Prove that $BPP^{BPP} = BPP$.
Obviously $BPP\subseteq BPP^{BPP}$. So we're left with proving $BPP^{BPP} \subseteq BPP$. Let's consider $L\in BPP^{BPP}$. Then, there's a PTM $M$ to decide $L$ for some $BPP(\alpha, \beta)$ which uses an oracle for some language $O\in BPP$. $O$ also has a $PTM\in BPP$, let's denote it $M_o$. So we could create $M'$, a PTM which behaves exactly as $M$, but everytime $M$ uses the oracle, it simulate $M_o$ on the given input for the oracle.
Now, I want to show that $M\in BPP$. By definition, $BPP=BPP(1/3, 1/3)$, so every time $M$ calls $M_o$ it takes a probabilistic risk basing a deciion upon $M_o$ answer.
We've learned in class that $BPP = BPP(1/3, 1,3) = BPP\left(\frac{1}{3^n}, \frac{1}{3^n} \right)$ and I think I shall explain that even though $M_o$ may return false answers, $M_o$ is still in $BPP$.
Can you help me with doing that?
Thanks
EDIT:
Basically, I want to give a lower bound for the probability of $M$, answering correctly. Let's assume that $M$ queries the oracle $n^c$ times for some $n\in\mathbb{N}$. I think the lower bound is $\frac{1}{3^{n^c}}$.
My explanation is that we're assuming the worst case. That is, if the oracle is returning a false answer then we are returning a false answer.
Is that right?
EDIT2
I'm not sure it's $100\%$ correct, since I didn't take into account the probabilistic behavior of $M$ itself.