Uniformly most powerful test

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.

Setting

$\phi (x)={\begin{cases}1&{\text{if }}x\in R\\0&{\text{if }}x\in A\end{cases}}$ Formal definition

$\sup _{\theta \in \Theta _{0}}\;\operatorname {E} _{\theta }\phi '(X)=\alpha '\leq \alpha =\sup _{\theta \in \Theta _{0}}\;\operatorname {E} _{\theta }\phi (X)\,$ we have

$\operatorname {E} _{\theta }\phi '(X)=1-\beta '\leq 1-\beta =\operatorname {E} _{\theta }\phi (X)\quad \forall \theta \in \Theta _{1}.$ The Karlin-Rubin theorem

The Karlin-Rubin theorem can be regarded as an extension of the Neyman-Pearson lemma for composite hypotheses. 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)={\begin{cases}1&{\text{if }}x>x_{0}\\0&{\text{if }}x where $x_{0}$ is chosen such that $\operatorname {E} _{\theta _{0}}\phi (X)=\alpha$ Important case: The exponential family

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.

Example

$f_{\theta }(X)=(2\pi )^{-MN/2}|R|^{-M/2}\exp \left\{-{\frac {1}{2}}\sum _{n=0}^{M-1}(X_{n}-\theta m)^{T}R^{-1}(X_{n}-\theta m)\right\}=$ $=(2\pi )^{-MN/2}|R|^{-M/2}\exp \left\{-{\frac {1}{2}}\sum _{n=0}^{M-1}(\theta ^{2}m^{T}R^{-1}m)\right\}$ $\exp \left\{-{\frac {1}{2}}\sum _{n=0}^{M-1}X_{n}^{T}R^{-1}X_{n}\right\}\exp \left\{\theta m^{T}R^{-1}\sum _{n=0}^{M-1}X_{n}\right\}$ 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)={\begin{cases}1&{\text{if }}T>t_{0}\\0&{\text{if }}T $\operatorname {E} _{\theta _{0}}\phi (T)=\alpha$ Further discussion

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.