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Many years ago, I needed a lower bound for a $(1- \alpha )100 \% $ one-sided confidence interval for $| \mu |$, assuming $X\sim N( \mu , \sigma^2I_n) $ with $ \sigma^2 $ known. Now $\bar{X}\sim N( \mu , \sigma^2/n) $. Let $\bar{x} $ denote the sample mean. At the time, I used a graphical approach to justify the argument that the lower bound $ \omega_ \alpha $ for this sample could be found as the solution for $| \mu |$ that satisfies

$$P_{|\mu |}( |\bar{X} |\leq |\bar{x} |) = 1- \alpha \tag 1.$$

That is, set $ \omega_ \alpha $ equal to the $| \mu |$ that satisfies (1). If (1) produces valid lower bounds then it is relatively straightforward (numerically) to find it in this case since the distribution of $| \bar{X} |$ is a folded normal, and the resulting bound seems correct. My problem is that I do not recall the theoretical justification for (1), even though I expect that the justification for (1) is quite elementary. For background, let’s see how (1)’s equivalent works correctly for determining the lower bound for the one-sided confidence interval for $\mu $.

One-sided 95% confidence lower bound for $\mu $ given $ \sigma^2 $ known.

The equivalent formulation to (1) in this case, with $1- \alpha = 0.95$, is

$$P_\mu ( \bar{X} \leq \bar{x} ) = 0.95. \tag 2$$

Now $P_\mu ( \bar{X} \leq \bar{x} )= P_0 (Z \leq \sqrt{n} (\bar{x}- \mu )/ \sigma ) $ and we know $ P_0 (Z \leq 1.645)=0.95$. Thus, from (2), $\sqrt{n} (\bar{x}- \omega_ \alpha)/ \sigma) =1.645$. Therefore $\omega_\alpha = \bar{x} -1.645 \sigma/\sqrt{n} $, which is the standard confidence lower bound for this case.

My question is: Can using (1) to find the lower bound for the one-sided confidence interval for $| \mu |$ be justified?

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3 Answers 3

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Yes, this technique works for $|\mu|$. The two key points that make it work are that

  1. the distribution of $|\bar{X}|$ depends on $\mu$ only through $|\mu|$, and
  2. the function $|\mu| \mapsto P_{|\mu|}(|\bar{X}| < y)$ is decreasing in $|\mu|$ for every $y > 0$.

To see (1.), notice that, if $Z \sim N(0, 1)$ is a standard Gaussian, \begin{align*} \bar{X}^2 &\stackrel{\mathrm{d}}{=} \frac{\sigma^2}{n} Z^2 + \frac{2\sigma}{\sqrt{n}}Z\mu + \mu^2 \\ &\stackrel{\mathrm{d}}{=} \frac{\sigma^2}{n} Z^2 + \frac{2\sigma}{\sqrt{n}}Z|\mu| + |\mu|^2, \end{align*} where the second equality holds because $Z \stackrel{\mathrm{d}}{=} \operatorname{sign}(\mu) Z$ by the symmetry of $Z$ around $0$.

Next, we show a general fact about inverting one-sided hypothesis tests. Specifically, suppose $\theta$ is a parameter and $T(X)$ a test statistic, the corresponding one-sided level-$\alpha$ test for $H_0\colon \theta = \theta_0$ rejects when $T(X) > t_{\theta_0}$, where $t_{\theta_0}$ solves $P_{\theta_0}(T(X) \leq t_{\theta_0}) = 1 - \alpha$.

The confidence set formed by inverting this test is then all those $\theta$ whose corresponding null hypotheses would not be rejected based on $T(X)$. That is, $$ \operatorname{CI}(x) = \{\theta : T(x) \leq t_\theta\}. $$

Writing $F_\theta(y) = P_\theta(T(X) \leq y)$ and assuming that $T(X)$ has a uniformly continuous distribution, the above then becomes \begin{align*} \operatorname{CI}(x) &= \{\theta : F_\theta(T(x)) \leq F_\theta(t_\theta)\} \\ &= \{\theta : F_\theta(T(x)) \leq 1 - \alpha\}. \end{align*}

If $F_\theta(y)$ is decreasing in $\theta$, the above is a one-sided interval $[\omega_\alpha, \infty)$, where $\omega_\alpha$ solves $$ F_{\omega_\alpha}(T(x)) = 1 - \alpha. $$

Now, in your case, $\theta = |\mu|$ and $T(x) = |\bar{x}|$, so $\omega_\alpha$ solves \begin{align*} 1 - \alpha &= F_{\omega_\alpha}(|\bar{x}|) \\ &= P_{\omega_\alpha}(|\bar{X}| \leq |\bar{x}|) \end{align*} as per your supposition.

