# Where is $\mathsf{BPP^{NP}}$ in the polynomial hierarchy?

We know that $$\mathsf{BPP}$$ is in $$\mathsf{\Sigma^P_2\cap \Pi^P_2}$$ by Sipser-Lautemann, as this proof relativizes we can get $$\mathsf{BPP^{NP} \subseteq \Sigma^P_3\cap \Pi^P_3}$$, but are there any result improving on this?

For several related class (e.g. $$\mathsf{ZPP^{NP}}$$) we know we don't have to increase the order of the polynomial hierarchy for containment ($$\mathsf{ZPP^{NP}\subseteq \Sigma^P_2\cap \Pi^P_2}$$). Are there similar results for $$\mathsf{BPP^{NP}}$$?

• I’m pretty sure nothing is known about the inclusion of $\mathrm{BPP^{NP}}$ in other classes other than what you get by relativizing the corresponding results for BPP with an NP-complete oracle. So, e.g., $\mathrm{BPP=ZPP^{promiseRP}}$ yields $\mathrm{BPP^{NP}=ZPP^{promiseRP^{NP}}}$. Apr 12 at 9:20
• Note that the inclusion for ZPP is, similarly, increased by one level: we have $\mathrm{ZPP\subseteq NP\cap coNP}$, which relativizes to $\mathrm{ZPP^{NP}\subseteq NP^{NP}\cap coNP^{NP}=\Sigma^P_2\cap\Pi^P_2}$. Apr 12 at 9:23
• @Marsh Does not seem to be. math.ucdavis.edu/~greg/zoology/diagram.xml Apr 13 at 4:52
• In the OQ: also for AM! (in only the second level of PH) Apr 14 at 21:26