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\title{Threshold parallel repetition for finite-dimensional entangled games}
\begin{document}
\maketitle
\begin{abstract}
Let $G$ be a finite two-player one-round game with entangled value
$v=\omega^*(G)<1$. Suppose two players play $k$ independent copies of
$G$ in parallel and may entangle their answers across all copies. We
prove that, for every $0<\delta<1-v$, the probability that they win at
least a $(v+\delta)$-fraction of the copies is at most
$\exp(-c\,\delta^5 k/(1+\log d))$, where $d=|\mathcal A|\,|\mathcal B|$
and $c>0$ is universal. The bound is uniform over all finite-dimensional
joint strategies, holds at every repetition length $k\ge1$ (including
small $k$, where ceiling effects matter), and allows arbitrary
correlated question distributions with zero entries or disconnected
support. For each fixed question distribution we also prove the
sharper-in-form bound
$\exp(-c'\,\delta^3 k/((1+\kappa_\mu)(1+\log d)))$, in which the
constant $\kappa_\mu$ depends only on the positives of the question
distribution. Either bound amplifies both completeness and soundness of
a one-round two-prover protocol without changing the number of provers,
the number of rounds, or the classical nature of the messages.
\end{abstract}
\tableofcontents
\clearpage
\section{Introduction}\label{sec:intro}
\subsection{The question, in plain words}
A \emph{one-round two-player game} has a referee, two players, and a
scoring rule. The referee draws a pair of questions $(x,y)$ from a known
distribution $\mu$, sends $x$ to the first player (Alice) and $y$ to the
second (Bob), and receives answers $(a,b)$ back. He then decides whether
to accept by applying a predicate $V(x,y,a,b)\in\{0,1\}$. The players
cannot speak during the game, but before it begins they may agree on any
shared strategy --- in particular on a shared quantum state of any finite
dimension, measured locally by each player. The best success probability
of such a strategy is the \emph{entangled value} $v=\omega^*(G)$.
Threshold parallel repetition asks a simple question. The referee now
plays $k$ independent copies of $G$ in parallel: questions
$(x_1,y_1),\ldots,(x_k,y_k)$ are sampled independently from $\mu$, and
each player is handed his entire list. The players may entangle their
$k$ answers globally, so their wins in different coordinates need not
behave like independent events. Can they nevertheless concentrate their
wins: win the game in substantially more than a $v$-fraction of the
coordinates with non-negligible probability?
The answer is no, and the decay is exponential. The main theorem makes
this precise. Three features make it nontrivial:
\begin{enumerate}
\item The wins are correlated across coordinates; the players can
choose a joint finite-dimensional strategy on the whole
$k$-fold game.
\item The bound must be uniform over all finite dimensions and over
all choices of $\mu$, including correlated $\mu$ with zeros or
disconnected support.
\item The bound must hold at every $k\ge1$, not only asymptotically.
\end{enumerate}
\subsection{The formal model}
A \emph{finite two-player one-round game} is a tuple
\[
G = (\mathcal X,\mathcal Y,\mathcal A,\mathcal B,\mu,V)
\]
consisting of:
\begin{itemize}
\item nonempty finite question alphabets $\mathcal X,\mathcal Y$;
\item nonempty finite answer alphabets $\mathcal A,\mathcal B$;
\item a probability distribution $\mu$ on $\mathcal X\times\mathcal Y$;
\item a Boolean predicate $V:\mathcal X\times\mathcal Y\times\mathcal A\times\mathcal B\to\{0,1\}$.
\end{itemize}
A \emph{finite-dimensional entangled strategy} for $G$ consists of a
unit vector $\psi\in\mathcal H_A\otimes\mathcal H_B$ in a
finite-dimensional tensor product of Hilbert spaces, together with:
\begin{itemize}
\item a positive operator-valued measurement (POVM) on $\mathcal H_A$
for each $x\in\mathcal X$, i.e.\ positive operators
$\{M_x^a\}_{a\in\mathcal A}$ with $\sum_a M_x^a = I$;
\item a POVM on $\mathcal H_B$ for each $y\in\mathcal Y$, i.e.\
positive operators $\{N_y^b\}_{b\in\mathcal B}$ with
$\sum_b N_y^b=I$.
\end{itemize}
The strategy wins with probability
\[
p_\psi(M,N)
=\sum_{x,y,a,b}\mu(x,y)\,V(x,y,a,b)\,
\langle\psi|M_x^a\otimes N_y^b|\psi\rangle .
\]
The \emph{entangled value} is
\[
v=\omega^*(G)=\sup_{\psi,M,N} p_\psi(M,N),
\]
where the supremum is over all finite dimensions and all such data. It
need not be attained. Taking $\psi$ to be pure causes no loss: any mixed
finite-dimensional state can be purified by adjoining a finite ancilla.
