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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 $q_2=1-q_1.$

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  • $\begingroup$ The criterion suggested in this answer I guess could also be used to choose a 'best' pivotal statistics to use for determining the CI ? $\endgroup$
    – Thomas
    Jan 16, 2022 at 17:45

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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.

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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 $q_1=\alpha/2.$

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  • $\begingroup$ This is a nice question (+1). Regarding this answer, how do you link in a general case the interval done with the quantiles of the pivotal statistics to the C.I. of the quantity you are jnterested? You would like to minimize the latter in the general case no? $\endgroup$
    – Thomas
    Jan 16, 2022 at 17:36

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