# $\mathsf{P}^\mathsf{BPP}$ vs $\mathsf{BPP}$ (Are they known to be equal)

Is it known if $\mathsf{P}^\mathsf{BPP}= \mathsf{BPP}$ ?

It's clear that $\mathsf{BPP} \subseteq \mathsf{P}^\mathsf{BPP}$. Now, since $\mathsf{BPP}$ is closed under complementation, union, and intersection, it seems to provide evidence that $\mathsf{BPP} = \mathsf{P}^\mathsf{BPP}$. However, I can't prove this.

• Is it well defined what it means to have a randomized complexity class (with no complete problems) as an oracle ? – Suresh Venkat Jan 1 '13 at 18:31
• Suresh: $P^{\mathcal C}$ is well defined for any set of languages $\mathcal{C}$. – Kristoffer Arnsfelt Hansen Jan 1 '13 at 18:42
• Can't you trivially simulate each BPP oracle call within your polynomial-time randomized algorithm? You'll just need to improve the error probability of each such simulation to, say, exponentially small so that the total chance of making an error throughout the algorithm stays small enough. – Mahdi Cheraghchi Jan 1 '13 at 20:22
• @Kristoffer would you mind giving the formal definition or confirming if the following is correct? "$P^{BPP}$ is the set of languages decidable in polytime by a deterministic TM with access to a yes/no oracle for some language in $BPP$." – usul Jan 1 '13 at 20:26
• @SureshVenkat: The motivation came from looking at oracle separation results and wondering ... what is the power of a BPP oracle? We know that P^P = P, we also know that P^NP probably != NP, probably != coNP; but for BPP, it's not clear at all to me how powerful it is as an oracle. I'll be sure to add more "motivation" in the future. – user13175 Jan 2 '13 at 17:19