Confidence Intervals guaranteed to be centered at 0?

Question

With most pivotal quantities that I'm thinking of, after we rearrange stuff to get an interval, the interval's "location" depends on a random point estimate. The example below has an interval with midpoint equal to the sample mean.

Is there a way to guarantee that we end up with an interval centered at $0$? By "centered at $0$" I mean an interval of the form $(-c,c)$ for some fixed $c \in \mathbb{R}^+$.

For example, with a normal random sample, we could start with the pivotal quantity $\sqrt{n}(\bar{x} - \mu)/s$, and rearrange the right hand side of

$$ 1 - \alpha = \mathbb{P}\left(-t_{\alpha/2,n-1} \le \sqrt{n}(\bar{x} - \mu)/s \le t_{\alpha/2,n-1} \right) $$

into

$$ \mathbb{P}\left(\bar{x}-\frac{s}{\sqrt{n}}t_{\alpha/2,n-1} \le \mu \le \bar{x}+\frac{s}{\sqrt{n}}t_{\alpha/2,n-1}\right). $$

Is there a way we could scoot it over $\bar{x}$ and then fix the (random) width so that we end up with $95$% coverage, still?

I suspect it isn't possible. If I'm not changing the pivotal quantity, all I can do is pick the quantiles. This example is an "equal-tailed" interval. But how can I deterministically pick quantiles, by picking two positive numbers that sum to $\alpha$, based on a random data set I haven't obtained yet?

Answer 1

I find this a strange question but one idea would be to construct the interval from a confidence distribution of the parameter of interest. In the iid normal case for $\mu$ the cdf of one such confidence distribution is $$ F(\mu) = F_{T_{n-1}}\left(\frac{\sqrt{n}(\mu-\bar x)}s\right). $$ where $F_{T_{n-1}}$ is the cdf of the student $t$-distribution with $n-1$ degrees of freedom.

Then, choosing $c$ such that $$ F(c)-F(-c)=1-\alpha, $$ we may hope to achieve coverage of the interval $(-c,c)$ close to the nominal level of $1-\alpha$. This indeed appears to happen for large $\mu/\sigma$ (black curve in plot below). However, as $\mu$ becomes small this procedure inevitably leads to an interval always containing $\mu$.

The interval $(-c,c)$ where $c=\max(|U|,|L|)$ and $(U,L)$ is the ordinary Student $t$-interval for $\mu$ (suggested by @whuber in the comments) appears to achieve a confidence level of $1-\alpha$ if the confidence level of the interval $(U,L)$ is $1-2\alpha$ (red curve in plot below). This implies that it is always shorter than the one derived via the confidence distribution. It also appears to have coverage always closer to the nominal level (red curve in plot below).

R code:

ci0 <- function(x, alpha=.05, upper=100) {
  n <- length(x)
  xbar <- mean(x)
  s <- sd(x)
  cdf <- function(mu) {
    pt(sqrt(n)*(mu - xbar)/s, df = n - 1)
  }
  f <- function(c) {
    cdf(c)-cdf(-c) - 1 + alpha
  }
  c <- uniroot(f, lower = 0, upper = upper)$root
  c(-c,c)
}
whuber <- function(x,alpha=.05) {
  ci <- t.test(x, conf.level=1-2*alpha)$conf.int
  c <- max(abs(ci))
  c(-c,c)
}
coverage <- function(fn, mu, sigma=1, n, nsim=1e+4, alpha=0.05) {
  hits <- 0
  for (i in 1:nsim) {
    x <- rnorm(n, mu, sigma)
    ci <- fn(x, alpha=alpha) 
    if (ci[1] < mu & ci[2] > mu) {
      hits <- hits + 1
    }
  }
  list(coverage=hits/nsim, binom.test(hits, nsim, p=1-alpha)$p.value)
}

m <- 40
mu <- seq(0, 3, length=m)
res1 <- res2 <- numeric(m)
for (i in 1:m) {
  res1[i] <- coverage(ci0, mu[i], n=10, nsim=1e+4)$coverage
  res2[i] <- coverage(whuber, mu[i], n=10, nsim=1e+4)$coverage
}
plot(mu, res1, xlab="mu/sigma", ylab="coverage", type="l")
lines(mu, res2, col="red")

abline(h=.95+sqrt(.95*.05/1e+4)*qnorm(c(.025,.975)), lty=3)
legend("topright",c("via conf.distr.","wbuber"),col=c("black","red"),lty=1)

Answer 2

I prefer visualizing frequentist inference using a confidence curve.

Without loss of generality we could consider a hypothesis $H_0: \theta=\theta_0$ (not necessarily zero). Above is a confidence curve for inference on a Bernoulli proportion $p$ from inverting the binomial CDF, $F_Y(y,n,p)$, based on $6$ events with a sample size of $n=10$. Considering the hypothesis $H_0:p=0.5$, for this experimental result the upper-tailed p-value is $0.38$. We are therefore $38\%$ confident the unknown fixed true $p$ is less than or equal to $0.5$ and $100(1-0.38)\%=62\%$ confident the unknown fixed true $p$ is greater than or equal to $0.5$. These confidence levels are nothing more than a restatement of the one-sided p-value testing $H_0: p= 0.5$. The vertical reference lines identify the two-sided equal-tailed $75\%$ confidence interval. This is the set of hypotheses that are most plausible given the observed data $-$ those hypotheses that when tested are not significant at the two-sided equal-tailed $25\%$ level. For each hypothesis in this interval the observed result is within a $75\%$ margin of error. Here is a related post.

For the given experimental result, by considering a margin of error that does not have equal tails you can construct a $95\%$ confidence interval that is centered at $H_0:p=0.5$.

For the same experimental result here is another confidence curve centered at $H_0: p=0.5$. The maximum likelihood estimate is still $\hat{p}=0.6$ but I have defined the peak of the curve at $p=0.5$. The $95\%$ confidence interval centered at $0.5$ is $(0.15, 0.85)$ The upper-tailed p-value testing $H_0: p=0.15$ is $0.001$ and the lower-tailed p-value testing $H_0: p=0.85$ is $0.049$. We are therefore $95\%$ confident the unknown fixed true $p$ is within $(0.15, 0.85)$.