\subsection{Repetition and the threshold event}
The $k$-fold repetition $G^k$ is the game in which the referee samples
$(X_1,Y_1),\ldots,(X_k,Y_k)$ independently from $\mu$, sends the entire
list $X_1,\ldots,X_k$ to Alice and $Y_1,\ldots,Y_k$ to Bob, and accepts
$(A_1,\ldots,A_k)$ and $(B_1,\ldots,B_k)$ when all $k$ coordinate pairs
satisfy $V$. A general \emph{$k$-coordinate strategy} is any
finite-dimensional entangled strategy for this game; it does not need to
be a tensor product of single-game strategies.
Write
\[
I_i = V(X_i,Y_i,A_i,B_i)\in\{0,1\},\qquad
W_k = I_1+\cdots+I_k,
\]
and set $d=|\mathcal A|\,|\mathcal B|$. All logarithms are natural.
\subsection{Main theorem}
\begin{theorem}[Threshold parallel repetition]\label{thm:main}
There is a universal constant $\kappa_0>0$ with the following property.
Let $G$ be a 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 the $k$-fold repetition of $G$,
\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 all such strategies.
\end{theorem}
Four features of \eqref{eq:main} deserve emphasis:
\begin{enumerate}
\item it is uniform over the dimension of the strategy;
\item $\mu$ may be any distribution on $\mathcal X\times\mathcal Y$,
including correlated $\mu$ with zeros or disconnected support;
\item the rate depends on $|\mathcal A|,|\mathcal B|$ only through
the product $d$; it does not involve $|\mathcal X|$ or
$|\mathcal Y|$;
\item the bound holds for every $k\ge1$, not just asymptotically.
The ceiling in $\lceil(v+\delta)k\rceil$ matters at small $k$
and is handled explicitly.
\end{enumerate}
\subsection{A distribution-dependent cubic bound}
We also prove a bound whose exponent is cubic in $\delta$, at the cost
of a constant that depends on the positives of $\mu$. Let
\[
H = \{(x,y)\in\mathcal X\times\mathcal Y:\mu(x,y)>0\}
\]
be the \emph{support} of $\mu$, and form its \emph{connection graph}:
its vertices are the pairs in $H$, and two vertices are joined when they
share a coordinate. Let $H_1,\ldots,H_J$ be the connected components,
$\lambda_j=\mu(H_j)$, $\mu_j=\mu(\cdot\mid H_j)$ the conditional law on
component $j$, and
\[
\mu_{j,\min}=\min_{z\in H_j}\mu_j(z).
\]
Define
\begin{equation}\label{eq:kappa-mu}
\kappa_\mu = 8\,\max_{1\le j\le J}\frac{|H_j|-1}{\mu_{j,\min}} .
\end{equation}
Singletons contribute $0$. The constant $\kappa_\mu$ is finite because
each component is finite and every positive mass is positive.
\begin{corollary}[Cubic bound for fixed $\mu$]\label{cor:support-threshold}
With the same hypotheses as Theorem~\ref{thm:main} and the constant
$\kappa_\mu$ of \eqref{eq:kappa-mu}, for every $0<\delta<1-v$, every
$k\ge1$, and every finite-dimensional joint strategy,
\begin{equation}\label{eq:support-threshold-rate}
\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).
\end{equation}
The same holds for the supremum over strategies.
\end{corollary}
Theorem~\ref{thm:main} is universal in $\mu$ and pays a fifth power of
$\delta$; Corollary~\ref{cor:support-threshold} is cubic in $\delta$ but
pays the distribution constant $\kappa_\mu$. Both bounds hold for the
same strategy, so one may use whichever rate is larger.
\subsection{Relation to prior work}
For the \emph{classical} value of a game, Raz \cite{raz1998} proved that
the all-wins probability decays exponentially, Holenstein
\cite{holenstein2009} improved the quantitative dependence on the gap
via correlated sampling, and Rao \cite{rao2011} proved concentration
above the classical value --- the threshold conclusion, which is
strictly stronger than an all-wins bound. For \emph{entangled}
strategies, structure-specific results came first: XOR games
\cite{csuu2008}, unique games \cite{krt2010}, projection games
\cite{dsv2015}, free games \cite{jpy2014,cs2015}, and complete-support
distributions \cite{cs2014}. Yuen \cite{yuen2016} proved polynomial
all-wins decay for arbitrary finite entangled games, and
Bavarian--Vidick--Yuen \cite{bvy2017,bvy2022} proved exponential
threshold decay after an anchoring transformation that modifies the
game. Chapter~6 of \cite{tenadvances2026} and Song \cite{song2026}
supply the analytic conditioning machinery (smooth samplers, resolvent
purifications) that we use. The contribution of the present paper is a
quantitative transfer from tested-coordinate conditioning to threshold
decay, together with the explicit universal fifth-power and
distribution-dependent cubic rates above, without any anchoring.
\subsection{Road map}
Section~\ref{sec:marginals} recalls notation and proves an elementary
warm-up bound. Section~\ref{sec:threshold} isolates the probabilistic
heart of the argument: a transfer from a conditioning bound to a
threshold bound. Section~\ref{sec:main-proof} states Song's conditioning
estimate and proves Theorem~\ref{thm:main} by plugging it into the
transfer. Sections~\ref{sec:operators}--\ref{sec:root-rounding} build
the analytic machinery needed for Corollary~\ref{cor:support-threshold}:
resolvent purifications, transport of measurements, exposure estimates,
and support-component rounding. Section~\ref{sec:amplification} derives
the two-sided amplification corollary. A reader who accepts Song's
estimate can read Sections~\ref{sec:intro}--\ref{sec:main-proof} and
\ref{sec:amplification} and stop there.
