MajoritySAT is PP-complete

On a well-formed formula with $n$ declared variables, a probabilistic machine chooses the $n$ assignment bits uniformly and accepts exactly on a satisfying assignment. Its acceptance probability is $\mathrm{N}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}\mathrm{S}\mathrm{A}\mathrm{T}(\phi )\mathrm{/}{2}^n$, so [L1] gives membership in PP; for $n=0$ the unique random string is the empty assignment.

For hardness, normalize a PP machine to $2^p$ paths with $p\ge 1$, $a$ of them accepting. By [L3] obtain a formula $\phi$ with exactly $a$ satisfying assignments. Add uniquely forced variables until its declared count $q$ obeys $q\ge p$, preserving its count. For $u=(u_{q-1},\dots ,u_0)$ let $T(u)=u_{q-1}\vee \cdots \vee u_{p-1}$; exactly $2^q-2^{p-1}$ assignments satisfy $T$. With a fresh selector $z$, put $\theta =(\mathrm{\neg }z\wedge \phi )\vee (z\wedge T)\mathrm{.}$ Then $\mathrm{\#}\theta =a+2^q-2^{p-1}>2^q⟺a>2^{p-1}\mathrm{.}$ The formula and its declared list are produced in polynomial time.

Step 1.2 is a membership-preserving reduction from every PP language to MajoritySAT, and step 1.1 gives membership; hence MajoritySAT is PP-complete.