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← Older revision		Revision as of 18:33, 26 September 2013
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	I'~~m Lesli~~ and I ~~live with my husband~~ and ~~our three children~~ in ~~Cambridge~~, in the ~~MA south part~~. ~~My hobbies~~ are ~~People watching~~, ~~Jewelry making~~ and ~~Microscopy~~.<br><br>~~Feel free~~ to ~~visit my web page~~ :: [http://~~014~~.~~me redis url shortener~~]		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>: ''θ'' = ''θ''<sub>0</sub> and ''H''<sub>1</sub>: ''θ'' = ''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 ''α'']] 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>.

			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 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]]

en>VolkovBot: r2.7.2) (Robot: Adding et:Helikoid

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I'm Lesli and I live with my husband and our three children in Cambridge, in the MA south part. My hobbies are People watching, Jewelry making and Microscopy.<br><br>Feel free to visit my web page :: [http://014.me redis url shortener]

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en>VolkovBot: r2.7.2) (Robot: Adding et:Helikoid