# Inequality for probability of an average in terms of individual probabilities

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 $$n > 2$$.

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.