[Solved]: Concrete understanding of difference between PP and BPP definitions

Problem Detail: 

I am confused about how PP and BPP are defined. Let us assume $\chi$ is the characteristic function for a language $\mathcal{L}$. M be the probabilistic Turing Machine. Are the following definitions correct:
$BPP =\{\mathcal{L} :Pr[\chi(x) \ne M(x)] \geq \frac{1}{2} + \epsilon \quad \forall x \in \mathcal{L},\ \epsilon > 0 \}$
$PP =\{\mathcal{L} :Pr[\chi(x) \ne M(x)] > \frac{1}{2} \}$

If the definition are wrong, please try to make minimal change to make them correct (i.e. do not give other equivalent definition which use counting machine or some modified model). I can not properly distinguish the conditions on probability on both the definitions.

Some concrete examples with clear insight into the subtle points would be very helpful.

Asked By : DurgaDatta

Answered By : adrianN

That looks correct to me. The difference between BPP and PP is that for BPP the probability has to be greater than $1/2$ by a constant, whereas for PP it could be $1/2+ 1/2^n$. So for BPP problems you can do probability amplification with a small number of repetitions, whereas for general PP problems you can't.

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Question Source : http://cs.stackexchange.com/questions/7848