change probability in definition of PP

Lets say a separation using \(p\in \mathbb {Q}\) exists for \(L\subseteq \Sigma ^*\) if there exists a polynomial probabilistic Turing machine \(M\) such that:

\(x\in L \Rightarrow \Pr \left[\text{M accepts x}\right]{\gt} p \qquad \) \(x\notin L \Rightarrow \Pr \left[\text{M accepts x}\right]{\lt} p\)

\(L\in PP\) iff there exists a separation for \(L\) using \(\frac{1}{2}\) (we can use a strong inequality in both cases). We proceed to show that \(\forall p,q\in (0,1)\cap \mathbb {Q}\), if there exists a separation for \(L\) using \(p\), then there exists a separation using \(q\). Now, given a machine \(M_p\) with separation \(p\) for \(L\), lets consider a machine \(M_q\), which starts by tossing an \(\alpha \)-biased coin. If the outcome is \(1\) then it accepts, otherwise it simulates \(M_p\) on the input and returns it’s answer.

Since we want \(M_q\) to be polynomial in the worst case, we restrict the simulation of the \(\alpha \)-biased coin to a certain number of iterations, such that the simulation halts with probability \(1-\epsilon \) (this can be done for any \(\epsilon {\gt}0\), simply use the fact that the simulation runs in expected constant time, and apply Markov’s inequality). If the simulation didn’t halt, \(M_q\) accepts. In this case we have:

\(\Pr \left[\)\(\)M_q\( accepts x\)=ε+(1-ε)α+(1-α)\(\)M_p\( accepts x\)\(\), so:

\(x\in L \Rightarrow \Pr \left[\)\(\)M_q\( accepts x\)> ε+(1-ε)α+(1-α)p \(\)

\(x\notin L \Rightarrow \Pr \left[\)\(\)M_q\( accepts x\)< ε+(1-ε)α+(1-α)p \(\) This means it is enough to require \(q=\epsilon +(1-\epsilon )\left(\alpha +(1-\alpha )p\right)\). Equivalently, \(\alpha =\frac{q-p-\epsilon (1-p)}{1-\epsilon }\).

Since \(p,q\in (0,1)\cap \mathbb {Q}\), if \(q{\gt}p\) we can find a small rational \(\epsilon \) and rational \(\alpha \in (0,1)\) to satisfy this. If \(p{\gt}q\) you can change \(M_q\) to reject when the outcome of the \(\alpha \)-biased coin is 1. When this equality holds, \(M_q\) achieves \(q\) separation for \(L\) and we are done.