers in the original
alphabets. All four effects also depend on $i,r_0$, which are suppressed
in the notation.
The subscripts in $F_y^\alpha,H_x^\gamma$ indicate the question on which
the conditional average depends. In particular,
\[
(F_{i-1}^\alpha,H_{i-1}^\gamma)=(F_y^\alpha,H_y^\gamma),
\qquad
(F_i^\alpha,H_i^\gamma)=(F_x^\alpha,H_x^\gamma).
\]
Here the first identity is evaluated at $Y_i=y$, and the second at
$X_i=x$. The mask depends only on $r_0,\alpha,\gamma$, not on $x,y$.
\subsection{The conditional vector and its measurements}
Apply the common finite-dimensional compression in
Lemma~\ref{lem:resolvent} to all four families
$F_x^\alpha,F_y^\alpha,H_x^\gamma,H_y^\gamma$, over every mode,
flag and positive-marginal question. This gives fixed finite-dimensional
target spaces $\mathcal K_A,\mathcal K_B$ for the maps $\mathcal F_T$,
including the maps used later for the approximations.
Let $R_i$ be the finite set of transcripts $r=(r_0,\alpha,\gamma)$,
where $r_0$ ranges over its positive-probability values and
$(\alpha,\gamma)\in\mathcal A^C\times\mathcal B^C$. Put
\begin{gather*}
w_i(r)=\nu_i(r_0)\theta(r),\qquad
\Psi_{xy}^r=(\mathcal F_{F_x^\alpha}\otimes
\mathcal F_{H_y^\gamma})\psi,\\
p_{xy}^r=\|\Psi_{xy}^r\|^2=\chi(F_x^\alpha,H_y^\gamma).
\end{gather*}
The last identity follows from $\mathcal F_T^*\mathcal F_T=T$.
The weights $w_i(r)$ use counting measure on the answer strings; they
are independent of $(x,y)$ and need not sum to one. Their sum is at
most $d_C$. The vector's squared norm supplies the remaining probability
of the tested answers.
We next exhibit measurements on $\Psi_{xy}^r$ reproducing the
conditional answers. Define the finer effects
\[
G_x^{\alpha,a}
=\mathbb E[M_{\mathbf X}^{\alpha,a_i=a}\mid r_0,X_i=x],\qquad
K_y^{\gamma,b}
=\mathbb E[N_{\mathbf Y}^{\gamma,b_i=b}\mid r_0,Y_i=y],
\]
where an unmentioned answer is summed out. They are positive and satisfy
$\sum_aG_x^{\alpha,a}=F_x^\alpha$ and
$\sum_bK_y^{\gamma,b}=H_y^\gamma$. Lemma~\ref{lem:measurement-transport}
provides POVMs $\{\widetilde M_{r,x}^a\}$ on $\mathcal K_A$ and
$\{\widetilde N_{r,y}^b\}$ on $\mathcal K_B$ such that
\[
\mathcal F_{F_x^\alpha}^*\widetilde M_{r,x}^a
\mathcal F_{F_x^\alpha}=G_x^{\alpha,a},\qquad
\mathcal F_{H_y^\gamma}^*\widetilde N_{r,y}^b
\mathcal F_{H_y^\gamma}=K_y^{\gamma,b}.
\]
If $p_{xy}^r>0$, write
$\varphi_{xy}^r=\Psi_{xy}^r/\sqrt{p_{xy}^r}$. Then
\begin{equation}\label{eq:exposure-fine-law}
\langle\varphi_{xy}^r|
\widetilde M_{r,x}^a\otimes\widetilde N_{r,y}^b
|\varphi_{xy}^r\rangle
=\frac{\chi(G_x^{\alpha,a},K_y^{\gamma,b})}{p_{xy}^r}.
\end{equation}
To identify this probability, condition the original experiment on
$r_0,X_i=x,Y_i=y$. The unfixed Alice questions $X_{>i}$ and Bob
questions $Y_{<i}$ are independent, with conditional laws determined
by $Y_{>i}$ and $X_{<i}$ respectively. More explicitly, for positive-mass
conditioning data their joint law is
\[
\Pr(X_{>i}=\mathbf u,Y_{<i}=\mathbf z\mid r_0,X_i=x,Y_i=y)
=\prod_{j>i}\mu(u_j\mid y_j)
\prod_{j<i}\mu(z_j\mid x_j).
\]
This is a statement about the original question law, before conditioning
on tested answers. Neither product depends on the fresh pair $(x,y)$.
Since Alice's effect depends only on $\mathbf X$, dropping the condition
$Y_i=y$ leaves its average unchanged; the analogous assertion holds for
Bob after dropping $X_i=x$. Hence averaging the joint effect
$M_{\mathbf X}^{\alpha,a_i=a}\otimes
N_{\mathbf Y}^{\gamma,b_i=b}$ gives exactly
$G_x^{\alpha,a}\otimes K_y^{\gamma,b}$. Its coarse-grained probability
is $p_{xy}^r$. When $\theta(r)=1$, conditioning further on the tested
answers $(\alpha,\gamma)$ already implies $E_C$. Thus
\eqref{eq:exposure-fine-law} is the conditional distribution of the
untested answers $(A_i,B_i)$ given $E_C,r,X_i=x,Y_i=y$. Values with
$\theta(r)p_{xy}^r=0$ have no mass in that conditional experiment.
In particular, the joint law of the transcript and fresh questions
conditioned on $E_C$ is
\begin{equation}\label{eq:exposure-law}
\lambda_i(r,x,y)=\frac{w_i(r)\mu(x,y)\|\Psi_{xy}^r\|^2}{p}.
\end{equation}
Thus $\lambda_i$ is a probability law even though the weights $w_i$
alone are not. The measurements in \eqref{eq:exposure-fine-law} are
local: Alice's depends only on $(i,r,x)$ and Bob's only on $(i,r,y)$.
The transcript is an analysis variable, not information given to the
players. A legal one-game strategy must start with a shared resource
independent of the fresh questions and obtain any transcript information
using only that resource and the player's own question.
\subsection{Local descriptions and their approximation error}
The conditional vector $\Psi_{xy}^r$ depends on both fresh questions.
Replace the effect on the opposite player's space by its conditional
average to obtain two vectors with one-question descriptions:
\[
\Psi_x^r=(\mathcal F_{F_x^\alpha}\otimes
\mathcal F_{H_x^\gamma})\psi,\qquad
\Psi_y^r=(\mathcal F_{F_y^\alpha}\otimes
\mathcal F_{H_y^\gamma})\psi.
