Inequality for probability of an average in terms of individual probabilities

Question

Let $X_1,\cdots,X_n$ be identically distributed nonnegative, real-valued random variables that are not necessarily independent. Note that we trivially have for any $\varepsilon > 0$:

$$ \mathbb{P}\left\{ \frac{1}{n} \sum_{i=1}^{n} X_i \geq \varepsilon \right\} \leq \sum_{i=1}^{n} \mathbb{P}\{X_i \geq \varepsilon\} = n \mathbb{P}\{X_1 \geq \varepsilon\}.$$

For $n \geq 3$, is there always an example when the above inequality holds with equality, or can we obtain a sharper bound for general $n$? I am especially interested in the setting where $n$ is large.

For $n = 2$, the example $X_1, X_2 \sim U[0,1]$ with $X_2 = 1 - X_1$ and $\varepsilon = 0.5$ presents an instance where the inequality holds with an equality, but it seems hard to generalize to $n > 2$.

Answer 1

Let $\Omega=\{1,2,3\}$ and $\mathcal F=2^\Omega$, $\mathbb P(\{1\})=\mathbb P(\{2\})=\mathbb P(\{3\})=\frac13$ be discrete probability space. Define three random variables $X_i(\omega)=\mathbb 1_{\{\omega=i\}}$.

Namely, $$ X_1(1)=1,\; X_2(1)=0,\; X_3(1)=0 $$ $$ X_1(2)=0,\; X_2(2)=1,\; X_3(2)=0 $$ and $$ X_1(3)=0,\; X_2(3)=0,\; X_3(3)=1. $$ Then $X_1+X_2+X_3=1$. Take $\varepsilon=\frac13$. Then $$ 1=\mathbb P\left(\frac13\sum_{i=1}^3 X_i \geq \frac13\right) = \sum_{i=1}^3 \mathbb P\left(X_i\geq \frac13\right) = 3\cdot \frac13=1. $$ This example easily extended to any number of random variables.