Asymmetric confidence intervals

Question

Suppose we have iid data $X_i$ with known variance $\sigma^2$, and wish to write an asymptotic $1-\alpha $ coverage CI for the population mean $\mu$. CLT implies that if $z_q$ represents the $q$ quantile of a standard normal, $$z_{\alpha/2}=-z_{1-\alpha/2}\leq \frac{\bar X-\mu}{\sigma/\sqrt n}\leq z_{1-\alpha/2}$$

occurs (asymptotically) with probability $1-\alpha$ and thus implies a CI for $\mu$ of $\bar X\pm z_{1-\alpha/2}\frac{\sigma}{\sqrt n}.$

Any particular reason we take symmetric bounds, or is this just a matter of simplicity? For instance, it seems to me we could have also used

$$ z_{q_1}\leq \frac{\bar X-\mu}{\sigma/\sqrt n}\leq z_{q_2}$$

for any $q_2-q_1=1-\alpha.$

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Update: By "symmetric," I mean using $q_2=1-q_1.$

Answer 1

Let's call your original CI a 'probability-symmetric' confidence interval. For a symmetrical distribution, such an interval may be the narrowest one.

However, the probability-symmetric 95% CI for normal $\sigma^2,$ based on pivoting $$\frac{(n-1)S^2}{\sigma^2}\sim\mathsf{Chisq}(\nu = n-1)$$ is not the shortest because chi-squared distributions are not symmetrical. For convenience, the probability-symmetric 95% CI is often used. (Also, 'minimum width' may not be the most important criterion, so the narrowest CI may not be the most useful.)

Example: Suppose a random normal sample of size $n=50$ has $S^2 = 13.52.$ Then the probability-symmetric 95% CI $(9.43,20.99)$ has width $11.56,$ while the 95% CI $(9.28,20.61)$ has width $11.33.$ [Using R below.]

CI=49*(13.52)/qchisq(c(.975, .025), 49);  CI
[1]  9.434025 20.994510
diff(CI)
[1] 11.56048

CI = 49*(13.52)/qchisq(c(.98, .03), 49);  CI
[1]  9.277642 20.611959
diff(CI)
[1] 11.33432

CI = 49*(13.52)/qchisq(c(1, .05), 49);  CI
[1]  0.00000 19.52473
diff(CI)
[1] 19.52473    # one-sided

In case width is especially important, one could search for the narrowest 95% CI.

Answer 2

Building on BruceET's point, I thought it would be interesting to include an addendum to my post, namely the idea that we may choose a confidence region to minimize its volume subject to meeting its coverage constraint.

For simplicity, I will work with the one dimensional case, letting $T_n$ denote some absolutely continuous pivotal statistic with known invertible CDF $F$, differentiable density $f$, and $q$ quantile given by $t_q\equiv F^{-1}(q).$

Then we wish to choose $q_1$ to minimize the length of the $1-\alpha$ CI:

$$t_{1-\alpha+q_1}-t_{q_1}=F^{-1}(1-\alpha+q_1)-F^{-1}(q_1),$$

giving first order condition

$$ (f(t_{1-\alpha+q_1}))^{-1}-(f(t_{q_1}))^{-1}=0\\ \implies f(t_{1-\alpha+q_1})=f(t_{q_1}),$$

and it suffices that $f'(t_{1-\alpha+q_1})<0<f'(t_{q_1})$ for the second order condition to be met.

For symmetric density, such as in the normal case, this implies $q_1=\alpha/2.$

Answer 3

The symmetric interval minimises interval length in this case

You can find a general exposition of optimal confidence intervals in this related answer. Here I will show you how to do the relevant optimisation for a confidence interval for the mean with known variance (via the normal approximation from the CLT). Let $0 \leqslant \theta \leqslant \alpha$ be the upper tail area for the interval and let $z_a$ denote the quantile of the standard normal distribution with upper tail area $a$. The general interval form is:

$$\text{CI}_\mu (1-\alpha|\theta) = \Bigg[ \bar{x}_n + \frac{z_{1-\alpha+\theta}}{\sqrt{n}} \cdot \sigma, \bar{x}_n + \frac{z_{\theta}}{\sqrt{n}} \cdot \sigma \Bigg].$$

The length of this confidence interval is:

$$\text{Length}(\theta) = (z_{\theta} - z_{1-\alpha+\theta}) \times \frac{\sigma}{\sqrt{n}}.$$

In order to obtain the optimal (minimum length) confidence interval of this form, we choose $\theta$ to minimise the length function. That is, we use the value:

$$\hat{\theta} = \underset{0 \leqslant \theta \leqslant \alpha}{\text{arg min}} \ \text{Length}(\theta) = \underset{0 \leqslant \theta \leqslant \alpha}{\text{arg min}} \ (z_{\theta} - z_{1-\alpha+\theta}) = \frac{\alpha}{2}.$$

Using this optimisation result, we see that the length of the confidence interval is minimised using the equal tail (symmetric) interval. Using the fact that $z_{1-\alpha/2} = - z_{\alpha/2}$ we can write this optimal confidence interval in the standard form:

$$\text{CI}_\mu (1-\alpha) = \Bigg[ \bar{x}_n - \frac{z_{\alpha/2}}{\sqrt{n}} \cdot \sigma, \bar{x}_n + \frac{z_{\alpha/2}}{\sqrt{n}} \cdot \sigma \Bigg].$$

Answer 4

There is no single best way of constructing a confidence interval.

In the below graph you see some different possibilities of a 95% confidence interval for a z-test.

In this graph the confidence interval is expressed as the boundaries in term of the amount of $\pm \sigma$ deviation from the observed $X$ and taken as a constant independent of $X$ (you could make confidence intervals with different shapes/sizes depending on $X$).

For instance for the percentage on the left tail, $q_1 = 0.025$, you get the middle interval which is $[\bar{x} -1.96\sigma, \bar{x}+1.96 \sigma]$.

The symmetric confidence intervals (in the middle on the graph, wich is based on a two-tailed hypothesis test), divides the attention equally to both sides. It is advantageous because it doesn't favor any side but also because it is a smaller confidence interval compared to the other choices.

On the extreme of the graph, we see confidence intervals based on one-tailed hypothesis tests. These confidence intervals are larger for the z-test, but the boundary on one side is less more close to the mean. So there are advantages and disadvantages and which is chosen depends on the application.