This is a question $9.1$ from Understanding Machine Learning Chapter 3. It goes like this:
Consider a variant of the PAC model in which there are two example oracles: one that generates positive examples and one that generates negative examples, both according to the underlying distribution $\mathcal{D}$ on $\mathcal{X}$. Formally, given a target function $f : \mathcal{X} \to {0,1}$, let $\mathcal{D}^+$ be the distribution over $\mathcal{X}^+ = \{x \in \mathcal{X}: f(x) = 1\}$ defined by $\mathcal{D}^+(A) = \frac{\mathcal{D}(A)}{\mathcal{D}(X^+)}$, for every $A \subset \mathcal{X}$. Similary $\mathcal{D}^-$ is the distribution over $\mathcal{X}^{-}$ induced by $\mathcal{D}$.
The definition of PAC learnability in the two-oracle model is the same as the standard definition of PAC learnability except that here the learner has access to $m^{+}_{\mathcal{H}}(\epsilon, \delta)$ i.i.d. examples from $\mathcal{D}^+$ and $m^{-}_{\mathcal{H}}(\epsilon, \delta)$ i.i.d. examples from $\mathcal{D}^{-}$. The learner’s goal is to output $h$ s.t. with probability at least $1-\delta$ (over the choice of the two training sets, and possibly over the nondeterministic decisions made by the learning algorithm), both $L_{(\mathcal{D}^+,f)}(h) \leq \epsilon$ and $L_{(\mathcal{D}^−,f)}(h) \leq \epsilon$
I am trying to prove that if $\mathcal{H}$ is PAC learnable in the standard one-oracle model, then $\mathcal{H}$ is PAC learnable in the two-oracle model. My attempt so far:
Note that $$L_{(D,f)}(h) = \mathcal{D}(\mathcal{X}^+)L_{(\mathcal{D}^+,f)}(h) + \mathcal{D}(\mathcal{X^{-}})L_{(\mathcal{D}^-,f)}(h).$$ Let $d = min \{ \mathcal{D^+}, \mathcal{D^-}\}$, then if $m\geq m_\mathcal{H}(\epsilon d, \delta)$, then it is clear that: $$\mathbb{P}[L_{(D,f)}(h)\leq \epsilon d] \geq 1-\delta \implies \mathbb{P}[L_{(D^+,f)}(h)\leq \epsilon] \geq 1-\delta$$ And, $$\mathbb{P}[L_{(D,f)}(h)\leq \epsilon d] \geq 1-\delta \implies \mathbb{P}[L_{(D^-,f)}(h)\leq \epsilon] \geq 1-\delta$$
So we know that if we have $m\geq m_{\mathcal{H}}(\epsilon d, \delta)$ samples drawn iid from $\mathcal{D}$, then we can guarantee $\mathbb{P}[L_{(D^+,f)}(h)\leq \epsilon] \geq 1-\delta$ and $\mathbb{P}[L_{(D^-,f)}(h)\leq \epsilon] \geq 1-\delta$.
How do I choose $m_{\mathcal{H}}^+(\epsilon, \delta)$ and $m_{\mathcal{H}}^-(\epsilon, \delta)$ such that if we have $m^+ \geq m_{\mathcal{H}}^-(\epsilon, \delta)$ samples iid according to $\mathcal{D}^+$ and $m^- \geq m_{\mathcal{H}}^-(\epsilon, \delta)$ drawn iid according to $\mathcal{D}^{-}$, then we can guarantee $\mathbb{P}[L_{(D^+,f)}(h)\leq \epsilon] \geq 1-\delta$ and $\mathbb{P}[L_{(D^-,f)}(h)\leq \epsilon] \geq 1-\delta$?
When is drawing $m^+$ samples according to $\mathcal{D}^+$ and $m^{-}$ samples according to $\mathcal{D}^-$ the same as drawing $(m^+ + m^-)$ samples according to $\mathcal{D}$?