- 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 thelikelihood-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 — seeLikelihood-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 thatalpha= 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 thatP(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]
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