Neyman-Pearson lemma

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In statistics, the Neyman-Pearson lemma states that when performing a hypothesis test between two point hypotheses H₀: θ=θ₀ and H₁: θ=θ₁, then the likelihood-ratio test which rejects H₀ in favour of H₁ when

$\Lambda(x)=\frac{ L( \theta _{0} \mid x)}{ L (\theta _{1} \mid x)} \leq \eta \mbox{ where } P(\Lambda(X)\leq \eta|H_0)=\alpha$

is the most powerful test of size α for a threshold η. If the test is most powerful for all $\theta_1 \in \Theta_1$ , it is said to be uniformly most powerful (UMP) for alternatives in the set $\Theta_1 \,$ .

In practice, the likelihood ratio is often used directly to construct tests — see Likelihood-ratio test. However it can also be used to suggest particular test-statistics that might be of interest or to suggest simplified tests — for this one considers algebraic manipulation of the ratio to see if there are key statistics in it is related to the size of the ratio (i.e. whether a large statistic corresponds to a small ratio or to a large one).

1 Example
2 See also
3 References
4 External links

[edit] Example

Let $X_1,\dots,X_n$ be a random sample from the $N (μ,σ 2)$ distribution where $μ$ is known, and suppose that we wish to test for $H_0:\sigma^2=\sigma_0^2$ against $H_1:\sigma^2=\sigma_1^2$ .

The likelihood for this set of normally distributed data is

$L\left(\sigma^2;\mathbf{x}\right)\propto \left(\sigma^2\right)^{-n/2} \exp\left\{-\frac{\sum_{i=1}^n \left(x_i-\mu\right)^2}{2\sigma^2}\right\}.$

We can compute the likelihood ratio to find the key statistic in this test and its effect on the test's outcome:

$\Lambda(\mathbf{x}) = \frac{L\left(\sigma_1^2;\mathbf{x}\right)}{L\left(\sigma_0^2;\mathbf{x}\right)} = \left(\frac{\sigma_1^2}{\sigma_0^2}\right)^{-n/2}\exp\left\{-\frac{1}{2}(\sigma_1^{-2}-\sigma_0^{-2})\sum_{i=1}^n \left(x_i-\mu\right)^2\right\}.$

This ratio only depends on the data through $\sum_{i=1}^n \left(x_i-\mu\right)^2$ . Therefore, by the Neyman-Pearson lemma, the most powerful test of this type of hypothesis for this data will depend only on $\sum_{i=1}^n \left(x_i-\mu\right)^2$ . Also, by inspection, we can see that if $\sigma_1^2>\sigma_0^2$ , then $\Lambda(\mathbf{x})$ is a decreasing function of $\sum_{i=1}^n \left(x_i-\mu\right)^2$ . So we should reject $H 0$ if $\sum_{i=1}^n \left(x_i-\mu\right)^2$ is sufficiently small. The rejection threshold depends on the size of the test.