The following is Hoeffding’s Inequality:
$P(|\nu - \mu| > \epsilon) \leq 2e^{-2\epsilon^2N}$
How can I use the above inequality to actually see how the upper bound changes as I change the sample size $N$ and tolerance $\epsilon$? $\mu$ is the population parameter and $\nu$ is the sample parameter which helps infer $\mu$ in the population.
For fixed $\epsilon > 0$ and $\alpha \in (0,1)$. Notice that: \begin{align*}2e^{-2\epsilon^2 N} \leq \alpha &\iff e^{-2\epsilon^2N} \leq \frac\alpha2\\ &\iff e^{2\epsilon^2N} \geq \frac2\alpha \\ &\iff 2\epsilon^2N\geq \log(\frac2\alpha) \\ &\iff N \geq \frac{\log(\frac2\alpha)}{2\epsilon^2}. \end{align*} Therefore, Hoeffding's inequality shows that the event $|\nu - \mu| \leq \epsilon$ will happen with probability greater than $1-\alpha$ if you choose $N \geq \dfrac{\log(\frac2\alpha)}{2\epsilon^2}$.