# What do stronger circuit lower bounds give in terms of derandomization?

We have $EXP\not\subseteq P/poly\implies BPP\subseteq io-DTIME(2^{n^\epsilon})$ at every $\epsilon>0$.

This is essentially $DTIME(2^{O(n)})\not\subseteq P/poly\implies BPP\subseteq io-DTIME(2^{n^\epsilon})$ at every $\epsilon>0$.

Is there any consequence of a stronger derandomization?

Let $a(n)$ be any time constructible superpolynomial.

1. What does $NTIME(a(n))\not\subseteq P/poly$ give?

2. What does $DTIME(a(n))\not\subseteq P/poly$ give?

Do these give $BPP\subseteq io-DTIME(poly(n))$ or some thing weaker?

It is known that if $E = DTIME(2^{O(n)})$ is not contained in $SIZE(2^{\varepsilon \cdot n})$ for some $\varepsilon>0$ then $BPP = P$ (https://dl.acm.org/citation.cfm?id=258590).
(Actually, a slightly stronger assumption is needed, namely, the separation between $E$ and $SIZE(2^{\varepsilon \cdot n})$ should hold for almost all input lengths - thanks to Ryan O'Donnel for pointing it out.)
• just for the record $P/poly$ is much smaller than $SIZE(2^{\epsilon\cdot n})$. right? – Brout Dec 7 '17 at 23:10