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Suppose $X_1,X_2,\ldots,X_n$ are iid $\mathcal{N}(\mu,\sigma^2)$, where $\sigma$ is known, but $\mu$ is not. We wish to construct a confidence interval of length $L$ (given) for $\mu$. Is it true that for each given pair $(\mu,\sigma)$, $P(|\mu-\hat{\mu}|<L)$ is maximized when $\hat{\mu}=\bar{X}$?

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    $\begingroup$ What is $\hat\mu$? How is it related to $\bar X$? $\endgroup$ – whuber Jan 28 '15 at 16:48
  • $\begingroup$ The question is whether the coverage probability viewed upon as being a function $\hat{\mu}$ is maximized when $\hat{\mu}=\bar{X}$. $\endgroup$ – Jeff Jan 28 '15 at 16:50
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    $\begingroup$ Are you trying to say that (1) $\hat \mu$ is assumed to be any measurable function of $(X_i)$ and, (2) among all such functions, you are looking for those that maximize the given probability? I'm reluctant to adopt that interpretation, because it is then trivial that any constant function $\hat \mu$ will maximize $P(|\mu-\hat\mu|\lt L)$ for all $\mu$ with $|\hat\mu - \mu|\lt L$, because then this probability is $1$. In fact, I don't even see what your "confidence interval" is: what are its upper and lower bounds? $\endgroup$ – whuber Jan 28 '15 at 16:59
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    $\begingroup$ Where is the minimax aspect involved? Do you mean solving $$\min_{\hat\mu}\max_{(\mu,\sigma)} \mathbb{P}(|\mu-\hat\mu|>L)$$which is associated with the loss function$$\mathbb{I}_{|\mu-\hat\mu|>L}$$ $\endgroup$ – Xi'an Jan 28 '15 at 17:11
  • $\begingroup$ Solving $\inf_{\mu}\sup_{\hat{\mu}}P(|\mu-\hat{\mu}|>L)$ assuming $\sigma$ is given. $\endgroup$ – Jeff Jan 28 '15 at 17:23
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If we take a frequentist decision-theoretic approach to the problem, it means evaluating estimators of $\mu$ by the zero-one loss function$$L(\mu,\hat\mu)=\mathbb{I}_{|\mu-\hat\mu|>L}=\begin{cases}1 &\text{if }|\mu-\hat\mu|>L\\0 &\text{if }|\mu-\hat\mu|\le L\\\end{cases}$$In the normal setting assumed here, sufficiency arguments lead to consider a single observation $X\sim\mathcal{N}(\mu,1)$. The estimator $\hat{\mu}_0(x)=x$ has a constant frequentist risk:$$R(\hat{\mu}_0,\mu)=\mathbb{P}_\mu(|\mu-X|>L)=\mathbb{P}_{\mu=0}(|X|>L)$$ If we can establish $\hat{\mu}_0$ is a Bayes estimator, then it will be minimax. This estimator is associated with the flat prior, since it is easy to show that the Bayes estimator associated with a posterior $\pi(\mu|x)$ satisfies$$\pi(\hat{\mu}+L|x)=\pi(\hat{\mu}-L|x)$$which means that the posterior density is the same on both values (and requires the posterior to be unimodal, I believe). When using the flat prior, $\pi(\mu|x)=\varphi(\mu-x)$, which leads to $\hat{\mu}^\pi(x)=x=\hat{\mu}_0(x)$. The sample mean is therefore a limit of Bayes estimators and the standard proof of minimaxity under quadratic loss extends to this case with the same type of arguments, namely to consider the sequence of normal priors $\mathcal{N}(0,n)$ denormalised by $\sqrt{n}$. See, e.g., my book, The Bayesian Choice, Chapter 2.

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  • $\begingroup$ Thank you. One question though: in general, $\sup_x\inf_y f(x,y)\le \inf_y\sup_x f(x,y)$. The above shows that $\sup_{\mu}R(\bar{X},\mu)=\inf_{\hat{\mu}}\sup_{\mu}R(\hat{\mu},\mu)$, i.e., that when $\hat{\mu}=\bar{X}$, the least coverage probability is maximized. That is, $\sup_{\hat{\mu}}\inf_{\mu}P(|\mu-\hat{\mu}|\le L)$ is attained when $\hat{\mu}=\bar{X}$. Is the same also true for $\inf_{\mu}\sup_{\hat{\mu}}P(|\mu-\hat{\mu}|\le L)$? $\endgroup$ – Jeff Jan 28 '15 at 19:31
  • $\begingroup$ I do not understand the question: are you asking whether the problem has a value, i.e. whether$$\sup_\pi\inf_{\hat\mu} \rho(\hat{\mu},\pi)=\inf_{\hat{\mu}}\sup_\mu R(\mu,\hat{\mu})\ ?$$This is a very hard question and the loss function is discontinuous and non-convex, so I cannot tell from the regular theorem. However, given a fixed risk Bayes estimator, I can deduce that it is minimax without applying this result. $\endgroup$ – Xi'an Jan 28 '15 at 20:21
  • $\begingroup$ No, the question is simpler: is it true that $\inf_{\mu}\sup_{\hat{\mu}}P(|\mu-\hat{\mu}|\le L)=\inf_{\mu}P(|\mu-\bar{X}|\le L)$? $\endgroup$ – Jeff Jan 28 '15 at 20:29
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    $\begingroup$ Of course not: the lhs is always equal to one if you take $\hat\mu=\mu$... This was also @whuber's point. $\endgroup$ – Xi'an Jan 28 '15 at 20:36
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    $\begingroup$ The book is "Examples and Problems in Mathematical Statistics" by S. Zacks. Formula (6.7.5) on p. 418 is incorrect: it states that $\inf_{\mu}\sup_{\hat{\mu}} P(|\mu-\hat{\mu}|\le L)=\inf_{\mu} P(|\mu-\bar{X}|\le L)$, which is false. Many thanks. $\endgroup$ – Jeff Jan 29 '15 at 10:24

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