\section{Warm-up: notation and the mean bound}\label{sec:marginals}
\subsection{Notation}
Symbols introduced so far:
\begin{center}
\begin{tabular}{ll}
$\mathcal X,\mathcal Y$ & question alphabets \\
$\mathcal A,\mathcal B$ & answer alphabets \\
$d = |\mathcal A|\,|\mathcal B|$ & product of answer alphabet sizes \\
$\mu$ & question distribution on $\mathcal X\times\mathcal Y$ \\
$V$ & acceptance predicate, $\{0,1\}$-valued \\
$v=\omega^*(G)$ & entangled value \\
$I_1,\ldots,I_k$ & coordinate win indicators of a joint strategy \\
$W_k=\sum_{i=1}^k I_i$ & total wins \\
$E_C$ & event that every coordinate in $C$ is won \\
$p_C=\Pr(E_C)$ & probability of $E_C$ \\
$\tau(C)$ & conditioning cost $(\log(1/p_C)+|C|\log d)/(k-|C|)$
\end{tabular}
\end{center}
The conditioning cost $\tau(C)$ will be the central bookkeeping device
of Sections~\ref{sec:threshold}--\ref{sec:main-proof}. Its numerator
combines the rarity of $E_C$ with the number of possible tested answer
strings; its denominator is the number of coordinates whose wins are
still ``free'' after testing $C$.
\subsection{The mean bound}
The following direct consequence of the definition of $v$ also covers
the case $v=0$, where the threshold bound becomes trivial.
\begin{lemma}[Marginal success probabilities]\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 joint strategy for the $k$-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 coordinates other than $i$.
On a fresh input $(x,y)$ to one copy of $G$, insert $x$ and $y$ at
coordinate $i$, use the fixed questions elsewhere, run the joint
strategy, and retain only the answers at coordinate $i$. What remains is
a finite-dimensional entangled strategy for $G$; its success probability
is at most $v$ by definition of $\omega^*(G)$. Averaging over the fixed
questions under $\mu^{\otimes(k-1)}$ gives $\mathbb E I_i\le v$. Summing
gives the second claim, and when $v=0$ the nonnegative indicator $I_i$
has expectation zero, hence is zero almost surely.
\end{proof}
\section{From conditioning bounds to threshold bounds}\label{sec:threshold}
This section contains the combinatorial heart of the argument. It
converts a single analytic input --- an upper bound on the average
conditional win rate outside a tested set $C$, in terms of the
conditioning cost $\tau(C)$ --- into the exponential threshold decay of
\eqref{eq:main}. The conversion uses only the joint distribution of the
binary win indicators, never the underlying quantum strategy. All
arithmetic below uses the symbols
\[
q=v+\delta,\quad q_0=v+\delta/2,\quad
\eta=\delta/16,\quad r=\log(q/q_0),\quad R=\log(d/q),
\]
and the target rate
\[
c=\frac{\delta\,T}{16(1+\log d)}.
\]
\subsection{Sampling a tested set from the upper tail}
Fix $0<q_0<q<1$ and suppose the upper tail $\{W_k\ge\lceil qk\rceil\}$
has positive probability $p_H$. We want one fixed set $C$ such that:
(i) $E_C$ has decent probability; (ii) conditioned on $E_C$, the
remaining coordinates still have win rate at least $q_0$ with
overwhelming probability. The following lemma shows that sampling
coordinates with replacement produces such a $C$.
\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
$p_H=\Pr(W_k\ge\lceil qk\rceil)>0$. Put
\[
\ell=\left\lceil \frac{\log(4/(\varepsilon p_H))}{\log(q/q_0)}\right\rceil .
\]
Then there exists $C\subseteq[k]$ with $|C|\le\ell$ such that
\begin{equation}\label{eq:sampled-set-guarantees}
p_C\ge\tfrac12 p_H\, q^\ell,
\qquad \Pr(W_k<q_0 k\mid E_C)\le\varepsilon.
\end{equation}
\end{lemma}
\begin{proof}
Draw $\ell$ indices $i_1,\ldots,i_\ell$ independently and uniformly from
$[k]$, and let $C$ be the set of distinct indices drawn. The draw is
independent of the indicators. For each possible value $C$, set
\[
a_C=\Pr(E_C),\quad b_C=\Pr(E_C,\,W_k<q_0 k),\quad A=p_H\,q^\ell.
\]
Two averaging identities hold over the random draw. First,
\[
\mathbb E_C a_C
=\mathbb E\big[(W_k/k)^\ell\big]
\ge p_H\,q^\ell=A ,
\]
because each drawn coordinate is won with conditional probability
$W_k/k$ given the draw, and the coordinates (with repeats) are
exchangeable. Second,
\[
\mathbb E_C b_C\le q_0^\ell\le\varepsilon A/4 ,
\]
because $W_k<q_0 k$ forces $(W_k/k)^\ell<q_0^\ell$, and $\ell$ was chosen
so that $q_0^\ell\le\varepsilon\,p_H q^\ell/4=\varepsilon A/4$.
