#StackBounty: #hypothesis-testing #confidence-interval #mean #asymptotics #small-sample What "nice" property of a confidence …

Bounty: 50

Suppose that I have $X_{i} overset{i.i.d.}{sim} P$ with $E[X_{i}]=mu$ and $V[X_{i}^{2}] = sigma^{2}<infty$.

Then by the central limit theorem I know that:
sqrt{n} (bar{X}{n} – mu) overset{d}{to} N(0,sigma^{2})

where $bar{X}
{n}$ is the sample average. Suppose for some silly reason I know the value of $sigma^{2}$. Then this asymptotic approximation allows me to justify confidence sets for $mu$ of the form:
bar{X}{n} pm q{alpha/2} sqrt{frac{sigma^{2}}{n}}

where $q_{alpha/2}$ is the $alpha/2^{th}$ quantile of the standard normal. In particular:
lim_{n to infty} P left(q_{alpha/2} leq sqrt{n} frac{(bar{X}{n} – mu)}{sigma} leq -q{alpha/2} right) = 1-alpha\
implies lim_{n to infty} P left(bar{X}{n} + q{alpha/2}frac{sigma}{sqrt{n}} leq mu leq bar{X}{n} -q{alpha/2}frac{sigma}{sqrt{n}} right) = 1-alpha\

For simplicity, let:
$$CI_{1} = left[bar{X}{n} + q{alpha/2}frac{sigma}{sqrt{n}} , bar{X}{n} -q{alpha/2}frac{sigma}{sqrt{n}} right]$$
Now suppose that I am a strange statistician, and that rather than the confidence interval constructed above, I prefer a confidence interval (for whatever reason) of my own making:
$$CI_{2} = left[bar{X}{n} + q{alpha/2}frac{sigma}{sqrt{n}}+b_{n} , bar{X}{n} -q{alpha/2}frac{sigma}{sqrt{n}} -b_{n}right]$$
where $b_{n} = o(n^{-1/2})$ is some vanishing deterministic sequence. Note that $CI_{2}$ also provides $1-alpha$ coverage probability asymptotically.

My question: is there any reason to prefer $CI_{1}$ to $CI_{2}$? Asymptotically they are the same, so I suspect any reason would need to appeal to finite sample arguments. For example, I can always construct the sequence $b_{n}$ such that $CI_{1}$ and $CI_{2}$ are VERY different in finite sample. So what statistical justification would lead someone to use $CI_{1}$ versus $CI_{2}$? Is there a name for the desirable property $CI_{1}$ possesses that $CI_{2}$ does not?

Thanks so much!

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