Is there anyone prove the results such like the follows?


If $NP\not\subseteq BP(2^{\Omega(n)}),$ then $BPP\subseteq P^{NP}$


In summary, my question it that, can we get some derandomized results based on some nondeterminitic assumptions.



Derandomize on nondeterministic assumptions
As for derandomization under nondeterministic assumptions, you can basically relativize the usual results such as Impagliazzo-Wigderson. Directly, this gives: if some language in $E^{NP}$ requires exponential circuits with an NP-oracle, then $BPP^{NP}=P^{NP}$. There are similar results by [Miltersen and Vinodchandran](http://dx.doi.org/10.1007/s00037-005-0197-7): if some language in $NE\cap coNE$ requires exponential-size nondeterministic circuits, then $AM = NP$.


