Uniformly most powerful test

From Wikipedia, the free encyclopedia
  (Redirected from UMP test)
Jump to: navigation, search

In statistical hypothesis testing, a uniformly most powerful (UMP) test is a hypothesis test which has the greatest power 1 − β among all possible tests of a given size α. For example, according to the Neyman–Pearson lemma, the likelihood-ratio test is UMP for testing simple (point) hypotheses.


Let X denote a random vector (corresponding to the measurements), taken from a parametrized family of probability density functions or probability mass functions f_{\theta}(x), which depends on the unknown deterministic parameter \theta \in \Theta. The parameter space \Theta is partitioned into two disjoint sets \Theta_0 and \Theta_1. Let H_0 denote the hypothesis that \theta \in \Theta_0, and let H_1 denote the hypothesis that \theta \in \Theta_1. The binary test of hypotheses is performed using a test function \phi(x).

\phi(x) = 
1 & \text{if } x \in R \\
0 & \text{if } x \in A

meaning that H_1 is in force if the measurement X \in R and that H_0 is in force if the measurement X \in A. Note that A \cup R is a disjoint covering of the measurement space.

Formal definition[edit]

A test function \phi(x) is UMP of size \alpha if for any other test function \phi'(x) satisfying

\sup_{\theta\in\Theta_0}\; \operatorname{E}_\theta\phi'(X)=\alpha'\leq\alpha=\sup_{\theta\in\Theta_0}\; \operatorname{E}_\theta\phi(X)\,

we have

 \forall \theta \in \Theta_1, \quad \operatorname{E}_\theta\phi'(X)=1-\beta'\leq 1-\beta=\operatorname{E}_\theta\phi(X).

The Karlin–Rubin theorem[edit]

The Karlin–Rubin theorem can be regarded as an extension of the Neyman–Pearson lemma for composite hypotheses.[1] Consider a scalar measurement having a probability density function parameterized by a scalar parameter θ, and define the likelihood ratio  l(x) = f_{\theta_1}(x) / f_{\theta_0}(x). If l(x) is monotone non-decreasing, in x, for any pair \theta_1 \geq \theta_0 (meaning that the greater x is, the more likely H_1 is), then the threshold test:

\phi(x) = 
1 & \text{if } x > x_0 \\
0 & \text{if } x < x_0
where x_0 is chosen such that \operatorname{E}_{\theta_0}\phi(X)=\alpha

is the UMP test of size α for testing  H_0: \theta \leq \theta_0 \text{ vs. } H_1: \theta > \theta_0 .

Note that exactly the same test is also UMP for testing  H_0: \theta = \theta_0 \text{ vs. } H_1: \theta > \theta_0 .

Important case: The exponential family[edit]

Although the Karlin-Rubin theorem may seem weak because of its restriction to scalar parameter and scalar measurement, it turns out that there exist a host of problems for which the theorem holds. In particular, the one-dimensional exponential family of probability density functions or probability mass functions with

f_\theta(x) = c(\theta)h(x)\exp(\pi(\theta)T(x))

has a monotone non-decreasing likelihood ratio in the sufficient statistic T(x), provided that \pi(\theta) is non-decreasing.


Let X=(X_0 , X_1 ,\dots , X_{M-1}) denote i.i.d. normally distributed N-dimensional random vectors with mean \theta m and covariance matrix R. We then have

f_\theta (X) = (2 \pi)^{-M N / 2} |R|^{-M / 2} \exp \left\{-\frac{1}{2} \sum_{n=0}^{M-1}(X_n - \theta m)^T R^{-1}(X_n - \theta m) \right\} =
 = (2 \pi)^{-M N / 2} |R|^{-M / 2} \exp \left\{-\frac{1}{2} \sum_{n=0}^{M-1}(\theta^2 m^T R^{-1} m) \right\}\exp \left\{-\frac{1}{2} \sum_{n=0}^{M-1}X_n^T R^{-1} X_n \right\} \exp \left\{\theta m^T R^{-1} \sum_{n=0}^{M-1}X_n \right\}

which is exactly in the form of the exponential family shown in the previous section, with the sufficient statistic being

T(X) = m^T R^{-1} \sum_{n=0}^{M-1}X_n.

Thus, we conclude that the test

\phi(T) = 
1 & \text{if } T > t_0 \\
0 & \text{if } T < t_0
\operatorname{E}_{\theta_0} \phi (T) = \alpha

is the UMP test of size \alpha for testing H_0: \theta \leq \theta_0 vs. H_1: \theta > \theta_0

Further discussion[edit]

Finally, we note that in general, UMP tests do not exist for vector parameters or for two-sided tests (a test in which one hypothesis lies on both sides of the alternative). Why is it so?

The reason is that in these situations, the most powerful test of a given size for one possible value of the parameter (e.g. for \theta_1 where \theta_1 > \theta_0) is different from the most powerful test of the same size for a different value of the parameter (e.g. for \theta_2 where \theta_2 < \theta_0). As a result, no test is uniformly most powerful.


  1. ^ Casella, G.; Berger, R.L. (2008), Statistical Inference, Brooks/Cole. ISBN 0-495-39187-5 (Theorem 8.3.17)

Further reading[edit]

  • L. L. Scharf, Statistical Signal Processing, Addison-Wesley, 1991, section 4.7.