It remains to show in detail (2.), but this is a straightforward calculus exercise.

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  • $\begingroup$ Thanks, that looks like the logic! I will give some more time for others, and probably tick tomorrow. (I guess you mean the absolute value of the mean in your point 1.) $\endgroup$ Commented Nov 1 at 6:41
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This is just a comment to the other excellent answer by Damian Pavlyshyn noting that when $|\bar x|$ is too small, the confidence set defined by $\{|\mu|:P_{|\mu|}(|\bar X|\le |\bar x|)\le 1-\alpha\}$ will include 0 since $$ \Phi\left(\frac{|\bar x|-|\mu|}{\sigma/\sqrt{n}}\right)-\Phi\left(\frac{-|\bar x|-|\mu|}{\sigma/\sqrt{n}}\right), $$ being the probability that $\bar X$ falls between $-\bar x$ and $\bar x$, will then always be smaller than $1-\alpha$. Still, the coverage appears to match the nominal level of $1-\alpha$ as indicated by the following simulation.

pfoldnorm <- function(x, mu, sigma) {
  pnorm((x-mu)/sigma) - pnorm((-x-mu)/sigma)
}
lowerbound <- function(x, alpha=0.05) {
  n <- length(x)
  absxbar <- abs(mean(x))
  f <- function(mu) {
    pfoldnorm(absxbar, mu, 1/sqrt(n)) - 1 + alpha
  }
  if (f(0)<=0) 
    return(0)
  else
    uniroot(f, c(0, 100))$root
}
coverage <- NULL
mu <- seq(0, 2, by=.1)
for (i in seq_along(mu)) {
  hits <- replicate(1e+5,{
    x <- rnorm(1, mu[i])
    l <- lowerbound(x, .05)
    l <= abs(mu[i])
  })
  coverage[i] <- mean(hits)
}
plot(mu, coverage)
abline(h=.95, col="green")

Created on 2025-11-01 with reprex v2.1.1

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    $\begingroup$ I think that in your equation you should have $|\bar{x}|$ instead of $\bar{x}$ in both normal CDFs. Making this change in your code shows the CI achieving the correct coverage for all values of $\mu$. $\endgroup$ Commented Nov 1 at 17:23
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    $\begingroup$ @DamianPavlyshyn You're absolutely right, my bad, the coverage indeed appears to exactly match the nominal level! $\endgroup$ Commented Nov 1 at 21:42
  • $\begingroup$ Thanks Jarle for checking that. $\endgroup$ Commented Nov 1 at 22:11
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Thank you, Damian Pavlyshyn for an excellent answer to my question. My ‘answer’ here is just to provide clarifying details on the calculation of confidence interval for $\mu $ to avoid any misunderstandings. As the distribution of $| \bar{X} |$ is a folded normal, we numerically solve for $ \omega_ \alpha $ the following equation: $$ \Phi\left(\frac{|\bar x|-\omega_ \alpha }{\sigma/\sqrt{n}}\right)-\Phi\left(\frac{-|\bar x|-\omega_ \alpha }{\sigma/\sqrt{n}}\right)=1-\alpha, $$ This can be done quite easily, using, for example, Excel Solver. Solutions to this equation are feasible for $|\bar{x} |> z_{\alpha /2} \sigma/\sqrt{n}$, where $z_{\alpha /2} $ defined by $\Phi (z_{\alpha /2} )=1- \alpha /2$. For $|\bar{x} |\leq z_{\alpha /2} \sigma/\sqrt{n}$, the CI lower bound $ \omega_ \alpha =0$.

Aside

As $\bar{x} >0$ increases, $ \omega_ \alpha $ rapidly approaches $\bar{x} -z_{\alpha } \sigma/\sqrt{n} $. This is not surprising since $\max(0, \bar{x} -z_{\alpha } \sigma/\sqrt{n} $ is the tighter $(1- \alpha )\times 100 \% $ one-sided confidence interval for $\mu $ that would have been able to be used had we known that $\mu \geq 0$.

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