\]
The first is determined by $(i,r,x)$ and the second by $(i,r,y)$.
Both lie in the same fixed space $\mathcal K_A\otimes\mathcal K_B$
as $\Psi_{xy}^r$. Put $p_x^r=\|\Psi_x^r\|^2$ and
$p_y^r=\|\Psi_y^r\|^2$. The identity
$\mathcal F_T^*\mathcal F_T=T$ and linearity of $\chi$ give
\begin{equation}\label{eq:exposure-norms}
\begin{gathered}
p_{xy}^r=\chi(F_x^\alpha,H_y^\gamma),\qquad
p_x^r=\sum_y\mu(y\mid x)p_{xy}^r,\qquad
p_y^r=\sum_x\mu(x\mid y)p_{xy}^r,\\
\mathbb E_{\nu_i\mu}\sum_{\alpha,\gamma}\theta p_{xy}^r
=\mathbb E_{\nu_i\mu}\sum_{\alpha,\gamma}\theta p_x^r
=\mathbb E_{\nu_i\mu}\sum_{\alpha,\gamma}\theta p_y^r=p.
\end{gathered}
\end{equation}
Here the total mass is $p$ by \eqref{eq:exposure-law}; the two averaging
identities give the same mass for the local descriptions.
Define the errors at mode $i$ by
\begin{equation}\label{eq:mode-error-definitions}
e_A(i)=\mathbb E_{\nu_i\mu}\sum_{\alpha,\gamma}\theta
\|\Psi_{xy}^r-\Psi_y^r\|^2,\qquad
e_B(i)=\mathbb E_{\nu_i\mu}\sum_{\alpha,\gamma}\theta
\|\Psi_{xy}^r-\Psi_x^r\|^2.
\end{equation}
The subscript names the tensor factor that changes: $e_A$ replaces
Alice's effect and compares with $\Psi_y$, which Bob can describe from
$(i,r,y)$; $e_B$ replaces Bob's effect and compares with $\Psi_x$, which
Alice can describe from $(i,r,x)$. We will bound their averages by an
entropy telescope.
\begin{lemma}[Mixed exposure]\label{lem:exposure}
For any finite-dimensional repeated-game strategy, proper tested set
$C$ with $p=\Pr(E_C)>0$, and fixed ordering of the $m=k-|C|$ untested
coordinates, the preceding construction gives finite-dimensional vectors
and local POVMs satisfying the exact conditional answer law
\eqref{eq:exposure-fine-law}, transcript law \eqref{eq:exposure-law},
and norm identities \eqref{eq:exposure-norms}. With
$\tau=(|C|\log(|\mathcal A||\mathcal B|)+\log(1/p))/m$,
\begin{equation}\label{eq:exposure-errors}
\mathbb E_i e_A(i)\le p\tau,\qquad
\mathbb E_i e_B(i)\le p\tau,\qquad
\mathbb E_i\bigl(e_A(i)+e_B(i)\bigr)\le2p\tau,
\end{equation}
where $i$ is uniform on $[m]$ and the errors are defined in
\eqref{eq:mode-error-definitions}. The question distribution may have
zero entries and disconnected support.
\end{lemma}
\begin{proof}
The construction and exact laws have been verified above. To prove the
error bounds, let $f(t)=t\log t$, with $f(0)=0$, as in
Lemma~\ref{lem:resolvent}. Use the flags $\Omega_j$ to define
\[
U_j=\mathbb E_{\Omega_j}\sum_{\alpha,\gamma}
\theta\,\chi(f(F_j^\alpha),H_j^\gamma),\qquad
Z_j=\mathbb E_{\Omega_j}\sum_{\alpha,\gamma}
\theta\,\chi(F_j^\alpha,f(H_j^\gamma)).
\]
We will show that moving from $\Omega_{i-1}$ to $\Omega_i$ increases
$U$ by at least $e_A(i)$ and decreases $Z$ by at least $e_B(i)$.
The common range bound
\begin{equation}\label{eq:exposure-potential-range}
-pL\le U_j,Z_j\le0\qquad(0\le j\le m)
\end{equation}
will then bound each sum of errors by $pL$.
For the lower bound on $U_j$, fix a flag and a pair of tested answer
strings. Set $F=F_j^\alpha$, $H=H_j^\gamma$,
$g=\chi(I,H)\le1$, and $h=\chi(F,H)$. The positive functional
$A\mapsto\chi(A,H)$ has mass $g$. Applying scalar concavity of
$h_0(t)=-t\log t$ to a spectral decomposition of $F$ gives
\[
-\chi(f(F),H)\le g h_0(h/g)
=h\log(g/h)\le h_0(h).
\]
The formula has value $0$ when $g=0$, since then $h=0$. Because
$\theta\in\{0,1\}$ and $h_0(0)=0$, Jensen's inequality over the flag
and the $d_C$ answer pairs, together with
\eqref{eq:exposure-mass}, gives
\[
-U_j\le\mathbb E\sum_{\alpha,\gamma}h_0(\theta h)
\le d_C h_0(p/d_C)=pL.
\]
The proof for $Z_j$ interchanges the two players. The upper bounds in
\eqref{eq:exposure-potential-range} follow from $f(T)\preceq0$ for
positive contractions $T$.
The changes when replacing $Y_i$ by $X_i$ in the flag are
\begin{align}
J_A(i):=U_i-U_{i-1}
&=\mathbb E_{\nu_i\mu_Y}\sum_{\alpha,\gamma}\theta\,
\chi\left(\sum_x\mu(x\mid y)f(F_x^\alpha)
-f(F_y^\alpha),H_y^\gamma\right),
\label{eq:exposure-jensen-A}\\
J_B(i):=Z_{i-1}-Z_i
&=\mathbb E_{\nu_i\mu_X}\sum_{\alpha,\gamma}\theta\,
\chi\left(F_x^\alpha,
\sum_y\mu(y\mid x)f(H_y^\gamma)-f(H_x^\gamma)\right).
\label{eq:exposure-jensen-B}
\end{align}
For example, linearity in the second argument rewrites $U_i$ as
$\mathbb E_{\nu_i\mu}\sum_{\alpha,\gamma}
\theta\chi(f(F_x^\alpha),H_y^\gamma)$; subtracting $U_{i-1}$ gives
\eqref{eq:exposure-jensen-A}. The other identity uses linearity in the
first argument.