Now decompose $\mathbb E_C a_C$ into three contributions: draws with
$a_C<A/2$; draws with $a_C\ge A/2$ and $b_C>\varepsilon a_C$; and all
remaining draws. The first class contributes at most $A/2$ trivially.
The second class contributes at most
\[
\varepsilon^{-1}\sum_C b_C
=\varepsilon^{-1}\,\mathbb E_C b_C
\le A/4 .
\]
Together they contribute strictly less than $\mathbb E_C a_C=A$, so the
remaining draws have positive total $a_C$-weight. Any such draw
satisfies $a_C\ge A/2$ and $b_C\le\varepsilon a_C$, i.e.,
\eqref{eq:sampled-set-guarantees}.
\end{proof}
\subsection{The conditioning-to-threshold transfer}
\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
\frac1{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\!\left[W_k\ge\lceil(v+\delta)k\rceil\right]
\le \exp\!\left(-\frac{\delta\,T}{16\,(1+\log d)}\,k\right).
\end{equation}
The same bound holds for the supremum over strategies.
\end{proposition}
The rest of this section is the proof.
\begin{proof}[Proof of Proposition~\ref{prop:conditioning-to-tail}]
Set
\[
q=v+\delta,\; q_0=v+\delta/2,\; \eta=\delta/16,\;
r=\log(q/q_0),\; R=\log(d/q),\;
c=\frac{\delta T}{16(1+\log d)}.
\]
The constraints $0<q_0<q<1$ and $r,R>0$ are immediate. The elementary
estimates $\log(1+u)\ge u/(1+u)$ for $u\ge0$ and $-q\log q\le1/e$ give
\begin{equation}\label{eq:rate-arithmetic}
\frac{c}{r}\le\frac{T}{8}\le\frac{\eta}{2},
\qquad c\left(1+\frac{R}{r}\right)\le\frac{3T}{16}.
\end{equation}
For the first inequality use $r\ge\delta/(2q)$, so
$c/r\le Tq/(8(1+\log d))\le T/8$; then $T/8\le\eta/2$ is $T\le\delta/4$,
which is our hypothesis. For the second, multiply by $\delta$ and
estimate $\delta(1+R/r)\le\delta+2q\log(d/q)\le3(1+\log d)$.
\paragraph{Large repetition lengths.}
We prove \eqref{eq:generic-threshold-rate} for all $k\ge k_0$, where
$k_0$ is a positive integer chosen below independently of the strategy.
Suppose, toward a contradiction, that
\[
p_H=\Pr(W_k\ge\lceil qk\rceil)\ge e^{-ck}.
\]
Apply Lemma~\ref{lem:tail-sampling} with $\varepsilon=\eta$. Set
\[
t=1+\frac{\log(4/\eta)}{r}, \qquad B=\log 2+R\,t .
\]
Since $p_H\ge e^{-ck}$,
\[
\ell\le1+\frac{\log(4/(\eta p_H))}{r}
\le1+\frac{\log(4/\eta)+ck}{r}=\frac{ck}{r}+t .
\]
Choose $k_0$ large enough that $t/k_0\le\eta/2$ and $B/k_0\le T/16$.
For $k\ge k_0$, \eqref{eq:rate-arithmetic} gives
$|C|\le\ell\le ck/r+t\le\eta k/2+\eta k/2=\eta k$, so $C\subsetneq[k]$
and $k-|C|>0$.
Write $u=|C|/k$, so $u\le\eta<q_0$. On the event $\{W_k<q_0 k\}$ we use
only that $W_k-|C|\ge0$. On its complement in $E_C$, $W_k-|C|\ge
q_0 k-|C|=(q_0-u)k$. Combining these two cases with
$\Pr(W_k<q_0 k\mid E_C)\le\eta$,
\[
\mathbb E[W_k-|C|\mid E_C]\ge(1-\eta)(q_0-u)k .
\]
Dividing by $k-|C|=k(1-u)$,
\[
\frac{\mathbb E[W_k-|C|\mid E_C]}{k-|C|}
\ge(1-\eta)\frac{q_0-u}{1-u} .
\]
The function $f(u)=(q_0-u)/(1-u)$ is decreasing on $[0,1)$ because
$f'(u)=(q_0-1)/(1-u)^2<0$. Since $u\le\eta$,
\[
\frac{q_0-u}{1-u}\ge\frac{q_0-\eta}{1-\eta},
\qquad\text{hence}\qquad
\frac{\mathbb E[W_k-|C|\mid E_C]}{k-|C|}\ge q_0-\eta .
\]
Since $q_0-\eta=v+\delta/2-\delta/16=v+7\delta/16$,
\begin{equation}\label{eq:conditional-high-rate}
\frac{\mathbb E[W_k-|C|\mid E_C]}{k-|C|}\ge v+\frac{7\delta}{16}.
