change probability in definition of PP

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

$x\in L\Rightarrow Pr\left[\text{M accepts x}\right]>p$ $x\notin L\Rightarrow Pr\left[\text{M accepts x}\right]<p$

$L\in PP$ 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 $\mathrm{\forall }p,q\in (0,1)\cap \mathbb{Q}$, 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 $1-ϵ$ (this can be done for any $ϵ>0$, 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:

$\text{Extra left or missing right}$$$M_q$acceptsx$=ε+(1-ε)α+(1-α)$$M_p$acceptsx$$$, so:

$\text{Extra left or missing right}$$$M_q$acceptsx$> ε+(1-ε)α+(1-α)p $$

$\text{Extra left or missing right}$$$M_q$acceptsx$< ε+(1-ε)α+(1-α)p $$ This means it is enough to require $q=ϵ+(1-ϵ)(\alpha +(1-\alpha )p)$. Equivalently, $\alpha =\frac{q-p-ϵ(1-p)}{1-ϵ}$.

Since $p,q\in (0,1)\cap \mathbb{Q}$, if $q>p$ we can find a small rational $ϵ$ and rational $\alpha \in (0,1)$ to satisfy this. If $p>q$ 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.