# Conditional coverage probability

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.

• just to check I understand... that final probability doesn't depend on epsilon? – Dave Jun 18 '19 at 23:20
• $\epsilon$ 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? – Metrics Man Jun 18 '19 at 23:26
• Check the definition of your upper limit – Glen_b Jun 19 '19 at 6:00