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

:$Lambda(x)=frac\{\; L(\; heta\; \_\{0\}\; mid\; x)\}\{\; L\; (\; heta\; \_\{1\}\; mid\; x)\}\; leq\; eta\; mbox\{\; where\; \}\; P(Lambda(X)leq\; eta|H\_0)=alpha$

is the most powerful test of size "α" for a threshold η. If the test is most powerful for all $heta\_1\; in\; Theta\_1$, 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 — 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).

Proof

If we define the rejection region of the null hypothesis, as $R\_\{NP\}=\{\; X:\; frac\{L(\; heta\_\{0\},X)\}\{L(\; heta\_\{1\},X)\}\; leq\; eta\}$ , and any other test will have a different rejection region that we define as $R\_\{A\}$. Furthermore define the function of region, and parameter $P(R,\; heta)=int\_\{R\}\; L(\; heta|x)\; dx,$ 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$alpha=\; P(R\_\{NP\},\; heta\_\{0\})=P(R\_\{A\},\; heta\_\{0\})$, however it is useful to break these down into integrals over distinct regions.

:$P(R\_\{NP\}\; cap\; R\_\{A\},\; heta)\; +\; P(R\_\{NP\}\; cap\; R\_\{A\}^\{c\},\; heta)\; =\; P(R\_\{NP\},\; heta)$and:$P(R\_\{NP\}\; cap\; R\_\{A\},\; heta)\; +\; P(R\_\{NP\}^\{c\}\; cap\; R\_\{A\},\; heta)\; =\; P(R\_\{A\},\; heta)$

Setting $heta=\; heta\_\{0\}$ and equating the above two expression, yields that$P(R\_\{NP\}\; cap\; R\_\{A\}^\{c\},\; heta\_\{0\})\; =\; P(R\_\{NP\}^\{c\}\; cap\; R\_\{A\},\; heta\_\{0\})$

Comparing the power of the two tests, which are $P(R\_\{NP\},\; heta\_\{1\})$ and $P(R\_\{A\},\; heta\_\{1\})$ one can see that

:$P(R\_\{NP\},\; heta\_\{1\})\; geq\; P(R\_\{A\},\; heta\_\{1\})\; mbox\{\; if,\; and\; only\; if,\; \}P(R\_\{NP\}\; cap\; R\_\{A\}^\{c\},\; heta\_\{1\})\; geq\; P(R\_\{NP\}^\{c\}\; cap\; R\_\{A\},\; heta\_\{1\})$.

Now by the definition of $R\_\{NP\}$

:$P(R\_\{NP\}\; cap\; R\_\{A\}^\{c\},\; heta\_\{1\})=\; int\_\{R\_\{NP\}cap\; R\_\{A\}^\{c\; L(\; heta\_\{1\}|x)dx\; geq\; frac\{1\}\{eta\}\; int\_\{R\_\{NP\}cap\; R\_\{A\}^\{c\; L(\; heta\_\{0\}|x)dx\; =\; frac\{1\}\{eta\}P(R\_\{NP\}\; cap\; R\_\{A\}^\{c\},\; heta\_\{0\})$:$=\; frac\{1\}\{eta\}P(R\_\{NP\}^\{c\}\; cap\; R\_\{A\},\; heta\_\{0\})\; =\; frac\{1\}\{eta\}int\_\{R\_\{NP\}^\{c\}\; cap\; R\_\{A\; L(\; heta\_\{0\}|x)dx\; geq\; int\_\{R\_\{NP\}^\{c\}cap\; R\_\{A\; L(\; heta\_\{1\}|x)dx\; =\; P(R\_\{NP\}^\{c\}\; cap\; R\_\{A\},\; heta\_\{1\})$

Hence the inequality holds.

Example

Let $X\_1,dots,X\_n$ be a random sample from the $mathcal\{N\}(mu,sigma^2)$ 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

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

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

:$Lambda(mathbf\{x\})\; =\; frac\{Lleft(sigma\_1^2;mathbf\{x\}\; ight)\}\{Lleft(sigma\_0^2;mathbf\{x\}\; ight)\}\; =\; left(frac\{sigma\_1^2\}\{sigma\_0^2\}\; ight)^\{-n/2\}expleft\{-frac\{1\}\{2\}(sigma\_1^\{-2\}-sigma\_0^\{-2\})sum\_\{i=1\}^n\; left(x\_i-mu\; ight)^2\; ight\}.$

This ratio only depends on the data through $sum\_\{i=1\}^n\; left(x\_i-mu\; ight)^2$. Therefore, by the Neyman-Pearson lemma, the most powerful test of this type of hypothesis for this data will depend only on $sum\_\{i=1\}^n\; left(x\_i-mu\; ight)^2$. Also, by inspection, we can see that if $sigma\_1^2>sigma\_0^2$, then $Lambda(mathbf\{x\})$ is a decreasing function of $sum\_\{i=1\}^n\; left(x\_i-mu\; ight)^2$. So we should reject $H\_0$ if $sum\_\{i=1\}^n\; left(x\_i-mu\; ight)^2$ 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]