Prove that we can change probability in definition of PP class
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:

$x\in L \Rightarrow \Pr\left[\text{M accepts x}\right]&gt; p$

$x\notin L \Rightarrow \Pr\left[\text{M accepts x}\right]&lt; p$

$L\in PP$ 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 $q$.

We shall use the fact that a probabilistic Turing machine with access to unbiased random coins, is able to simulate an $\alpha$-biased coin for any rational (this can be strengthened) $\alpha\in [0,1]$, in $O(1)$ expected time. See lemma 7.14 in Arora and Barak's book for a proof. 

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:

$\Pr\left[\text{$M_q$ accepts x}\right]=\epsilon+(1-\epsilon)\left(\alpha+(1-\alpha)\Pr\left[\text{$M_p$ accepts x}\right]\right)$, so:

$x\in L \Rightarrow \Pr\left[\text{$M_q$ accepts x}\right]&gt;
\epsilon+(1-\epsilon)\left(\alpha+(1-\alpha)p\right)
$

$x\notin L \Rightarrow \Pr\left[\text{$M_q$ accepts x}\right]&lt;
\epsilon+(1-\epsilon)\left(\alpha+(1-\alpha)p\right)
$

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.