Suppose $X_1,X_2,\ldots,X_n$ is a random sample drawn from the distribution


It can be shown that there exist some $c_{\alpha}, d_{\alpha}$ such that


and $$P_{\theta}(\overline X+d_{\alpha}<\theta<\overline X)=1-\alpha\quad\forall\,\theta\tag{2}$$

I am asked to justify, "without any formal derivation", which statement among $(1)$ and $(2)$ leads to UMAU confidence interval for $\theta$ with confidence coefficient $1-\alpha$.

I know that a Uniformly Most Accurate Unbiased (UMAU) region for $\theta$ can be obtained from a Uniformly Most Powerful Unbiased (UMPU) test if I consider the problem of testing $H:\theta=\theta_0$ against some $K:\theta\ne \theta_0$.

But how can I justify which of the two tests, one based on $X_{(1)}$ and another based on $\overline X$, is UMPU for testing $H$ without actually deriving the tests?

The CI obtained from $(1)$ has the shortest length based on the pivot $X_{(1)}-\theta$, but is it UMAU (or equivalently, is the test based on $X_{(1)}$ UMPU)?

I have this following definition of UMAU confidence set at hand which is somewhat difficult to work with directly:

$S_0(\mathbf X)$ for varying $\mathbf X\in \mathscr{X}$ is said to be a family of UMAU confidence sets of $\theta$ with confidence coefficient $1-\alpha$ if

(i) $P_{\theta}(\theta\in S_0(\mathbf X))=1-\alpha\quad\forall\,\theta\in\Theta$

(ii) $P_{\theta'}(\theta\in S_0(\mathbf X))\le 1-\alpha\quad\forall\,\theta,\,\theta'(\ne \theta)$

(iii) $P_{\theta'}(\theta\in S_0(\mathbf X))\le P_{\theta'}(\theta\in S(\mathbf X))\quad\forall\,\theta,\,\theta'(\ne \theta)$, for whatever $S$ satisfying (i) and (ii).

Any hint would be great rather than a full answer.


1 Answer 1


This is where you need the Karlin-Rubin theorem. This theorem says that if the likelihood-ratio function is a monotone function of a statistic, then the Uniformly Most Powerful Test (UMPT) is a threshold test based on that statistic. In this case, you can see from your density function that the likelihood ratio statistic depends on the data only through the minimum value $x_{(1)}$ (the data values in the density function cancel out in the ratio). Hence, it should be the case that the UMPT is a threshold test on $x_{(1)}$. Relating that back to the region you are looking for, your UMAU region should then be based on that statistic, so the UMAU region is (1), not (2).

Have a look into this theorem, and hypothesis testing problems that involve distributions with a monotone log-likelihood ratio function. Understanding these types of problems will give you a bit more insight into some common forms of test/confidence intervals that occur in a wide class of cases. Since you are asked to obtain this answer "without any formal derivation", all you can really do here is to try to give an explanation of the problem based on the fact that it involves a monotone likelihood ration function.

Just in case what I have said about the likelihood-ratio function is unclear, let me back that up with some derivation. (Naughty me!) Letting $x_{(1)} \equiv \min \{ x_1,...,x_n \}$ denote the minimum order statistic, your log-likelihood function in this problem is:

$$\ell_\mathbf{x}(\theta) = \begin{cases} n(\theta - \bar{x}) & & & \text{for } \theta < x_{(1)}, \\[6pt] -\infty & & & \text{for } \theta \geqslant x_{(1)}. \\[6pt] \end{cases}$$

For any parameter values $\theta_0 < \theta_1$ you have:

$$\ell_\mathbf{x}(\theta_1) - \ell_\mathbf{x}(\theta_0) = \begin{cases} \text{undefined} & & & \text{for } \theta_0 \geqslant x_{(1)}, \\[6pt] -\infty & & & \text{for } \theta_0 < x_{(1)} \leqslant \theta_1, \\[6pt] n(\theta_1 - \theta_0) & & & \text{for } \theta_1 < x_{(1)}. \\[6pt] \end{cases}$$

We can see that this log-likelihood difference is non-decreasing in $x_{(1)}$, so the Uniformly Most Powerful Test (UMPT) of $H_0: \theta \leqslant \theta_0$ against $H_0: \theta > \theta_0$ is a threshold test in $x_{(1)}$ (higher values being more conducive to the alternative hypothesis).


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