For fixed $r_0,y,\alpha,\gamma$, apply the Jensen estimate of
Lemma~\ref{lem:resolvent} to the ensemble $F_x^\alpha$ with probabilities
$\mu(x\mid y)$ and then evaluate its operator inequality against
$H_y^\gamma$ in the state $\psi$. This bounds the corresponding
squared distance between $\Psi_{xy}^r$ and $\Psi_y^r$, and hence
$e_A(i)\le J_A(i)$. Applying the same argument to the ensemble
$H_y^\gamma$ with probabilities $\mu(y\mid x)$ gives
$e_B(i)\le J_B(i)$. In particular, both potential changes are
nonnegative. Telescoping along the fixed ordering now yields
\begin{equation}\label{eq:exposure-telescope}
\frac1m\sum_{i=1}^mJ_A(i)\le p\tau,\qquad
\frac1m\sum_{i=1}^mJ_B(i)\le p\tau,
\end{equation}
by \eqref{eq:exposure-potential-range} and $L=m\tau$.
These inequalities prove \eqref{eq:exposure-errors}. The ordering was
arbitrary and fixed throughout; no averaging over orderings is required.
\end{proof}
\subsection{The change in the sampled question distribution}
The vectors describe the experiment after conditioning on $E_C$.
A one-game strategy, however, receives questions with the original law
$\mu$. The following estimate bounds the resulting discrepancy. We use
$\|\cdot\|_1$ for the full $\ell_1$ distance of probability distributions.
\begin{lemma}[Question drift]\label{lem:question-drift}
Under the preceding assumptions, put
\[
\Delta_i=\|\operatorname{law}(X_i,Y_i\mid E_C)-\mu\|_1.
\]
For a uniformly chosen untested coordinate,
\[
\mathbb E_i\Delta_i
\le\sqrt{\frac{2\log(1/p)}{m}}\le\sqrt{2\tau}.
\]
\end{lemma}
\begin{proof}
Let $Q=((X_j,Y_j))_{j\in[k]}$ be the full question vector and let
$P=\operatorname{law}(Q\mid E_C)$. On the support of $\mu^{\otimes k}$,
\[
\frac{P(Q)}{\mu^{\otimes k}(Q)}
=\frac{\Pr(E_C\mid Q)}p\le\frac1p.
\]
It follows that $D(P\|\mu^{\otimes k})\le\log(1/p)$, where $D$ denotes
relative entropy with natural logarithms. Relative entropy to a product
measure is at least the sum of the relative entropies of its marginals:
this follows by expanding the entropies and using subadditivity.
Retaining only the $m$ untested coordinates, Pinsker's inequality and
Cauchy--Schwarz give
\[
\frac1m\sum_{i=1}^m\Delta_i
\le \frac1m\sum_{i=1}^m
\sqrt{2D(P_i\|\mu)}
\le\sqrt{\frac2m\sum_{i=1}^mD(P_i\|\mu)}
\le\sqrt{\frac{2\log(1/p)}m}.
\]
\end{proof}
==== END Threshold-parallel-repetition-for-finite-dimensional-entangled-games//sections/02-exposure.tex ====
==== BEGIN Threshold-parallel-repetition-for-finite-dimensional-entangled-games//sections/04-threshold.tex ====
\section{From conditioning to a threshold bound}
\label{sec:threshold}
Fix a repeated strategy and write $v=\omega^*(G)$ and
$d=|\mathcal A||\mathcal B|$. For a set $C\subseteq[k]$, put
\[
E_C=\{I_i=1\text{ for every }i\in C\},
\qquad p_C=\Pr(E_C),\qquad W=\sum_{i=1}^k I_i.
\]
For $C\subsetneq[k]$ with $p_C>0$, define its conditioning cost by
\[
\tau(C)=\frac{\log(1/p_C)+|C|\log d}{k-|C|}.
\]
For a target excess $0<\delta<1-v$, the analytic task is to find a budget
$T>0$ such that $\tau(C)\le T$ forces the average win rate outside $C$,
conditioned on $E_C$, to be at most $v+\delta/4$.
We prove that this implication, uniformly over repetition lengths and
strategies, yields exponential rate $\delta T/(16(1+\log d))$ whenever
$T\le\delta/4$. Thus the quantitative role of a conditioning estimate
is entirely determined by the budget it permits.
The proof has two steps. If a large upper tail persists, sampling a short
list of indices selects a tested set of sufficiently large probability
whose conditional win rate is too high. This gives the bound for all
sufficiently large repetition lengths. Independent tensor copies then
extend the same rate to every length. The selection itself uses only the
joint law of the binary win indicators.
The passage from tested-coordinate conditioning to concentration has
precedents in Rao's classical proof \cite[Section~8]{rao2011} and in the
anchored entangled-game theorem \cite[Section~5.4]{bvy2017}. Sampling
tested indices with replacement also appears in Yuen's proof of
Proposition~9 in the full version of~\cite{yuen2016}. The transfer below
separates this probabilistic step from the choice of a conditioning bound.
\input{sections/00b-marginals}
\subsection{Selecting the tested coordinates}
The next lemma identifies one fixed tested set. Its hypotheses impose no
independence on the win indicators.
\begin{lemma}[Sampling the upper tail]\label{lem:tail-sampling}
Let $I_1,\ldots,I_k$ be arbitrary jointly distributed $\{0,1\}$-valued
random variables. Suppose $0<q_0<q<1$, $0<\varepsilon<1$, and
$p_H=\Pr(W\ge\lceil qk\rceil)>0$. Put
\[
\ell=\left\lceil
\frac{\log(4/(\varepsilon p_H))}{\log(q/q_0)}
\right\rceil.
\]
There is a set $C\subseteq[k]$, with $|C|\le\ell$, such that
\begin{equation}\label{eq:sampled-set-guarantees}
p_C\ge\tfrac12p_Hq^\ell,
\qquad \Pr(W<q_0k\mid E_C)\le\varepsilon.
\end{equation}
\end{lemma}
\begin{proof}
Draw $\ell$ indices independently and uniformly from $[k]$, and let $C$ be
their distinct set. The draw is independent of the indicators. For each
draw define
\[
a_C=\Pr(E_C),\qquad b_C=\Pr(E_C,\ W<q_0k),
\qquad A=p_Hq^\ell.
