# On $BPP$ in $P^{NP}$ and $SETH$

It is believed showing $$BPP$$ in $$P$$ involves good $$PRG$$s and faces lower bound barriers.

1. Does showing $$BPP$$ in $$P^{NP}$$ which would mean $$BPP\neq EXP^{NP}$$ face similar $$PRG$$ and give lower bounds?

2. Does $$BPP$$ in $$P^{NP}$$ and $$SAT$$ is in $$2^{n}/f(n)$$ at a superpolynomial $$f(n)$$ give anything for $$BPP$$ other than $$BPP\neq NEXP$$ ($$NEXP\not\subseteq P/poly$$ would hold)?

3. $$BPP\neq EXP^{NP}$$ and $$BPP\neq NEXP$$ mean $$BPP\neq EXP^{NP}\cup NEXP$$. What does this mean for lower bounds?

4. How far would we be away from $$BPP\neq EXP$$? Does this truly need good $$PRG$$s if $$BPP\neq EXP^{NP}\cup NEXP$$ is achieved without constructing good $$PRG$$s?

• I think that by "derandomizing to P^NP" the OP means showing that BPP is in P^NP. – Huck Bennett Feb 19 at 21:29