\end{equation}
On the other hand, the first conclusion of the sampling lemma gives
\[
\log(1/p_C)+|C|\log d
\le\log(1/p_H)+\log 2+|C|R
\le ck+\log 2+\ell R .
\]
Using $\ell\le ck/r+t$,
\[
\log(1/p_C)+|C|\log d\le ck(1+R/r)+B .
\]
Divide by $k-|C|\ge(1-\eta)k$ and apply \eqref{eq:rate-arithmetic}:
\[
\frac{\log(1/p_C)+|C|\log d}{k-|C|}
\le\frac{c(1+R/r)+B/k}{1-\eta}
\le\frac{3T/16+T/16}{1-\eta}
\le T .
\]
The hypothesis \eqref{eq:threshold-input} now contradicts
\eqref{eq:conditional-high-rate}. This establishes
\eqref{eq:generic-threshold-rate} for $k\ge k_0$.
\paragraph{All repetition lengths.}
Fix any $k$-coordinate strategy with $k<k_0$ and let
$p_H=\Pr(W_k\ge\lceil qk\rceil)$. Take $N$ independent tensor copies of
the strategy; this is a legal $(Nk)$-coordinate finite-dimensional
strategy. 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 already-proved bound yields
$p_H^N\le\exp(-cNk)$, i.e., $p_H\le\exp(-ck)$. This is exactly
\eqref{eq:generic-threshold-rate} for the original $k$.
\paragraph{Suprema.}
The bound \eqref{eq:generic-threshold-rate} was derived for an
arbitrary finite-dimensional strategy with constants independent of
dimension. Monotonicity of the supremum over strategies preserves the
inequality.
\end{proof}
\section{The universal threshold bound}\label{sec:main-proof}
\subsection{Song's conditioning estimate}
The sole external analytic input in the proof of Theorem~\ref{thm:main}
is the following estimate.
\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 any
$C\subsetneq[k]$ 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}
The estimate is proved in \cite{song2026} using a smooth quantum
sampler together with the analytic conditioning machinery of Chapter~6
of \cite{tenadvances2026}; the proof does not use the
Dinur--Steurer--Vidick sampling theorem. Everything in
Theorem~\ref{thm:main} except the elementary algebra that converts
\eqref{eq:song-conditioning} into \eqref{eq:main} is due to
\cite{song2026}. The rest of this section makes that algebra explicit.
\subsection{Proof of Theorem~\ref{thm:main}}
\begin{proof}[Proof of Theorem~\ref{thm:main}]
Fix $0<\delta<1-v$ and set
\[
T=\left(\frac{\delta}{4B_4}\right)^4 .
\]
Because $B_4\ge1$ and $0<\delta<1$, $T\le\delta/4<1$. Suppose a
conditioning cost satisfies $\tau\le T$. Then
Proposition~\ref{prop:song-conditioning} gives
\[
\frac1{k-|C|}\sum_{i\notin C}\Pr(I_i=1\mid E_C)
\le v+B_4\,T^{1/4}=v+\frac{\delta}{4}.
\]
So hypothesis \eqref{eq:threshold-input} of
Proposition~\ref{prop:conditioning-to-tail} holds at this $T$. Applying
that proposition,
\[
\Pr(W_k\ge\lceil(v+\delta)k\rceil)
\le\exp\!\left(-\frac{\delta T}{16(1+\log d)}\,k\right)
=\exp\!\left(-\frac{\kappa_0\,\delta^5}{1+\log d}\,k\right)
\]
with $\kappa_0=1/(16(4B_4)^4)$. Because the bound is uniform in the
strategy, the same inequality holds for the supremum over strategies.
\end{proof}
The remaining sections prove Corollary~\ref{cor:support-threshold}. That
proof does not use Song's sampler; instead it builds an alternative
conditioning estimate from the exposure machinery of
Sections~\ref{sec:operators}--\ref{sec:exposure}, which yields a
square-root (rather than fourth-root) error and hence a cubic rate in
$\delta$.
\section{A common purification of effects}\label{sec:operators}
Conditioning on answers replaces the shared quantum state by local
positive operators applied to it. To compare the resulting vectors, we
need purifications of those 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
construction below furnishes all purifications in one common
finite-dimensional space. This preserves the answer measurements on the
purified side, not just the norms of the conditioned vectors. The
formulation and its Jensen estimate appear in
\cite[Chapter~6, Definition~4.2, Lemma~4.3, and Appendix~A.4]{tenadvances2026};
the proof is included below, along with the finite-dimensional
compression.
For a finite-dimensional complex Hilbert space $\mathcal H$, an
\emph{effect} is a self-adjoint operator $T$ on $\mathcal H$ 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 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
$B=\sum_j q_j A_j$, then
\begin{equation}\label{eq:resolvent-jensen}
\sum_j q_j(\mathcal F_{A_j}-\mathcal F_B)^*
(\mathcal F_{A_j}-\mathcal F_B)
\preceq \sum_j q_j f(A_j)-f(B).
