PCP soundness amplification by independent repetition

Define $V^{(k)}$ to use $k$ independent blocks of $r(n)$ random bits, run $V$ once on each block with the original fixed proof, and accept exactly when every run accepts. Concatenating the query lists gives at most $kq(n)$ symbol queries, all determined by the input and full random tape before answers are read; the proof length and alphabet are unchanged, and the verifier remains uniform polynomial time for fixed $k$.

Fix an input $x$ and proof $\pi$, let $p=\mathrm{Pr}[V^{\pi }(x)\text{ accepts}]$, and let $A_i$ be the event that run $i$ accepts. The $k$ coin blocks form the product space in [F2], each $A_i$ depends only on coordinate $i$, and [F3] makes them mutually independent in the sense of [F4]. Therefore $\mathrm{Pr}[V^{(k),\pi }(x)\text{ accepts}]=\mathrm{Pr}[{\bigcap }_{i=1}^k{A}_i]={\prod }_{i=1}^{k}p=p^k$. This remains valid for zero random bits, where each coordinate space is a singleton.

If $x$ is a yes input, [F1] supplies one fixed proof with $p\ge c$, so step 1.2 gives acceptance $p^k\ge c^k$. If $x$ is a no input, every fixed proof has $p\le s$, so every repeated run on that same proof has acceptance $p^k\le s^k$. In particular $c=1$ gives completeness one.

For $0<s<1$, put $u=1-s>0$. For each $k\ge 1$, $s^{-k}=(1+u\mathrm{/}s{)}^k\ge 1+ku\mathrm{/}s\ge 1+ku$, hence $s^k\le (1+ku{)}^{-1}$. Choosing a fixed integer $k\ge ({\tau }^{-1}-1)\mathrm{/}u$ gives $s^k\le \tau$; when $s=0$, step 2.1 already gives soundness zero with $k=1$. Because $s,\tau$ are constants independent of $n$, this $k$ is fixed, so multiplying $r(n)$ and $q(n)$ by $k$ preserves logarithmic randomness and constant query bounds.