How can I perform Union operation into two events?

Question

Assume there are two events A and B such as $A := \frac{1}{n} \sum_{i=1}^{n}\left(f\left(x_{i}\right)-\mathbb{E}[f]\right) z_{i} \geq \frac{\epsilon}{8}$

and $B := \exists f \in \mathcal{F}: \frac{1}{n} \sum_{i=1}^{n} \mathbb{E}[f] z_{i} \geq \frac{\epsilon}{8}$

and P(A) and P(B) is given to me! I need to Calculate the Upperbound of the P(C) where C is defiend as $$ C:= \mathbb{P}\left(\exists f \in \mathcal{F}: \frac{1}{n} \sum_{i=1}^{n} f\left(x_{i}\right) z_{i} \geq \frac{\epsilon}{4}\right)$$

I have used the upper bound $P(A \cup B) \leq P(A) + P(B)$

Further I have reached upto

Given to me := \begin{align} \mathbb{P}\left (\frac{1}{n}\left| \sum_{i=1}^n (f(x_i) - \mathbb{E}[f])z_i\right| \geq \frac{\epsilon}{8} \right ) \leq 2\exp\left (-\frac{\epsilon^2nd}{9^4cL^2} \right) ... (1) \end{align}

\begin{align} \mathbb{P}\left(\exists f \in \mathcal{F}: \frac{1}{n} \sum_{i=1}^{n} \mathbb{E}[f] z_{i} \geq \frac{\epsilon}{8}\right) \leq \mathbb{P}\left(\left|\frac{1}{n} \sum_{i=1}^{n} z_{i}\right| \geq \frac{\epsilon}{8}\right) ...(2)\end{align}

\begin{align} \mathbb{P}(A \cup B)\\ &= \mathbb{P} \left( \underbrace{ \left( \frac{1}{n} \left|\sum_{i=1}^{n}\left(f\left(x_{i}\right)-\mathbb{E}[f]\right) z_{i}\right| \geq \frac{\epsilon}{8} \right)}_{A} \bigcup \underbrace{\left( \exists f \in \mathcal{F}: \frac{1}{n} \sum_{i=1}^{n} \mathbb{E}[f] z_{i} \geq \frac{\epsilon}{8}\right)}_{B}\right)\\ &\leq \mathbb{P}\left ( \frac{1}{n}\left| \sum_{i=1}^n (f(x_i) - \mathbb{E}[f])z_i\right| \geq \frac{\epsilon}{8} \right ) + \mathbb{P}\left(\exists f \in \mathcal{F}: \frac{1}{n} \sum_{i=1}^{n} \mathbb{E}[f] z_{i} \geq \frac{\epsilon}{8}\right)\\ &\leq 2 \exp\left (-\frac{\epsilon^2nd}{9^4cL^2} \right) + \mathbb{P}\left(\left|\frac{1}{n} \sum_{i=1}^{n} z_{i}\right| \geq \frac{\epsilon}{8}\right) && \text{(Using Inequality 1 and 2)}\\ &\leq 2 \exp\left (-\frac{\epsilon^2nd}{9^4cL^2} \right) + 2 \exp \left(\frac{-n \epsilon^{2}}{ 8^{3}}\right) && \text{(By Hoeffding's inequality)} \end{align}

Can I proof LHS as with event C. I am not getting clue? Can anyone please help!

My question is := can I reduce $P(A \cup B) = P(C)$ if yes then how?

N.B > I am trying to understand the proof of Theorem 2(Page -$7$) in the paper "A Universal Law of Robustness via isoperimetry" by Bubeck and Sellke.

Answer 1

For two events $E,F$ one always has that if "$E \implies F$" then $P(E) \leq P(F)$. To see this note \begin{align} P(F) &= P(E \text{ and } F) + P(F \text{ and not } E) \\ &= P(F|E)P(E) + P(F \text{ and not } E) \\ &= 1 P(E) + P(F \text{ and not } E) \\ &\geq P(E) \end{align} where we have used that $P(F|E) = 1$ since $E \implies F$, and also $P(F \text{ and not } E) \geq 0$.

If we apply this with $E = C$ and $F = A \cup B$ then we will have $P(C) \leq P(A \cup B)$ which is what you need above. So all we need to do is check that $C \implies A \cup B$.

Suppose that $C$ holds and let $f \in \mathcal{F}$ be a function such that $\frac{1}{n}\sum_{i=1}^n f(x_i)z_i \geq \epsilon /4$. Suppose for contradiction that neither $A$ nor $B$ occurs so that we have both \begin{align} \frac{1}{n}\left | \sum_{i=1}^n (f(x_i) - \mathbb{E}f)z_i \right | &< \epsilon / 8 \\ \frac{1}{n}\sum_{i=1}^n\mathbb{E}[f]z_i &< \epsilon /8. \end{align} By the first inequality we have \begin{align} \frac{1}{n}\left ( \sum_{i=1}^n (f(x_i) - \mathbb{E}f)z_i \right ) &< \frac{\epsilon}{8} \\ \implies \frac{1}{n}\sum_{i=1}^n f(x_i)z_i&< \frac{\epsilon}{8} + \frac{1}{n}\sum_{i=1}^n \mathbb{E}[f]z_i \end{align} and by the second we have \begin{align} \frac{1}{n}\sum_{i=1}^n f(x_i)z_i &< \frac{\epsilon}{8} + \frac{1}{n}\sum_{i=1}^n \mathbb{E}[f]z_i \\ &< \frac{\epsilon}{8} + \frac{\epsilon}{8} = \frac{\epsilon}{4} \end{align} but this is a contradiction since we assumed that $\frac{1}{n}\sum_{i=1}^n f(x_i)z_i \geq \frac{\epsilon}{4}$.

In particular this implies that when $C$ holds at least one of $A$ or $B$ must hold, which is to say $A \cup B$ holds. This confirms that indeed $C \implies A \cup B$ and completes the proof.

Answer 2

I have answered here : Union Bound of two events?

But I am not getting How that |F| is coming in the RHS.