Consequences of $\mathsf{NP}$ containing $\mathsf{BPP}$

Question

Many believe that $\mathsf{BPP} = \mathsf{P} \subseteq \mathsf{NP}$. However we only know that $\mathsf{BPP}$ is in the second level of polynomial hierarchy, i.e. $\mathsf{BPP}\subseteq \Sigma^ \mathsf{P}_2 \cap \Pi^ \mathsf{P}_2$. A step towards showing $\mathsf{BPP} = \mathsf{P}$ is to first bring it down to the first level of the polynomial hierarchy, i.e. $\mathsf{BPP} \subseteq \mathsf{NP}$.

The containment would mean that nondeterminism is at least as powerful as randomness for polynomial time.

It also means that if for a problem we can find the answers using efficient (polynomial time) randomized algorithms then we can verify the answers efficiently (in polynomial time) .

Are there any known interesting consequences for $\mathsf{BPP} \subseteq \mathsf{NP}$?

Are there any reasons to believe that proving $\mathsf{BPP} \subseteq \mathsf{NP}$ is out of reach right now (e.g. barriers or other arguments)?

Answer 1

For one, proving $BPP \subseteq NP$ would easily imply that $NEXP \neq BPP$, which already means that your proof can't relativize.

But let's look at something even weaker: $coRP \subseteq NTIME[2^{n^{o(1)}}]$. If that is true, then polynomial identity testing for arithmetic circuits is in nondeterministic subexponential time. By Impagliazzo-Kabanets'04, such an algorithm implies circuit lower bounds: either the Permanent does not have poly-size arithmetic circuits, or $NEXP \not\subset P/poly$.

I personally don't know why it looks "out of reach" but it does seem hard to prove. Certainly some genuinely new tricks will be needed to prove it.