# When does a UMP test fail to exist?

I have a sample $X=(X_1, ...,X_n)\sim N(\mu,\sigma^2)$ with $\sigma^2$ known. The hypotheses are $H_0: \mu=\mu_0, H_1:\mu \neq \mu_0$. I know that in such a case an UMP test does not exist and so that I should proceed using a LR test, in order to find the rejection rule.

My professor also told me that for a sample distribution that belongs to the Exponential Family in the case of simple vs bilateral hypotheses an UMP test does not exist.

Thus, my question is theoretic: why does an UMP test does not exist in such cases? Which are the conditions under which an UMP test does not exist?

EDIT: I have found an example in which, instead, although the alternative hypothesis is bilateral, the UMP test exists.

A sample $X\sim U(0,\theta)$. The hypotheses are $H_0:\theta=\theta_0, H_1:\theta\neq \theta_0$.

• Simply put, a UMP test does not exist because the critical regions turn out to be different for $\theta> \theta_0$ and $\theta<\theta_0$. This means that there are only UMP tests for one-sided hypotheses, where you may still use the NP lemma. Commented Dec 30, 2015 at 21:54
• @JohnK Ok, that's clear. But how can I apply this idea to the two examples above? How can I show that it is actually the case? Commented Dec 31, 2015 at 8:57
• I am not familiar with a general proof. Perhaps you can work something out starting from the definition of an exponential density. My guess is that it would be non-trivial. Commented Dec 31, 2015 at 10:57
• @Alessandro: it depends on how deep you want to dive into it. I think the paper is interesting to learn which conditions are needed and why
– user83346
Commented Jan 2, 2016 at 12:10
• I thought you were interested in exponential densities. The uniform distribution does not belong to the exponential family so there is no general rule there. Commented Jan 3, 2016 at 16:51

In the example that you have provided, go on to calculate the likelihood ratio and you will find that it comes out to be a function of the order statistics, X(1) and X(2). Question 8.33 of Statistical Inference by George and Casella will help. The solution is provided in the link below: http://www.ams.sunysb.edu/~zhu/ams570/Solutions-Casella-Berger.pdf

Coming back to the existence of a UMP test, Karlin Rubin Theorem tells that the MLR should exist, so that the inverse operation can be applied to get the test. The example on the link below will surely help. http://web.eecs.umich.edu/~cscott/past_courses/eecs564w11/25_ump.pdf

To prove the non-existence of a UMP test for this two-sided hypotheses based on the normal distribution family, use proof by contradiction. The idea is that if such a UMP test for testing the two-sided hypotheses existed, then it would also become UMP tests for testing two one-sided hypotheses, which enabled us to arrive at a contradiction.

Without loss of generality, assume $$\sigma^2 = 1$$. In order to show that no size $$\alpha$$ UMP test (abbreviated as UMP-$$\alpha$$ test hereafter) exist for testing \begin{align*} H: \mu = \mu_0 \text{ versus } K: \mu \neq \mu_0, \tag{1}\label{1} \end{align*} consider the following two related one-sided hypotheses testing problem:
\begin{align*} & H: \mu = \mu_0 \text{ versus } K_+: \mu > \mu_0, \tag{2}\label{2} \\ & H: \mu = \mu_0 \text{ versus } K_-: \mu < \mu_0. \tag{3}\label{3} \end{align*} Given an i.i.d. sample $$(X_1, \ldots, X_n) \sim N(\mu, 1)$$, it is well-known that (cf. Corollary 3.4.1 of Testing Statistical Hypotheses (Third Edition) by Lehmann and Romano) the UMP-$$\alpha$$ test for testing $$\eqref{2}$$ exists, and takes the form \begin{align*} & \tau_+(\bar{x}) = \begin{cases} 1 & \bar{x} > \mu_0 + \frac{1}{\sqrt{n}}z_\alpha, \\ 0 & \bar{x} < \mu_0 + \frac{1}{\sqrt{n}}z_\alpha. \end{cases} \tag{4}\label{4} \end{align*} Similarly, the UMP-$$\alpha$$ test for testing $$\eqref{3}$$ exists, and takes the form \begin{align*} & \tau_-(\bar{x}) = \begin{cases} 1 & \bar{x} < \mu_0 - \frac{1}{\sqrt{n}}z_\alpha, \\ 0 & \bar{x} > \mu_0 - \frac{1}{\sqrt{n}}z_\alpha. \end{cases} \tag{5}\label{5} \end{align*} In $$\eqref{4}$$ and $$\eqref{5}$$, $$z_\alpha$$ denotes the upper-$$\alpha$$ quantile of a standard normal distribution.

Now suppose if a UMP-$$\alpha$$ test $$\tau$$ exists for testing $$\eqref{1}$$, then it must also be a UMP-$$\alpha$$ test for testing $$\eqref{2}$$ and $$\eqref{3}$$. As a result, we have found two UMP-$$\alpha$$ tests $$\tau$$ and $$\tau_+$$ for testing $$\eqref{2}$$, which implies that power functions of $$\tau$$ and $$\tau_+$$ evaluated at each parameter in $$K_+$$ must be identical, i.e., $$E_\mu(\tau(\bar{X})) = E_\mu(\tau_+(\bar{X}))$$ for each $$\mu \in (\mu_0, +\infty)$$. That is, if denoting $$\tau - \tau_+$$ by $$g$$, then $$E_\mu(g(Y)) = 0$$ holds for all $$\mu \in (\mu_0, +\infty)$$, where $$Y \sim N(\mu, n^{-1})$$. Since the family $$\{N(\mu, n^{-1}): \mu \in (\mu_0, +\infty)\}$$ is complete (check this link for a short proof), $$E_\mu(g(Y)) = 0$$ for all $$\mu \in (\mu_0, +\infty)$$ implies that $$P_\mu(g(Y) = 0) = 1$$, i.e., $$g = 0$$ a.e., or $$\tau = \tau_+$$ a.e.. In the same manner, it can be shown that $$\tau = \tau_-$$ a.e.. Therefore, $$\tau_+ = \tau_-$$ a.e., which is in clear contradiction with $$\eqref{4}$$ and $$\eqref{5}$$, as they clearly don't agree almost everywhere on the real line.

Note that this question is just a special case of the following more general result (Problem 3.54 in Testing Statistical Hypotheses). A proof to this general result, which is of the same spirit as the proof above, can be found in this answer.

Let $$\theta$$ be a real number, and let $$X$$ have probability density (with respect to some measure $$\mu$$) \begin{align*} p_\theta(x) = C(\theta)e^{Q(\theta)T(x)}h(x), \end{align*} where $$Q$$ is strictly monotone. Given $$\theta_1 < \theta_2$$, prove there does not exist UMP test of size $$\alpha$$ (where $$0 < \alpha < 1$$) for testing \begin{align*} H: \theta = \theta_0 \text{ versus } K: \theta \neq \theta_0. \tag{\star}\label{star} \end{align*}

In words, there exists no UMP test for two-sided hypotheses $$\eqref{star}$$ if sample is from an one-parameter exponential family. This is in consistent with what your professor told you. The uniform family you mentioned in the end of your post, does NOT make this statement invalid because the uniform distribution family is not an exponential family!