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

In general, one may choose a confidence set so that the probability of coverage is at least $1-\alpha$. As you mention, for quantifying the uncertainty of $\mu$, this can be done with an asymmetric confidence interval.

In other cases, the notion of a "symmetric confidence interval" doesn't make much sense; e.g. consider the problem of finding a confidence interval for $\sigma$, when performing parameter estimation on $N(\mu, \sigma^2)$.

However, the reason why we choose to use a symmetric interval for $\mu$ is because that gives the confidence interval of smallest length, which is a desirable property when doing uncertainty quantification.

Answer 2

Building on Jose'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 $q_1=\alpha/2.$