\]
Expectations over the draw satisfy
\[
\mathbb E_C a_C=\mathbb E(W/k)^\ell\ge A,
\qquad
\mathbb E_C b_C\le q_0^\ell\le\varepsilon A/4.
\]
The draws with $a_C<A/2$ contribute at most $A/2$ to
$\mathbb E_C a_C$. Those with $b_C>\varepsilon a_C$ contribute at most
$\varepsilon^{-1}\mathbb E_C b_C\le A/4$. Thus the remaining draws have
positive total $a_C$-weight. Any such draw satisfies
\eqref{eq:sampled-set-guarantees}.
\end{proof}
\subsection{An exponential rate for every repetition length}
The next statement isolates the quantitative use of a conditioning
estimate. All logarithms are natural.
\begin{proposition}[Conditioning-to-threshold transfer]
\label{prop:conditioning-to-tail}
Let $G$ be a finite two-prover game with nonempty answer alphabets,
$d=|\mathcal A||\mathcal B|$, and $v=\omega^*(G)<1$.
Fix $0<\delta<1-v$ and $0<T\le\delta/4$. Suppose the following holds for
every $k\ge1$, every finite-dimensional strategy for $k$ repetitions,
and every $C\subsetneq[k]$
with $p_C>0$:
\begin{equation}\label{eq:threshold-input}
\frac{\log(1/p_C)+|C|\log d}{k-|C|}\le T
\quad\Longrightarrow\quad
\frac{1}{k-|C|}\sum_{i\notin C}\Pr(I_i=1\mid E_C)
\le v+\delta/4.
\end{equation}
Then, for every $k\ge1$ and every finite-dimensional strategy,
\begin{equation}\label{eq:generic-threshold-rate}
\Pr\bigl(W\ge\lceil(v+\delta)k\rceil\bigr)
\le \exp\left(-\frac{\delta T}{16(1+\log d)}\,k\right).
\end{equation}
The same bound holds for the supremum over those strategies.
\end{proposition}
\begin{proof}
We first derive a bound for large repetition lengths and then use tensor
copies to remove that restriction. Set
\[
q=v+\delta,\quad q_0=v+\delta/2,\quad
\eta=\delta/16,\quad r=\log(q/q_0),\quad R=\log(d/q),
\quad c=\frac{\delta T}{16(1+\log d)}.
\]
Here $0<q_0<q<1$, $r>0$, and $R>0$. The elementary inequalities
$r\ge\delta/(2q)$ and $q\log(1/q)\le1/e$ give
\begin{equation}\label{eq:rate-arithmetic}
\frac c r\le\frac{T}{8}\le\frac{\eta}{2},
\qquad
c\left(1+\frac Rr\right)\le\frac{3T}{16}.
\end{equation}
For the second inequality, multiply by $\delta$ and use
\[
\delta\left(1+\frac Rr\right)
\le\delta+2q\log d+2q\log(1/q)
\le3(1+\log d).
\]
\emph{Large repetition lengths.}
We prove \eqref{eq:generic-threshold-rate} for all sufficiently large $k$,
with the lower bound on $k$ independent of the strategy.
Suppose, to the contrary, that
$p_H=\Pr(W\ge\lceil qk\rceil)\ge e^{-ck}$. Apply
Lemma~\ref{lem:tail-sampling} with $\varepsilon=\eta$.
With
\[
t=1+\frac{\log(4/\eta)}r,
\qquad B=\log2+Rt,
\]
the number of draws satisfies $\ell\le ck/r+t$. Choose $k_0$ large
enough that
\[
\frac{t}{k_0}\le\frac\eta2,
\qquad \frac{B}{k_0}\le\frac T{16}.
\]
For $k\ge k_0$, \eqref{eq:rate-arithmetic} implies $|C|\le\ell\le\eta k$,
so $C\subsetneq[k]$. On $E_C$ precisely $|C|$ tested wins are already
known. The second conclusion of the sampling lemma therefore gives
\begin{align}
\frac{\mathbb E[W-|C|\mid E_C]}{k-|C|}
&\ge (1-\eta)\frac{q_0-\eta}{1-\eta}
=v+\frac{7\delta}{16}.
\label{eq:conditional-high-rate}
\end{align}
Indeed, the function $(q_0-u)/(1-u)$ decreases for $0\le u<1$, and
$|C|/k\le\eta<q_0$; outside $\{W\ge q_0k\}$ the remaining number of
wins is still nonnegative.
On the other hand, the first conclusion of the sampling lemma yields
\[
\log(1/p_C)+|C|\log d
\le \log(1/p_H)+\log2+\ell R.
\]
Using \eqref{eq:rate-arithmetic} once more, we obtain
\[
\frac{\log(1/p_C)+|C|\log d}{k-|C|}
\le
\frac{c(1+R/r)+B/k}{1-\eta}
\le\frac{T}{4(1-\eta)}<T.
\]
The hypothesis \eqref{eq:threshold-input} now contradicts
\eqref{eq:conditional-high-rate}. This establishes the claimed
exponential rate for every $k\ge k_0$.
\emph{All repetition lengths.}
To retain the same rate for smaller $k$, fix any $k$-coordinate strategy
and take $N$ independent tensor copies of it. This is a finite-dimensional
strategy for $Nk$ coordinates. The event that every block has at least
$\lceil qk\rceil$ wins has probability $p_H^N$, and it implies the global
threshold event because
\[
N\lceil qk\rceil\ge\lceil qNk\rceil.
\]
Choose $N$ with $Nk\ge k_0$. The bound already proved gives
$p_H^N\le e^{-cNk}$, hence $p_H\le e^{-ck}$. This proves
\eqref{eq:generic-threshold-rate} for every $k$. All constants are
independent of the strategy and its dimension, so taking suprema
preserves the inequality without any attainment assumption.
\end{proof}
\subsection{Proof of the universal threshold bound}
We use the following conditioning estimate as an external theorem.
It supplies a budget uniform in the question distribution and the
strategy dimension.
\begin{proposition}[Song's conditioning estimate,
{\cite[Definition~4.3 and Lemma~4.4]{song2026}}]
\label{prop:song-conditioning}
There is a universal constant $B_4\ge1$ with the following property.
