# Asymmetric confidence intervals

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

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

• It's certainly possible, see this paper (section 5.2). Briefly, if $(-\infty, U(X)]$ is a $100(1 - \alpha_1)\%$ upper confidence bound and $[L(X), \infty)$ is a $100(1 - \alpha_2)\%$ lower confidence bound, then $[L(X), U(X)]$ is a $100(1 - \alpha_1 - \alpha_2)\%$ confidence interval. Commented Jan 1, 2022 at 8:45
• You are assuming that $X$ is a (approximately) normal distributed variable with a constant variance as function of $\mu$. That symmetric distribution leads to a symmetric confidence interval. But if you do not have a symmetric distribution, or if the variance is not constant, then this will be different. See for instance this question: the basic logic of constructing a confidence interval. Commented Jan 1, 2022 at 11:05
• @Sextus Empiricus, I didn't mean symmetric in that way. Normality was only for illustration. I meant using $q_2=1-q_1.$ Let me update the question. Commented Jan 1, 2022 at 11:07
• @ COOLSerdash, thanks Commented Jan 1, 2022 at 11:15

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.

• "Also, 'minimum width' may not be the most important criterion" BruceET, can you edit the answer to provide some characteristics which may be more important, or that one may have to make tradeoffs with narrowness of width to achieve? Commented Jan 1, 2022 at 21:04
• Sometimes you'd like for the point estimate to be at least roughly at the center of the CI, sometimes you like probabilities above and below the point estimate to be about the same, sometimes one-sided CIs are required and they are hardly ever the shortest, // Finding narrowest CIs requires extra effort, so it's easier to use probability-symmetric intervals; and because they are so often used, people get used to interpreting them in practice. // Giving specific examples of each scenario would make the A needlessly long--at least until more context is given in the Q. Commented Jan 1, 2022 at 21:26
• Oh, I did not mean examples of each, I just meant clarify… like "Also, 'minimum width' may not be the most important criterion, computational ease, agreement between actual vs nominal interval coverage, yadda, blah, and such may be more important" Something like that Commented Jan 1, 2022 at 21:31
• Computational ease is bound to be popular. For an exact CI based on parameter(s) of continuous distributions (such as the one for normal variance in my example) actual and nominal coverage are often the same. Commented Jan 1, 2022 at 21:40

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 for the second order condition to be met.

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

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 closer to the mean. So there are advantages and disadvantages and which is chosen depends on the application. (For more about those disadvantages and advantages, see also Why does $\mu > 0$ (or even $\mu > \epsilon$) "seem easier” to substantiate than $\mu \neq 0$?).

### 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:

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