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Suppose I'm interested in a population mean $\mu$ of some finite discrete variable $Y \sim P$ defined on $\mathbb{R}$. I have a $100(1-\alpha)\%$ confidence interval for the sample estimate $\hat{\mu}$, \begin{equation*}[C^L, \, C^H] = [\hat{\mu} - Z_{1-\alpha/2} \, \sigma / \sqrt{n}, \, \hat{\mu} + Z_{1-\alpha/2} \, \sigma / \sqrt{n}] \end{equation*} where $Z_{1-\alpha/2}$ is the $(1-\alpha/2)$ quantile of the normal distribution and $\sigma$ is the population standard deviation. For simplicity I'll assume, \begin{equation*} \text{Pr}\left[\mu \in [C^L, \, C^H]\right] = 1 - \alpha \end{equation*} So the confidence interval has exact coverage over repeated sampling. But suppose I pick some parameter $\epsilon > 0$ and I define an event, \begin{equation*} A \equiv \left\lbrace \sup_y \vert \hat{P}(y) - P(y) \vert < \epsilon \right\rbrace \end{equation*} where $\hat{P}$ is the empirical mass of $y$. Is it true that, \begin{equation*} \text{Pr}\left[\mu \in [C^L, \, C^H] \, | \, A \right] \geq 1 - \alpha? \end{equation*} I feel that this must be true but I'm having a very hard time formalising it. Any thoughts on how to prove this (or an explanation of why it's not true in general) would be very appreciated.

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  • $\begingroup$ just to check I understand... that final probability doesn't depend on epsilon? $\endgroup$ – Dave Jun 18 '19 at 23:20
  • $\begingroup$ $\epsilon$ is just a fixed constant. The last probability will depend on $\epsilon$ in the sense that $\epsilon$ characterises the 'width' of $A$, though. If $\epsilon = 0$, for example, then we will trivially have 100% coverage since $\hat{P} = P$. Does that make sense? $\endgroup$ – Metrics Man Jun 18 '19 at 23:26
  • $\begingroup$ Check the definition of your upper limit $\endgroup$ – Glen_b Jun 19 '19 at 6:00

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