Let $G$ be a finite two-player game with entangled value $v<1$, and fix
a finite-dimensional strategy for $k\ge1$ repetitions. For
$C\subsetneq[k]$ with $p_C=\Pr(E_C)>0$, put $m=k-|C|$ and
\[
\tau=\frac{\log(1/p_C)+|C|\log(|\mathcal A||\mathcal B|)}m.
\]
If $\tau\le1$, then
\begin{equation}\label{eq:song-conditioning}
\frac1m\sum_{i\notin C}\Pr(I_i=1\mid E_C)
\le v+B_4\tau^{1/4}.
\end{equation}
\end{proposition}
Song proves this estimate using a smooth sampler and the analytic
conditioning machinery of Chapter~6 of~\cite{tenadvances2026}; the
proof does not use the Dinur--Steurer--Vidick sampling theorem.
\begin{proof}[Proof of Theorem~\ref{thm:main}]
For $0<\delta<1-v$, take
\[
T=\left(\frac\delta{4B_4}\right)^4.
\]
Since $B_4\ge1$ and $0<\delta<1$, we have $T\le\delta/4<1$.
Whenever the conditioning cost is at most $T$,
Proposition~\ref{prop:song-conditioning} bounds the excess conditional
win rate by $B_4T^{1/4}=\delta/4$.
Set $\kappa_0=1/(16(4B_4)^4)$.
Proposition~\ref{prop:conditioning-to-tail} therefore gives
\begin{equation}\label{eq:universal-threshold-rate}
\Pr\bigl(W\ge\lceil(v+\delta)k\rceil\bigr)
\le\exp\left(-\frac{\kappa_0\delta^5}{1+\log d}\,k\right).
\end{equation}
This holds for every $k\ge1$ and every finite-dimensional strategy,
and hence also for their supremum.
\end{proof}
The next sections develop the common purification and exposure
construction needed for the fixed-distribution cubic bound. That argument
does not use Song's sampler.
==== END Threshold-parallel-repetition-for-finite-dimensional-entangled-games//sections/04-threshold.tex ====
==== BEGIN Threshold-parallel-repetition-for-finite-dimensional-entangled-games//sections/05-root-support.tex ====
\section{A cubic bound for a fixed question distribution}
\label{app:root-rounding}
The exposure estimates also give a direct route to threshold repetition
when constants may depend on the question distribution. Within each
connected component of its support, we replace the conditional vector by
one reference vector. Adjacent question pairs have a common local
approximation, so the cost of this replacement is controlled along paths
in the support. The resulting conditioning error is of square-root order,
which gives a cubic threshold rate. The construction uses the exposure
vectors and transported answer measurements of Section~\ref{sec:exposure}.
The connection-graph viewpoint of Dinur, Harsha, Venkat, and Yuen
\cite{dhvy2017} helped motivate this fixed-support approach. The
quantum common-vector replacement by a component root and the explicit
cubic coefficient are proved here, rather than imported from their
classical parallel-repetition theorem.
For a finite question distribution $\mu$, let
\[
H=\{(x,y):\mu(x,y)>0\}.
\]
Join two vertices of $H$ when they share either question. Let $H_j$ be
the connected components, and define
\begin{equation}\label{eq:root-support-constant}
\lambda_j=\mu(H_j),\qquad
\mu_j=\mu(\,\cdot\mid H_j),\qquad
\mu_{j,\min}=\min_{z\in H_j}\mu_j(z),\qquad
\kappa_\mu=8\max_j\frac{|H_j|-1}{\mu_{j,\min}}.
\end{equation}
This constant depends on the positive masses of $\mu$, as well as its
support. In particular, $\kappa_\mu=0$ when every component is a singleton.
\begin{proposition}[Conditioning through support components]
\label{prop:root-conditioning}
Let $G$ be a finite two-prover game with question distribution $\mu$ and
$v=\omega^*(G)$. For any finite-dimensional strategy for $k$ copies and
$C\subsetneq[k]$, suppose that
\[
E_C=\{I_j=1\text{ for every }j\in C\},\qquad p=\Pr(E_C)>0.
\]
Put $m=k-|C|$ and
$\tau=(|C|\log(|\mathcal A||\mathcal B|)+\log(1/p))/m$.
Then
\begin{equation}\label{eq:root-conditioning}
\frac1m\sum_{i\notin C}\Pr(I_i=1\mid E_C)
\le v+\sqrt{\frac{\log(1/p)}{2m}}
+4\sqrt{\kappa_\mu\tau}+2\kappa_\mu\tau.
\end{equation}
The constants are independent of the strategy dimension and of $k$.
\end{proposition}
\begin{proof}
Write $G_j$ for the game with distribution $\mu_j$ and the original
predicate, and put $v_j=\omega^*(G_j)$. Every positive-probability
question $x$ identifies a unique component, because all support pairs
with first entry $x$ are adjacent. The same holds for each $y$. Thus
\begin{equation}\label{eq:root-component-values}
\sum_j\lambda_jv_j\le v.
\end{equation}
Indeed, tensor together finite-dimensional strategies approaching the
finitely many component suprema, and let each player use the factor
identified by the local question. Their weighted success approaches
the left side. No supremum need be attained.
We must control two changes. Replacing the vectors within a component
will produce a legal strategy for $G_j$. Conditioning on $E_C$ can also
change the frequencies of the components; after constructing those
strategies, we will compare the new component weights with $\lambda_j$.
Use Lemma~\ref{lem:exposure} for the fixed set $C$. In this proof we
restore the mode $i\in[m]$ in the notation for its vectors and
measurements. For $r=(r_0,\alpha,\gamma)$ put
\[
w_{i,r}=\frac1m w_i(r)=\frac1m\nu_i(r_0)\theta(r),\qquad
\Psi_{i,z}^r=\Psi_{xy}^r\quad(z=(x,y)).
\]
These finite nonnegative weights do not depend on the fresh pair $z$.
They need not sum to one. Let $\Psi_{i,x}^r,\Psi_{i,y}^r$ be the local
vectors from the exposure lemma, and let
\[
W_{i,r,z}=\sum_{a,b}V(x,y,a,b)
\widetilde M_{r,x}^a\otimes\widetilde N_{r,y}^b.
