Yet Another Proof that $BPP \subseteq PH$

Oded's comments

I agree that this proof that BPP is in PH is simpler and more intuitive than previous ones, but note that it places BPP in AM0 (perfect completeness version of AM) rather than in MA0 (perfect completeness version of MA). (Recall that Lautemann's proof uses shifts sent by the prover, whereas the proof of GZ relies on a construction of randomness extractors.)

Original abstract

We present a new, simplified proof that the complexity class BPP is contained in the Polynomial Hierarchy (PH), using $k$-wise independent hashing as the main tool. We further extend this approach to recover several other previously known inclusions between complexity classes. Our techniques are inspired by the work of Bellare, Goldreich, and Petrank (Information and Computation, 2000).

See ECCC TR26-004.