QMA error reduction
= QMA error reduction
{c}
{title2=$\epsilon\leq(2\sqrt2/3)^r$}
Starting from completeness $2/3$ and soundness $1/3$, run an odd number $r$ of parallel copies and take majority. The <QMA parallel repetition with entangled witnesses> argument proves soundness for arbitrary repeated witnesses. Exponential <Markov's inequality> gives majority error at most $(2\sqrt2/3)^r$. Hence $O(\log(1/\epsilon))$ copies suffice for error $\epsilon$, with polynomial overhead when the target error is inverse-polynomial or exponentially small in input length.