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We know that BPP is described as $\{L\mid \exists \text{ TM }M, \text{ s.t. }\Pr[M(x)=L(x)]\geq2/3\}$. I saw a proof which uses Chernoff bound to prove that any probability larger than $1/2$ can be turned to any probability in $(1/2,1)$. My question is what about probabilities below $1/2$. Do they fall on a different class?

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Let $\mathsf{BPP}_p$ be the class of languages $L$ such that some polytime Turing machine satisfies $\Pr[M(x) = L(x)] \geq p$ for all $x$. For any fixed $p \in (1/2,1)$, we have $\mathsf{BPP}_p = \mathsf{BPP}$. On the other hand:

  • $\mathsf{BPP}_1 = \mathsf{P}$.
  • If $p \leq 1/2$ then $\mathsf{BPP}_p$ consists of all languages. This is because the machine $M$ which flips an unbiased coin satisfies $\Pr[M(x) = L(x)] = 1/2 \geq p$ for all $x$.
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  • $\begingroup$ I did not get your last point. How $\Pr[M(x) = L(x)] = 1/2 \geq p$ for all $x$ applicable for all other language? How every language is coming in that category? $\endgroup$ – roydiptajit Nov 8 '20 at 12:31
  • $\begingroup$ I described a machine $M$ with this property. $\endgroup$ – Yuval Filmus Nov 8 '20 at 14:26
  • $\begingroup$ If $L$ has a deterministic algorithm, then how a probabilistic TM, $M$ decides $L$? $\endgroup$ – roydiptajit Nov 8 '20 at 15:21
  • $\begingroup$ Every deterministic algorithm can be viewed as a probabilistic algorithm which uses no randomness. $\endgroup$ – Yuval Filmus Nov 8 '20 at 15:39
  • $\begingroup$ Ok, I get it. Now Suppose we have $L$, s.t. $\exists M$, $Pr[M(x)=L(x)]=p$, $p\leq 1/2$. Now we can construct $TM$ $M'$, s.t. it repeats $M$. By this we get a probability amplification. Now, if $p>1/2$, after amplifying, does that comes under BPP? So $L\in BPP$? $\endgroup$ – roydiptajit Nov 8 '20 at 16:44

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