\]
Here the measurements also have their mode-$i$ dependence, and
$0\le W_{i,r,z}\le I$. The norm, error and fine-answer identities give
\begin{align}
\sum_{i,r,z}w_{i,r}\mu(z)\|\Psi_{i,z}^r\|^2&=p,
\label{eq:root-input-mass}\\
\sum_{i,r,z}w_{i,r}\mu(z)c_{i,r,z}&\le2p\tau,
\label{eq:root-input-error}\\
\sum_{i,r,z}w_{i,r}\mu(z)
\langle\Psi_{i,z}^r,W_{i,r,z}\Psi_{i,z}^r\rangle
&=p\bar s,\qquad
\bar s=\frac1m\sum_{i\notin C}\Pr(I_i=1\mid E_C),
\label{eq:root-input-win}
\end{align}
where
\[
c_{i,r,z}=\|\Psi_{i,z}^r-\Psi_{i,x}^r\|^2
+\|\Psi_{i,z}^r-\Psi_{i,y}^r\|^2.
\]
Choose a root $\rho_j\in H_j$. We first bound the error of replacing
$\Psi_{i,z}^r$ by $\Psi_{i,\rho_j}^r$ on its entire component. For
fixed $i,r$, abbreviate these vectors by $\Psi_z$ and the costs by
$c_z$. If $z,z'$ share $y$, their common anchor $\Psi_y$ gives
$\|\Psi_z-\Psi_{z'}\|^2\le2(c_z+c_{z'})$; the same assertion follows
from $\Psi_x$ when they share $x$. A simple path to the root has length
at most $|H_j|-1$. Cauchy--Schwarz along that path, and the fact that
each vertex lies on at most two of its edges, give
\[
\|\Psi_z-\Psi_{\rho_j}\|^2
\le4(|H_j|-1)\sum_{a\in H_j}c_a
\le\frac{4(|H_j|-1)}{\mu_{j,\min}}
\sum_{a\in H_j}\mu_j(a)c_a.
\]
For a singleton component the left side is zero. Define
\[
E_j=\sum_{i,r,z\in H_j}w_{i,r}\mu_j(z)
\|\Psi_{i,z}^r-\Psi_{i,\rho_j}^r\|^2,\qquad
E=\sum_j\lambda_jE_j.
\]
Summing the path estimate and applying \eqref{eq:root-input-error}
shows
\begin{equation}\label{eq:root-path-error}
E\le\kappa_\mu p\tau.
\end{equation}
The root vectors now define legal component strategies. We compare their
total squared norms and accepting masses with the corresponding masses
before replacement. Define
\begin{align*}
P_j&=\sum_{i,r,z\in H_j}w_{i,r}\mu_j(z)\|\Psi_{i,z}^r\|^2,\\
\widehat P_j&=\sum_{i,r}w_{i,r}\|\Psi_{i,\rho_j}^r\|^2,\\
T_j&=\sum_{i,r,z\in H_j}w_{i,r}\mu_j(z)
\langle\Psi_{i,z}^r,W_{i,r,z}\Psi_{i,z}^r\rangle,\\
\widehat T_j&=\sum_{i,r,z\in H_j}w_{i,r}\mu_j(z)
\langle\Psi_{i,\rho_j}^r,W_{i,r,z}\Psi_{i,\rho_j}^r\rangle.
\end{align*}
If $\widehat P_j>0$, the shared pure state
\[
|\Xi_j\rangle=\widehat P_j^{-1/2}
\sum_{i,r}\sqrt{w_{i,r}}\,
|i,r\rangle_A|i,r\rangle_B\otimes\Psi_{i,\rho_j}^r
\]
is normalized and finite dimensional. Both players use the original
transported POVMs, controlled by their shared flag $(i,r)$.
The state and the flag distribution are independent of the fresh
questions: the vector arguments are fixed at $\rho_j$, and the weights
exclude the new pair. Each controlled measurement depends only on its
own player's input. The resulting success in $G_j$ is
$\widehat T_j/\widehat P_j$. If $\widehat P_j=0$, then
$\widehat T_j=0$. Therefore, in all cases,
\begin{equation}\label{eq:root-legal-strategy}
\widehat T_j\le v_j\widehat P_j.
\end{equation}
For vectors $a,b$ and an effect $0\le W\le I$,
\[
|\langle a,Wa\rangle-\langle b,Wb\rangle|
\le(\|a\|+\|b\|)\|a-b\|.
\]
Weighted Cauchy--Schwarz, with $W=I$ or $W=W_{i,r,z}$, consequently
gives
\[
|P_j-\widehat P_j|,\ |T_j-\widehat T_j|
\le e_j:=(\sqrt{P_j}+\sqrt{\widehat P_j})\sqrt{E_j}.
\]
Write $\widehat P=\sum_j\lambda_j\widehat P_j$. The triangle
inequality in the weighted direct sum of the vector spaces yields
$\sqrt{\widehat P}\le\sqrt p+\sqrt E$. A further Cauchy--Schwarz
inequality therefore gives
\[
\sum_j\lambda_je_j
\le(\sqrt p+\sqrt{\widehat P})\sqrt E
\le2\sqrt{pE}+E.
\]
By \eqref{eq:root-legal-strategy} and $v_j\le1$,
$T_j\le v_jP_j+2e_j$. Since
$\sum_j\lambda_jP_j=p$ and $\sum_j\lambda_jT_j=p\bar s$, it follows
that
\begin{equation}\label{eq:root-conditional-components}
\bar s\le\sum_j\widetilde\lambda_jv_j
+4\sqrt{E/p}+2E/p,
\qquad\widetilde\lambda_j=\lambda_jP_j/p.
\end{equation}
It remains to compare the conditional component weights with
$\lambda_j$. Let $J_i$ be the support component of the $i$-th untested
question pair. The exact norm identity identifies
$\widetilde\lambda$ with the average of
$\operatorname{law}(J_i\mid E_C)$ over $i\in[m]$. Unconditionally,
$(J_1,\ldots,J_m)$ has law $\lambda^{\otimes m}$. After conditioning
on $E_C$, its density relative to that law is at most $1/p$, so
\[
D\bigl(\operatorname{law}(J_1,\ldots,J_m\mid E_C)
\,\|\,\lambda^{\otimes m}\bigr)\le\log(1/p).
\]
As in Lemma~\ref{lem:question-drift}, product relative entropy bounds
the sum of its marginal relative entropies. Pinsker's inequality,
convexity of total variation and Cauchy--Schwarz imply
\[
\operatorname{TV}(\widetilde\lambda,\lambda)
\le\sqrt{\frac{\log(1/p)}{2m}},
\qquad \operatorname{TV}(a,b)=\tfrac12\|a-b\|_1.