\end{equation}
Finally, for any finite collection $\mathscr T$ of effects, all maps
$\mathcal F_T$, $T\in\mathscr T$, can be compressed to a single
finite-dimensional subspace of $L^2((0,\infty),ds;\mathcal H)$,
preserving every product $\mathcal F_S^*\mathcal F_T$ for
$S,T\in\mathscr T$.
\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
$\mathcal F_T^*\mathcal F_T=T$.
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 therefore restrict to
$\operatorname{supp}B$, where $B>0$.
The scalar integral
\[
f(t)=\int_0^\infty
\left(\frac{t}{1+s}-\frac{t}{t+s}\right) ds \qquad (t\ge0)
\]
converges (including at $t=0$, where the integrand is zero). Functional
calculus, together with $\sum_j q_j D_j=0$, gives
\begin{align}
\sum_j q_j f(A_j)-f(B)
&=\int_0^\infty s\sum_j q_j
\big( (A_j+sI)^{-1} - (B+sI)^{-1}\big)\,ds \notag\\
&=\int_0^\infty s\sum_j q_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*}
and 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 yields
\[
A_j(A_j+sI)^{-1}-B(B+sI)^{-1}
= s(A_j+sI)^{-1}D_j(B+sI)^{-1}.
\]
Consequently the left-hand side of \eqref{eq:resolvent-jensen} has
integrand
\[
s^2\sum_j q_j(B+sI)^{-1}D_j(A_j+sI)^{-2}D_j(B+sI)^{-1}.
\]
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.
For the finite-dimensional compression, 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
images, hence every stated product. An unused one-dimensional summand
can 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
applications the number of effects is finite, although their number and
the resulting dimension need not have a uniform bound.
\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 $\mathcal A$ is a nonempty finite set and
positive operators $(G_a)_{a\in\mathcal A}$ satisfy $\sum_a G_a=T$.
Then there is a POVM $(\widetilde G_a)_{a\in\mathcal A}$ on $\mathcal K$
with
\begin{equation}\label{eq:measurement-pullback}
F^*\widetilde G_a F = G_a\qquad (a\in\mathcal A).
\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 $V_T=F T^{-1/2}$. 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
\[
\widetilde G_a
=V_T T^{-1/2}G_a T^{-1/2}V_T^*
+\mathbf{1}_{\{a=a_*\}}(I_{\mathcal K}-V_TV_T^*).
\]
These operators are positive and sum to $I_{\mathcal K}$. Since
$F=V_T T^{1/2}$, their pullbacks are $G_a$, as claimed. The same
formula applies when $T=0$: all $G_a$ and $V_T$ vanish, and the POVM
always outputs $a_*$.
\end{proof}
In particular, if $\Psi=(\mathcal F_T\otimes\mathcal F_S)\psi$ and
$\sum_a G_a=T$, $\sum_b H_b=S$, the transported measurements satisfy
\begin{equation}\label{eq:joint-measurement-pullback}
\langle\Psi|\widetilde G_a\otimes\widetilde H_b|\Psi\rangle
=\langle\psi|G_a\otimes H_b|\psi\rangle.
\end{equation}
Thus the common purification preserves the exact answer probabilities,
not just the norms of the conditioned vectors.
\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 $m=k-|C|\ge1$, 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 one untested coordinate. Two approximating
vectors, each determined by one player's question and a transcript, will
have total squared error at most $2p\tau$ on average. These comparisons
let Section~\ref{sec: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 a fixed
ordering of the untested coordinates.
\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. 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 \emph{mode} is an index $i\in[m]$. Delete its question from the flag
and set
\[
r_0=(X_C,Y_C,X_{<i},Y_{>i}),\qquad \nu_i=\operatorname{law}(r_0).
\]
At $\Omega_{i-1}$ the questions $X_{\le i-1}$ and $Y_{\ge i}$ are
exposed; at $\Omega_i$ the questions $X_{\le i}$ and $Y_{>i}$ are
exposed. The transition from $\Omega_{i-1}$ to $\Omega_i$ reveals
$X_i$ and hides $Y_i$, and this is the entitlement to which the entropy
telescope of Lemma~\ref{lem:exposure} applies.
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 we suppress 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 $Y_i=y$, 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 yields fixed finite-dimensional
target spaces $\mathcal K_A,\mathcal K_B$ for the maps $\mathcal F_T$,
including the maps used later for the local approximations.
Let $R_i$ be the finite set of transcripts $\mathbf r=(r_0,\alpha,\gamma)$,
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(\mathbf r)=\nu_i(r_0)\,\theta(\mathbf r),\qquad
\Psi_{xy}^{\mathbf r}
=(\mathcal F_{F_x^\alpha}\otimes\mathcal F_{H_y^\gamma})\psi,\\
p_{xy}^{\mathbf r}=\|\Psi_{xy}^{\mathbf 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 $w_i(\mathbf r)$ 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 $\Psi_{xy}^{\mathbf r}$ that reproduce
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 $\sum_a G_x^{\alpha,a}=F_x^\alpha$ and
$\sum_b K_y^{\gamma,b}=H_y^\gamma$. Lemma~\ref{lem:measurement-transport}
provides POVMs $\{\widetilde M_{\mathbf r,x}^a\}$ on $\mathcal K_A$ and
$\{\widetilde N_{\mathbf r,y}^b\}$ on $\mathcal K_B$ such that
\[
\mathcal F_{F_x^\alpha}^*\widetilde M_{\mathbf r,x}^a
\mathcal F_{F_x^\alpha}=G_x^{\alpha,a},
\qquad
\mathcal F_{H_y^\gamma}^*\widetilde N_{\mathbf r,y}^b
\mathcal F_{H_y^\gamma}=K_y^{\gamma,b}.
