May be this is trivial but I do not know the answer.As far as we know $$\mathsf{BPP}\subseteq\mathsf{\Sigma}_2\cap\mathsf{\Pi}_2$$ holds.As far as we know $$\mathsf{NP}\cup\mathsf{coNP}\subseteq\mathsf{BPP}\subseteq\mathsf{\Sigma}_2\cap\mathsf{\Pi}_2$$ could hold. Am I correct in this?If so is there a problem that is currently known to be in $$\mathsf{BPP}\backslash\mathsf{NP}\cup\mathsf{coNP}=\mathsf{BPP}\cap\overline{\mathsf{NP}}\cap\overline{\mathsf{coNP}}$$ but conjectured to be in $\mathsf{P}$ just because of our belief $$\mathsf{P}=\mathsf{BPP}$$ should be true essential verdict?That is is there a natural problem (not amalgamated ones) with a randomized poly algorithm nevertheless does not have a short yes or no certificate? A problem that comes close is PIT which is not known to be in $\mathsf{NP}$ but is in $\mathsf{coNP}$ which is the basis of the amalgamated problem (non-natural) below by Ricky Demer.
Is there a problem currently known to be outside class $\mathsf{NP}\cup\mathsf{coNP}$ but inside $\mathsf{BPP}$?
cc.complexity theory;complexity classes;big picture;randomness;polynomial hierarchy
In your initial use of it, there was no need for As far as we know - We do know thatpromiseBPP  promiseS2P    promise - (2 2)holds.If  ZPP = NP  thenNP  coNP  =  ZPP  coZPP  =  ZPP  ZPP  =  ZPP    BPP    S2P    2 2, so as far as we knowNP  coNP    BPP    2 2could hold.If  NP = coNP  then    BPP \ (NP  coNP)  =  BPP \ S2P    BPP \ BPP  =  { }  ,in which case there are no problems outside class  NP  coNP  but inside BPP.In particular, no problems are currently known to be outside class  NP  coNP  but inside BPP.I'm also not aware of any languages L known to be in BPP for whichit's also known that if L is in  NP  coNP  then    BPP    NP  coNP  .Is exactly 1 of these 2 polynomials identically zero?  is a problem currently known to beinside BPP but not known to be in  NP  coNP .    More generally, non-monotonecombinations of [problems in RP that aren't known to be in coNP] yield such problems.There is an oracle relative to which BPP is not a subset of PNP,so presumably  BPP  PNP  is not known to hold.If that containment is indeed not known to hold, then there pretty much can't be any knownresult to the effect that All problems in BPP are reducible to amalgamations of NP problems..