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In [[statistics]], the '''Neyman–Pearson [[lemma (mathematics)|lemma]]''', named after [[Jerzy Neyman]] and [[Egon Pearson]], states that when performing a [[statistical hypothesis testing|hypothesis test]] between two point  hypotheses ''H''<sub>0</sub>:&nbsp;''θ''&nbsp;=&nbsp;''θ''<sub>0</sub> and ''H''<sub>1</sub>:&nbsp;''θ''&nbsp;=&nbsp;''θ''<sub>1</sub>, then the [[likelihood-ratio test]] which rejects ''H''<sub>0</sub> in favour of ''H''<sub>1</sub> when


:<math>\Lambda(x)=\frac{ L( \theta _0 \mid x)}{ L (\theta _1 \mid x)} \leq \eta</math>
where
:<math>P(\Lambda(X)\leq \eta\mid H_0)=\alpha </math>


is the '''most [[Statistical power|powerful]] test''' of [[Type I and type II errors|size ''&alpha;'']] for a threshold η. If the test is most powerful for all <math>\theta_1 \in \Theta_1</math>, it is said to be [[uniformly most powerful]] (UMP) for alternatives in the set <math>\Theta_1 \, </math>.
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In practice, the [[likelihood ratio]] is often used directly to construct tests &mdash; 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 &mdash; for this one considers algebraic manipulation of the ratio to see if there are key statistics in it related to the size of the ratio (i.e. whether a large statistic corresponds to a small ratio or to a large one).
 
==Proof==
Define the rejection region of the null hypothesis for the NP test as
:<math>R_{NP}=\left\{ x: \frac{L(\theta_{0}|x)}{L(\theta_{1}|x)} \leq \eta\right\} .</math>
 
Any other test will have a different rejection region that we define as <math>R_A</math>. Furthermore, define the probability of the data falling in region R, given parameter <math>\theta</math> as
 
:<math>P(R,\theta)=\int_R L(\theta|x)\, dx, </math>
 
For both tests to have size <math>\alpha</math>, it must be true that
 
:<math>\alpha= P(R_{NP}, \theta_0)=P(R_A, \theta_0) \,.</math>
 
It will be useful to break these down into integrals over distinct regions:
 
:<math>P(R_{NP},\theta) = P(R_{NP} \cap R_A, \theta) + P(R_{NP} \cap R_A^c, \theta),</math>
 
and
 
:<math>P(R_A,\theta) = P(R_{NP} \cap R_A, \theta) + P(R_{NP}^c \cap R_A, \theta).</math>
 
Setting <math>\theta=\theta_0</math> and equating the above two expression yields that
 
:<math>P(R_{NP} \cap R_A^c, \theta_0) =  P(R_{NP}^c \cap R_A, \theta_0).</math>
 
Comparing the powers of the two tests, <math>P(R_{NP},\theta_1)</math> and <math>P(R_A,\theta_1)</math>, one can see that
 
:<math>P(R_{NP},\theta_1) \geq P(R_A,\theta_1) \iff
P(R_{NP} \cap R_A^c, \theta_1) \geq P(R_{NP}^c \cap R_A, \theta_1). </math>
 
Now by the definition of <math>R_{NP}</math>,
 
:<math> P(R_{NP} \cap R_A^c, \theta_1)= \int_{R_{NP}\cap R_A^c} L(\theta_{1}|x)\,dx \geq \frac{1}{\eta} \int_{R_{NP}\cap R_A^c} L(\theta_0|x)\,dx = \frac{1}{\eta}P(R_{NP} \cap R_A^c, \theta_0)</math>
:<math> = \frac{1}{\eta}P(R_{NP}^c \cap R_A, \theta_0) = \frac{1}{\eta}\int_{R_{NP}^c \cap R_A} L(\theta_{0}|x)\,dx \geq \int_{R_{NP}^c\cap R_A} L(\theta_{1}|x)dx  = P(R_{NP}^c \cap R_A, \theta_1).</math>
 
Hence the [[Inequality (mathematics)|inequality]] holds.
 
==Example==
 
Let <math>X_1,\dots,X_n</math> be a random sample from the <math>\mathcal{N}(\mu,\sigma^2)</math> distribution where the mean <math>\mu</math> is known, and suppose that we wish to test for <math>H_0:\sigma^2=\sigma_0^2</math> against <math>H_1:\sigma^2=\sigma_1^2</math>. The likelihood for this set of [[normally distributed]] data is
 
:<math>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\}.</math>
 
We can compute the [[likelihood ratio]] to find the key statistic in this test and its effect on the test's outcome:
 
:<math>\Lambda(\mathbf{x}) = \frac{L\left(\sigma_0^2;\mathbf{x}\right)}{L\left(\sigma_1^2;\mathbf{x}\right)} =
\left(\frac{\sigma_0^2}{\sigma_1^2}\right)^{-n/2}\exp\left\{-\frac{1}{2}(\sigma_0^{-2}-\sigma_1^{-2})\sum_{i=1}^n \left(x_i-\mu\right)^2\right\}.</math>
 
This ratio only depends on the data through <math>\sum_{i=1}^n \left(x_i-\mu\right)^2</math>. Therefore, by the Neyman–Pearson lemma, the most [[Statistical power|powerful]] test of this type of [[Statistical hypothesis testing|hypothesis]] for this data will depend only on <math>\sum_{i=1}^n \left(x_i-\mu\right)^2</math>. Also, by inspection, we can see that if <math>\sigma_1^2>\sigma_0^2</math>, then <math>\Lambda(\mathbf{x})</math> is a [[decreasing function]] of <math>\sum_{i=1}^n \left(x_i-\mu\right)^2</math>. So we should reject <math>H_0</math> if <math>\sum_{i=1}^n \left(x_i-\mu\right)^2</math> is sufficiently large. The rejection threshold depends on the [[Type I and type II errors|size]]  of the test. In this example, the test statistic can be shown to be a scaled Chi-square distributed random variable and an exact critical value can be obtained.
 
==See also==
* [[Statistical power]]
 
==References==
* {{cite doi|10.1098/rsta.1933.0009}}
*[http://cnx.org/content/m11548/latest/ cnx.org: Neyman–Pearson criterion]
 
==External links==
* Cosma Shalizi, a professor of statistics at Carnegie Mellon University, gives an intuitive derivation of the Neyman–Pearson Lemma [http://cscs.umich.edu/~crshalizi/weblog/630.html using ideas from economics]
 
{{DEFAULTSORT:Neyman-Pearson Lemma}}
[[Category:Statistical theorems]]
[[Category:Statistical tests]]
[[Category:Articles containing proofs]]

Latest revision as of 15:41, 5 September 2014


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