\]
Because $0\le v_j\le1$, this bounds the increase of their weighted
average. Substituting \eqref{eq:root-component-values} and
\eqref{eq:root-path-error} into
\eqref{eq:root-conditional-components} proves
\eqref{eq:root-conditioning}.
\end{proof}
For a fixed $\mu$, \eqref{eq:root-conditioning} is
$O_\mu(\sqrt\tau+\tau)$. Song's estimate in
Proposition~\ref{prop:song-conditioning} has fourth-root error with a
universal constant. Where both apply, one may use the smaller error.
The proof above does not use Song's sampler.
\subsection{The resulting threshold rate}
\begin{corollary}\label{cor:support-threshold}
Let $G$ be a finite two-player game with Boolean acceptance predicate,
question distribution $\mu$, and entangled value $v<1$. Put
$d=|\mathcal A||\mathcal B|$, and define $\kappa_\mu$ by
\eqref{eq:root-support-constant}. For every $0<\delta<1-v$, every
$k\ge1$, and every finite-dimensional joint strategy for $k$ repetitions,
\begin{equation}\label{eq:support-threshold-rate}
\Pr\bigl(W_k\ge\lceil(v+\delta)k\rceil\bigr)
\le\exp\left(-\frac{\delta^3}
{16384(1+\kappa_\mu)(1+\log d)}\,k\right).
\end{equation}
The same bound holds for the supremum over those strategies.
\end{corollary}
\begin{proof}
Proposition~\ref{prop:root-conditioning} bounds the conditioning error by
\[
\sqrt{\tau/2}+4\sqrt{\kappa_\mu\tau}+2\kappa_\mu\tau.
\]
At $T=\delta^2/(1024(1+\kappa_\mu))$, this is at most
\[
\frac\delta{32}+\frac\delta8+\frac\delta{512}
=\frac{81\delta}{512}<\frac\delta4.
\]
Since also $T\le\delta/4$, Proposition~\ref{prop:conditioning-to-tail}
gives \eqref{eq:support-threshold-rate}.
\end{proof}
The larger of the exponents in
\eqref{eq:universal-threshold-rate} and \eqref{eq:support-threshold-rate}
is valid. Lemma~\ref{lem:mean-bound} gives the sharper probability zero
when $v=0$.
==== END Threshold-parallel-repetition-for-finite-dimensional-entangled-games//sections/05-root-support.tex ====
==== BEGIN Threshold-parallel-repetition-for-finite-dimensional-entangled-games//sections/06-consequences.tex ====
\section{Two-sided error amplification}
\label{sec:amplification}
A threshold between completeness and soundness amplifies both guarantees
without changing the number of players or rounds. Tensor copies of a
single-game strategy give completeness; Theorem~\ref{thm:main} controls
soundness even for joint strategies across all repetitions.
\begin{corollary}[Two-sided error amplification]
\label{cor:error-amplification}
Let $G$ be a finite two-player one-round game in the model above, and
fix $0\le s<c\le1$. Put $\Delta=c-s$, $t=(c+s)/2$, and
$d=|\mathcal A||\mathcal B|$. For an integer $k\ge1$, let $H_k$ sample $k$
independent question pairs from $G$, send each player the entire local
question string, and accept exactly when $W_k\ge\lceil tk\rceil$.
Then
\begin{align*}
\omega^*(G)\ge c
&\quad\Longrightarrow\quad
\omega^*(H_k)\ge1-\exp(-\Delta^2k/8),\\
\omega^*(G)\le s
&\quad\Longrightarrow\quad
\omega^*(H_k)\le
\exp\!\left(-\frac{\kappa_0(\Delta/2)^5}{1+\log d}\,k\right),
\end{align*}
where $\kappa_0$ is the universal constant in Theorem~\ref{thm:main}.
\end{corollary}
\begin{proof}
For completeness, the supremum definition supplies a finite-dimensional
strategy with success $p>c-\Delta/4=t+\Delta/4$, whether or not the
value is attained. Its $k$ tensor copies give independent Bernoulli win
indicators of mean $p$. Since $W_k$ is integer valued,
$\{W_k<\lceil tk\rceil\}=\{W_k<tk\}$, so Hoeffding's inequality
\cite[Theorem~2]{hoeffding1963} bounds rejection by
$\exp(-2(p-t)^2k)\le\exp(-\Delta^2k/8)$.
For soundness, set $v=\omega^*(G)\le s$ and $\delta=t-v$.
Then $\delta\ge\Delta/2>0$ and $\delta<1-v$, since $t<1$.
Theorem~\ref{thm:main} applies at exactly the threshold
$\lceil(v+\delta)k\rceil=\lceil tk\rceil$, uniformly over all
finite-dimensional joint strategies; taking their supremum proves the
claim.
\end{proof}
For polynomial-time verifier families with efficiently computable rational
bounds $c,s$, inverse-polynomial gap $c-s$, and polynomial answer
bit-length, polynomially many repetitions therefore make both errors
exponentially small in the input length, while preserving two provers,
one round, and classical messages. Indeed, $1+\log d$ is polynomially
bounded and the ceiling threshold is efficiently computable.
