# Neyman-Pearson lemma

Neyman-Pearson lemma

In statistics, the Neyman-Pearson lemma states that when performing a hypothesis test between two point hypotheses "H"0: "θ"="θ"0 and "H"1: "θ"="θ"1, then the likelihood-ratio test which rejects "H"0 in favour of "H"1 when

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

is the most powerful test of size "α" for a threshold η. If the test is most powerful for all $heta_1 in Theta_1$, 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 &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 is related to the size of the ratio (i.e. whether a large statistic corresponds to a small ratio or to a large one).

Proof

If we define the rejection region of the null hypothesis, as $R_\left\{NP\right\}=\left\{ X: frac\left\{L\left( heta_\left\{0\right\},X\right)\right\}\left\{L\left( heta_\left\{1\right\},X\right)\right\} leq eta\right\}$ , and any other test will have a different rejection region that we define as $R_\left\{A\right\}$. Furthermore define the function of region, and parameter $P\left(R, heta\right)=int_\left\{R\right\} L\left( heta|x\right) dx,$ hence this is the probability of the data falling in region R, given parameter $heta$.

For both tests to have significance level $alpha$, it must be true that$alpha= P\left(R_\left\{NP\right\}, heta_\left\{0\right\}\right)=P\left(R_\left\{A\right\}, heta_\left\{0\right\}\right)$, however it is useful to break these down into integrals over distinct regions.

:$P\left(R_\left\{NP\right\} cap R_\left\{A\right\}, heta\right) + P\left(R_\left\{NP\right\} cap R_\left\{A\right\}^\left\{c\right\}, heta\right) = P\left(R_\left\{NP\right\}, heta\right)$and:$P\left(R_\left\{NP\right\} cap R_\left\{A\right\}, heta\right) + P\left(R_\left\{NP\right\}^\left\{c\right\} cap R_\left\{A\right\}, heta\right) = P\left(R_\left\{A\right\}, heta\right)$

Setting $heta= heta_\left\{0\right\}$ and equating the above two expression, yields that$P\left(R_\left\{NP\right\} cap R_\left\{A\right\}^\left\{c\right\}, heta_\left\{0\right\}\right) = P\left(R_\left\{NP\right\}^\left\{c\right\} cap R_\left\{A\right\}, heta_\left\{0\right\}\right)$

Comparing the power of the two tests, which are $P\left(R_\left\{NP\right\}, heta_\left\{1\right\}\right)$ and $P\left(R_\left\{A\right\}, heta_\left\{1\right\}\right)$ one can see that

:$P\left(R_\left\{NP\right\}, heta_\left\{1\right\}\right) geq P\left(R_\left\{A\right\}, heta_\left\{1\right\}\right) mbox\left\{ if, and only if, \right\}P\left(R_\left\{NP\right\} cap R_\left\{A\right\}^\left\{c\right\}, heta_\left\{1\right\}\right) geq P\left(R_\left\{NP\right\}^\left\{c\right\} cap R_\left\{A\right\}, heta_\left\{1\right\}\right)$.

Now by the definition of $R_\left\{NP\right\}$

:$P\left(R_\left\{NP\right\} cap R_\left\{A\right\}^\left\{c\right\}, heta_\left\{1\right\}\right)= int_\left\{R_\left\{NP\right\}cap R_\left\{A\right\}^\left\{c L\left( heta_\left\{1\right\}|x\right)dx geq frac\left\{1\right\}\left\{eta\right\} int_\left\{R_\left\{NP\right\}cap R_\left\{A\right\}^\left\{c L\left( heta_\left\{0\right\}|x\right)dx = frac\left\{1\right\}\left\{eta\right\}P\left(R_\left\{NP\right\} cap R_\left\{A\right\}^\left\{c\right\}, heta_\left\{0\right\}\right)$:$= frac\left\{1\right\}\left\{eta\right\}P\left(R_\left\{NP\right\}^\left\{c\right\} cap R_\left\{A\right\}, heta_\left\{0\right\}\right) = frac\left\{1\right\}\left\{eta\right\}int_\left\{R_\left\{NP\right\}^\left\{c\right\} cap R_\left\{A L\left( heta_\left\{0\right\}|x\right)dx geq int_\left\{R_\left\{NP\right\}^\left\{c\right\}cap R_\left\{A L\left( heta_\left\{1\right\}|x\right)dx = P\left(R_\left\{NP\right\}^\left\{c\right\} cap R_\left\{A\right\}, heta_\left\{1\right\}\right)$

Hence the inequality holds.

Example

Let $X_1,dots,X_n$ be a random sample from the $mathcal\left\{N\right\}\left(mu,sigma^2\right)$ distribution where the mean $mu$ 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

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

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

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

This ratio only depends on the data through $sum_\left\{i=1\right\}^n left\left(x_i-mu ight\right)^2$. Therefore, by the Neyman-Pearson lemma, the most powerful test of this type of hypothesis for this data will depend only on $sum_\left\{i=1\right\}^n left\left(x_i-mu ight\right)^2$. Also, by inspection, we can see that if $sigma_1^2>sigma_0^2$, then $Lambda\left(mathbf\left\{x\right\}\right)$ is a decreasing function of $sum_\left\{i=1\right\}^n left\left(x_i-mu ight\right)^2$. So we should reject $H_0$ if $sum_\left\{i=1\right\}^n left\left(x_i-mu ight\right)^2$ is sufficiently small. The rejection threshold depends on the size of the test.

ee also

* Statistical power
* Receiver operating characteristic

References

* cite journal
title=On the Problem of the Most Efficient Tests of Statistical Hypotheses
author=Jerzy Neyman, Egon Pearson
journal=Philosophical Transactions of the Royal Society of London. Series A, Containing Papers of a Mathematical or Physical Character
volume=231
year=1933
pages=289–337
url=http://links.jstor.org/sici?sici=0264-3952%281933%29231%3C289%3AOTPOTM%3E2.0.CO%3B2-X
doi=10.1098/rsta.1933.0009

* [http://cnx.org/content/m11548/latest/ cnx.org: Neyman-Pearson criterion]

External links

* MIT OpenCourseWare lecture notes: [http://ocw.mit.edu/NR/rdonlyres/Mathematics/18-443Fall2003/18B765F6-A398-48BF-A893-49A4965DED98/0/lec19.pdf most powerful tests] , [http://ocw.mit.edu/NR/rdonlyres/Mathematics/18-443Fall2003/D6F12E47-A9A2-4FE0-AC3C-588B6A5EE5B6/0/lec20.pdf uniformly most powerful tests]

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