\]
If $p_{xy}^{\mathbf r}>0$, write
$\varphi_{xy}^{\mathbf r}=\Psi_{xy}^{\mathbf r}/\sqrt{p_{xy}^{\mathbf r}}$.
Then
\begin{equation}\label{eq:exposure-fine-law}
\langle\varphi_{xy}^{\mathbf r}|
\widetilde M_{\mathbf r,x}^a\otimes\widetilde N_{\mathbf r,y}^b
|\varphi_{xy}^{\mathbf r}\rangle
=\frac{\chi(G_x^{\alpha,a},K_y^{\gamma,b})}{p_{xy}^{\mathbf 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 $X_i=x$. Averaging the joint effect
$M_{\mathbf X}^{\alpha,\,a_i=a}\otimes N_{\mathbf Y}^{\gamma,\,b_i=b}$
therefore gives exactly $G_x^{\alpha,a}\otimes K_y^{\gamma,b}$. Its
coarse-grained probability is $p_{xy}^{\mathbf r}$. When
$\theta(\mathbf r)=1$, 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 $E_C,\mathbf r,X_i=x,Y_i=y$. Values
with $\theta(\mathbf r)p_{xy}^{\mathbf r}=0$ 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(\mathbf r,x,y)
=\frac{w_i(\mathbf r)\,\mu(x,y)\,\|\Psi_{xy}^{\mathbf 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 $(i,\mathbf r,x)$ and Bob's only on
$(i,\mathbf r,y)$. 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 $\Psi_{xy}^{\mathbf r}$ 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:
\[
\Psi_x^{\mathbf r}=(\mathcal F_{F_x^\alpha}\otimes
\mathcal F_{H_x^\gamma})\psi,\qquad
\Psi_y^{\mathbf r}=(\mathcal F_{F_y^\alpha}\otimes
\mathcal F_{H_y^\gamma})\psi.
\]
The first is determined by $(i,\mathbf r,x)$ and the second by
$(i,\mathbf r,y)$. Both lie in the same fixed space
$\mathcal K_A\otimes\mathcal K_B$ as $\Psi_{xy}^{\mathbf r}$. Put
$p_x^{\mathbf r}=\|\Psi_x^{\mathbf r}\|^2$ and
$p_y^{\mathbf r}=\|\Psi_y^{\mathbf r}\|^2$. The identity
$\mathcal F_T^*\mathcal F_T=T$ and linearity of $\chi$ give
\begin{equation}\label{eq:exposure-norms}
\begin{gathered}
p_{xy}^{\mathbf r}=\chi(F_x^\alpha,H_y^\gamma),\qquad
p_x^{\mathbf r}=\sum_y\mu(y\mid x)p_{xy}^{\mathbf r},\qquad
p_y^{\mathbf r}=\sum_x\mu(x\mid y)p_{xy}^{\mathbf r},\\
\mathbb E_{\nu_i\mu}\sum_{\alpha,\gamma}\theta\,p_{xy}^{\mathbf r}
=\mathbb E_{\nu_i\mu}\sum_{\alpha,\gamma}\theta\,p_x^{\mathbf r}
=\mathbb E_{\nu_i\mu}\sum_{\alpha,\gamma}\theta\,p_y^{\mathbf 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}^{\mathbf r}-\Psi_y^{\mathbf r}\|^2,\qquad
e_B(i)=\mathbb E_{\nu_i\mu}\sum_{\alpha,\gamma}\theta
\|\Psi_{xy}^{\mathbf r}-\Psi_x^{\mathbf 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
$(i,\mathbf r,y)$; $e_B$ replaces Bob's effect and compares with
$\Psi_x$, which Alice can describe from $(i,\mathbf r,x)$. 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\Big(\sum_x\mu(x\mid y)f(F_x^\alpha)
-f(F_y^\alpha),H_y^\gamma\Big),
\label{eq:exposure-jensen-A}\\
J_B(i):=Z_{i-1}-Z_i
&=\mathbb E_{\nu_i\mu_X}\sum_{\alpha,\gamma}\theta\,
\chi\Big(F_x^\alpha,
\sum_y\mu(y\mid x)f(H_y^\gamma)-f(H_x^\gamma)\Big),
\label{eq:exposure-jensen-B}
\end{align}
where $\mu_X,\mu_Y$ are the two marginals of $\mu$. 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 $\Psi_{xy}^{\mathbf r}$ and
$\Psi_y^{\mathbf r}$, 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}^m J_A(i)\le p\tau,\qquad
\frac1m\sum_{i=1}^m J_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 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}
\section{A cubic bound for a fixed question distribution}\label{sec: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} motivated this fixed-support
approach; the quantum common-vector replacement by a component root and
the explicit cubic coefficient are proved here.
