Let $X_1, \ldots, X_n$ be i.i.d. random variables from the standard normal distribution and let $\bar{X} = \frac{1}{n}\sum_{i=1}^n X_i$ be their sample mean.
I'm interested in the distribution of the $Y_i = \bar{X}-X_i$. Specifically, assuming that there exists a $p(\varepsilon)$ such that for any $i$ $$\mathbb{P}(|Y_i|\le\varepsilon) \le p(\varepsilon),$$ I'm trying to bound $\mathbb{P}(\max_i|Y_i|\le\varepsilon)$ as a function of $p(\epsilon)$ and $n$. If the $Y_i$ were independent, we could readily write $$\mathbb{P}(\max_i|Y_i|\le\varepsilon) = \prod_{i=1}^n \mathbb{P}(|Y_i|\le\varepsilon) \le p(\varepsilon)^n.$$ Unfortunately, the $Y_i$ are not independent. I've tried another approach and I'm wondering whether it is correct.
Step 1. The $Y_i$ are jointly normally distributed and they are all independent from $\bar{X}$.
Step 2. Therefore, the conditional probability distribution of $Y_i$ given $\bar{X}$ is equal to its unconditional distribution, and so $\mathbb{P}(|Y_i|\le\varepsilon) = \mathbb{P}(|Y_i|\le\varepsilon \mid \bar{X}=\bar{x})$.
Step 3. When $\bar{X}=\bar{x}$ is fixed, $Y_i = \bar{X} - X_i = \bar{x} - X_i$ and since the $X_i$ are independent, the $Y_i$ are all mutually conditionally independent given $\bar{X}$. Therefore, $$\mathbb{P}(\max_i|Y_i|\le\varepsilon \mid \bar{X}=\bar{x}) = \prod_{i=1}^n \mathbb{P}(|Y_i|\le\varepsilon \mid \bar{X}=\bar{x}).$$
It seems to me that by combining Steps 1-3 we can conclude $$\mathbb{P}(\max_i|Y_i|\le\varepsilon) = \mathbb{P}(\max_i|Y_i|\le\varepsilon \mid \bar{X}=\bar{x}) = \prod_{i=1}^n \mathbb{P}(|Y_i|\le\varepsilon \mid\bar{X}=\bar{x})= \prod_{i=1}^n \mathbb{P}(|Y_i|\le\varepsilon) \le p(\varepsilon)^n.$$
I find this result "too good to be true" since this is what we would get if the $Y_i$ were independent. Could someone confirm whether the line of reasoning above is correct or, if not, point out where the error lies?