==== END Threshold-parallel-repetition-for-finite-dimensional-entangled-games//sections/06-consequences.tex ====
==== BEGIN Threshold-parallel-repetition-for-finite-dimensional-entangled-games//references.bib ====
@misc{tenadvances2026,
author = {{OpenAI}},
title = {Ten Advances in Mathematics and Theoretical Computer Science},
year = {2026},
note = {Updated August 6, 2026. Chapter 6: Quantum parallel repetition, pp. 154--182. \url{https://cdn.openai.com/pdf/ten-proofs-oai.pdf}}
}
@article{dsv2015,
author = {Irit Dinur and David Steurer and Thomas Vidick},
title = {A parallel repetition theorem for entangled projection games},
journal = {Computational Complexity},
volume = {24},
number = {2},
pages = {201--254},
year = {2015},
doi = {10.1007/s00037-015-0098-3},
note = {Lemma numbering follows arXiv:1310.4113v2, March 2, 2015. \url{https://arxiv.org/abs/1310.4113v2}}
}
@inproceedings{yuen2016,
author = {Henry Yuen},
title = {A Parallel Repetition Theorem for All Entangled Games},
booktitle = {43rd International Colloquium on Automata, Languages, and Programming (ICALP 2016)},
series = {Leibniz International Proceedings in Informatics},
volume = {55},
pages = {77:1--77:13},
publisher = {Schloss Dagstuhl--Leibniz-Zentrum f{\"u}r Informatik},
year = {2016},
doi = {10.4230/LIPIcs.ICALP.2016.77},
note = {\url{https://doi.org/10.4230/LIPIcs.ICALP.2016.77}. Full version: \url{https://arxiv.org/abs/1604.04340v1}}
}
@article{holenstein2009,
author = {Thomas Holenstein},
title = {Parallel Repetition: Simplifications and the No-Signaling Case},
journal = {Theory of Computing},
volume = {5},
number = {8},
pages = {141--172},
year = {2009},
doi = {10.4086/toc.2009.v005a008},
note = {\url{https://theoryofcomputing.org/articles/v005a008/}}
}
@article{bvy2022,
author = {Mohammad Bavarian and Thomas Vidick and Henry Yuen},
title = {Anchored Parallel Repetition for Nonlocal Games},
journal = {SIAM Journal on Computing},
volume = {51},
number = {2},
pages = {214--253},
year = {2022},
doi = {10.1137/21M1405927},
note = {\url{https://doi.org/10.1137/21M1405927}}
}
@techreport{song2026,
author = {Song, Zhao},
title = {Parallel Repetition for Entangled Games with Gap Exponent Three},
institution = {Electronic Colloquium on Computational Complexity},
number = {TR26-199},
year = {2026},
url = {https://eccc.weizmann.ac.il/report/2026/199/},
note = {Manuscript dated August 5, 2026. \url{https://eccc.weizmann.ac.il/report/2026/199/}}
}
@article{rao2011,
author = {Rao, Anup},
title = {Parallel Repetition in Projection Games and a Concentration Bound},
journal = {SIAM Journal on Computing},
volume = {40},
number = {6},
pages = {1871--1891},
year = {2011},
doi = {10.1137/080734042},
url = {https://homes.cs.washington.edu/~anuprao/pubs/parrep.pdf},
note = {\url{https://doi.org/10.1137/080734042}}
}
@inproceedings{bvy2017,
author = {Bavarian, Mohammad and Vidick, Thomas and Yuen, Henry},
title = {Hardness Amplification for Entangled Games via Anchoring},
booktitle = {Proceedings of the 49th Annual ACM SIGACT Symposium on Theory of Computing},
pages = {303--316},
year = {2017},
doi = {10.1145/3055399.3055433},
url = {https://par.nsf.gov/servlets/purl/10026355},
note = {\url{https://doi.org/10.1145/3055399.3055433}}
}
@article{hoeffding1963,
author = {Wassily Hoeffding},
title = {Probability Inequalities for Sums of Bounded Random Variables},
journal = {Journal of the American Statistical Association},
volume = {58},
number = {301},
pages = {13--30},
year = {1963},
doi = {10.1080/01621459.1963.10500830},
note = {\href{https://doi.org/10.1080/01621459.1963.10500830}{Published article}}
}
@article{raz1998,
author = {Ran Raz},
title = {A Parallel Repetition Theorem},
journal = {SIAM Journal on Computing},
volume = {27},
number = {3},
pages = {763--803},
year = {1998},
doi = {10.1137/S0097539795280895},
note = {\url{https://doi.org/10.1137/S0097539795280895}}
}
@article{csuu2008,
author = {Richard Cleve and William Slofstra and Falk Unger and Sarvagya Upadhyay},
title = {Perfect Parallel Repetition Theorem for Quantum {XOR} Proof Systems},
journal = {Computational Complexity},
volume = {17},
number = {2},
pages = {282--299},
year = {2008},
doi = {10.1007/s00037-008-0250-4},
note = {\url{https://doi.org/10.1007/s00037-008-0250-4}}
}
@article{krt2010,
author = {Julia Kempe and Oded Regev and Ben Toner},
title = {Unique Games with Entangled Provers Are Easy},
journal = {SIAM Journal on Computing},
volume = {39},
number = {7},
pages = {3207--3229},
year = {2010},
doi = {10.1137/090772885},
note = {\url{https://doi.org/10.1137/090772885}}
}
@misc{jpy2014,
author = {Rahul Jain and Attila Pereszl{\'e}nyi and Penghui Yao},
title = {A parallel repetition theorem for entangled two-player one-round games under product distributions},
year = {2014},
note = {CCC 2014 camera-ready version, arXiv:1311.6309v2. \url{https://arxiv.org/abs/1311.6309v2}}
}
@misc{cs2015,
author = {Andr{\'e} Chailloux and Giannicola Scarpa},
title = {Parallel Repetition of Free Entangled Games: Simplification and Improvements},
year = {2015},
note = {arXiv:1410.4397v2. \url{https://arxiv.org/abs/1410.4397v2}}
}
@misc{cs2014,
author = {Andr{\'e} Chailloux and Giannicola Scarpa},
title = {Parallel Repetition of Entangled Games with Exponential Decay via the Superposed Information Cost},
year = {2014},
note = {Corrected version, arXiv:1310.7787v3, October 2, 2014. \url{https://arxiv.org/abs/1310.7787v3}}
}
@article{petz2007,
author = {D{\'e}nes Petz},
title = {Bregman divergence as relative operator entropy},
journal = {Acta Mathematica Hungarica},
volume = {116},
pages = {127--131},
year = {2007},
doi = {10.1007/s10474-007-6014-9},
note = {\url{https://doi.org/10.1007/s10474-007-6014-9}}
}
@article{kim2014,
author = {Isaac H. Kim},
title = {Modulus of convexity for operator convex functions},
journal = {Journal of Mathematical Physics},
volume = {55},
number = {8},
pages = {082201},
year = {2014},
doi = {10.1063/1.4890292},
note = {\url{https://doi.org/10.1063/1.4890292}. Preprint: \url{https://arxiv.org/abs/1310.0746v3}}
}
@misc{dhvy2017,
author = {Dinur, Irit and Harsha, Prahladh and Venkat, Rakesh and Yuen, Henry},
title = {Multiplayer parallel repetition for expander games},
year = {2017},
note = {arXiv:1610.08349v2. \url{https://arxiv.org/abs/1610.08349v2}},
url = {https://arxiv.org/abs/1610.08349v2}
}
==== END Threshold-parallel-repetition-for-finite-dimensional-entangled-games//references.bib ====