For a finite question distribution $\mu$, let
\[
H=\{(x,y):\mu(x,y)>0\}.
\]
Join two vertices of $H$ when they share either coordinate. Let $H_j$ be
the connected components, and let $\lambda_j,\mu_j,\mu_{j,\min}$,
$\kappa_\mu$ be as in \eqref{eq:kappa-mu}. This constant depends on the
positive masses of $\mu$, as well as on 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_j v_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 $\mathbf r=(r_0,\alpha,\gamma)$ put
\[
w_{i,\mathbf r}=\frac1m w_i(\mathbf r)=\frac1m\nu_i(r_0)\theta(\mathbf r),
\qquad
\Psi_{i,z}^{\mathbf r}=\Psi_{xy}^{\mathbf r}\quad(z=(x,y)).
\]
These finite nonnegative weights do not depend on the fresh pair $z$.
They need not sum to one. Let $\Psi_{i,x}^{\mathbf r},\Psi_{i,y}^{\mathbf r}$
be the local vectors from the exposure lemma, and let
\[
W_{i,\mathbf r,z}=\sum_{a,b}V(x,y,a,b)\,
\widetilde M_{\mathbf r,x}^a\otimes
\widetilde N_{\mathbf r,y}^b .
\]
Here the measurements also have their mode-$i$ dependence, and
$0\le W_{i,\mathbf r,z}\le I$. The norm, error, and win identities give
\begin{align}
\sum_{i,\mathbf r,z}w_{i,\mathbf r}\mu(z)\|\Psi_{i,z}^{\mathbf r}\|^2
&=p, \label{eq:root-input-mass}\\
\sum_{i,\mathbf r,z}w_{i,\mathbf r}\mu(z)c_{i,\mathbf r,z}
&\le2p\tau, \label{eq:root-input-error}\\
\sum_{i,\mathbf r,z}w_{i,\mathbf r}\mu(z)
\langle\Psi_{i,z}^{\mathbf r},
W_{i,\mathbf r,z}\Psi_{i,z}^{\mathbf 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
\[
c_{i,\mathbf r,z}=\|\Psi_{i,z}^{\mathbf r}-\Psi_{i,x}^{\mathbf r}\|^2
+\|\Psi_{i,z}^{\mathbf r}-\Psi_{i,y}^{\mathbf r}\|^2.
\]
Choose a root $\rho_j\in H_j$. We first bound the error of replacing
$\Psi_{i,z}^{\mathbf r}$ by $\Psi_{i,\rho_j}^{\mathbf r}$ on its entire
component. For fixed $i,\mathbf r$, 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
\[
E_j=\sum_{i,\mathbf r,z\in H_j}w_{i,\mathbf r}\mu_j(z)
\|\Psi_{i,z}^{\mathbf r}-\Psi_{i,\rho_j}^{\mathbf r}\|^2,
\qquad
E=\sum_j\lambda_jE_j.
\]
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,\mathbf r,z\in H_j}w_{i,\mathbf r}\mu_j(z)
\|\Psi_{i,z}^{\mathbf r}\|^2,\\
\widehat P_j&=\sum_{i,\mathbf r}w_{i,\mathbf r}
\|\Psi_{i,\rho_j}^{\mathbf r}\|^2,\\
T_j&=\sum_{i,\mathbf r,z\in H_j}w_{i,\mathbf r}\mu_j(z)
\langle\Psi_{i,z}^{\mathbf r},
W_{i,\mathbf r,z}\Psi_{i,z}^{\mathbf r}\rangle,\\
\widehat T_j&=\sum_{i,\mathbf r,z\in H_j}w_{i,\mathbf r}\mu_j(z)
\langle\Psi_{i,\rho_j}^{\mathbf r},
W_{i,\mathbf r,z}\Psi_{i,\rho_j}^{\mathbf r}\rangle.
\end{align*}
If $\widehat P_j>0$, the shared pure state
\[
|\Xi_j\rangle=\widehat P_j^{-1/2}
\sum_{i,\mathbf r}\sqrt{w_{i,\mathbf r}}\,
|i,\mathbf r\rangle_A|i,\mathbf r\rangle_B\otimes
\Psi_{i,\rho_j}^{\mathbf r}
\]
is normalized and finite dimensional. Both players use the original
transported POVMs, controlled by their shared flag $(i,\mathbf r)$.
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 $W=W_{i,\mathbf r,z}$,
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{proof}[Proof of Corollary~\ref{cor:support-threshold}]
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}
(cf.\ \eqref{eq:threshold-input}) gives \eqref{eq:support-threshold-rate}.
\end{proof}
The larger of the exponents in \eqref{eq:main} and
\eqref{eq:support-threshold-rate} is valid. Lemma~\ref{lem:mean-bound}
gives the sharper probability zero when $v=0$.
\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.
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\end{document}