If there exists a threshold k 0 such that. then we can take k = k 0 γ =0 and achieve a test of size α. c 2004 by Mark R. Bell,

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1 Recall The Neyman-Pearson Lemma Neyman-Pearson Lemma: Let Θ = {θ 0, θ }, and let F θ0 (x) be the cdf of the random vector X under hypothesis and F θ (x) be its cdf under hypothesis. Assume that the cdfs F θi (x) have corresponding pdfs or pmfs f θi (x), i =0,. Then a test of the form φ(x) =, for f θ (x) kf θ0 (x), γ, for f θ (x) =kf θ0 (x), 0, for f θ (x) kf θ0 (x), for some k 0 and some 0 γ is the most powerful test of size α for testing hypothesis : θ = θ 0 versus : θ = θ. Choosing the Threshold for the Neyman-Pearson Test To choose a threshold k and parameter γ to produce a N-P test of the form, for f θ (x) kf θ0 (x), φ(x) = γ, for f θ (x) =kf θ0 (x), 0, for f θ (x) kf θ0 (x), with the desired size α, we note that P θ0 ({f θ (X) kf θ0 (X)}) α =E θ0 [φ(x)] = P θ0 ({f θ (X) kf θ0 (X)}) + γp θ0 ({f θ (X) =kf θ0 (X)}) = P θ0 ({f θ (X) kf θ0 (X)}) + γp θ0 ({f θ (X) =kf θ0 (X)}).

2 α = P θ0 ({f θ (X) kf θ0 (X)}) + γp θ0 ({f θ (X) =kf θ0 (X)}). If there exists a threshold k 0 such that This term can be made equal to zero (e.g., γ =0) P θ0 ({f θ (X) k 0 f θ0 (X)}) = α then we can take k = k 0 γ =0 and achieve a test of size α. P θ0 ({f θ (X) k 0 f θ0 (X)}) α P θ0 ({f θ (X) k 0 f θ0 (X)}) Inclusion of k 0 results in strict bracketing of α This can only occur when P θ0 ({f θ (X) =k 0 f θ0 (X)}) 0. In this case, we select k = k 0 such that the bracketing occurs and then solve for γ to achieve a size α test. The resulting value of γ is γ = P θ 0 ({f θ (X) k 0 f θ0 (X)}) ( α). P θ0 ({f θ (X) =k 0 f θ0 (X)})

3 The Likelihood Ratio Test We can rewrite the Neyman-Pearson decision rule in terms of the Likelihood Ratio L(x) = f θ (x) f θ0 (x). The Neyman-Pearson test can be rewritten as 8, for L(x) k, (x) =, for for L(x) = k, : 0, for L(x) k. (x) = 8, for L(x) k,, for for L(x) = k, : 0, for L(x) k. If there is a k 0 such that take k = k 0. P θ0 ({L(X) k 0 (X)}) = α If not, then find a k 0 such that P θ0 ({L(X) k 0 }) α P θ0 ({L(X) k 0 )}) and take k = k 0 and γ = P θ 0 ({L(X) k 0 }) ( α) P θ0 ({L(X) =k 0 })

4 Because L(X) is a function of a random vector X, it is itself a scalar random variable, and it takes on only nonnegative values. If P θ0 ({L(X) =k}) = 0, then the threshold k achieving false alarm probability α can be found by solving α = P θ0 ({L(X) k}) = for k, where f L,θ0 (l) is the density function of L(X) under k f L,θ0 (l) dl, We will find it convenient to use the log-likelihood ratio l(x) = log (L(X)). Because log( ) is a monotonically increasing function on (0, ), the most powerful test of size α equivalent to the likelihood ratio test will take the form φ(x) = where the threshold l 0 = log k., for l(x) l 0, γ, for l(x) =l 0, 0, for l(x) l 0, Working with l(x) often yields simpler results than L(X). Notation: L(X) k or l(x) l 0,

5 Example Let X be a Gaussian random variable. Under : X N [0, σ 2 ] Under : X N [µ, σ 2 ] Find the most powerful test of size α, and determine an expression for the power β as a function of α, µ, and σ. The likelihood ratio is L(X) = f (X) f 0 (X) = { } 2πσ exp (X µ) 2 2σ 2 2πσ exp { } X 2 2σ 2 2Xµ µ 2 µx =exp =exp µ k The log-likelihood ratio is l(x) = 2µX µ2 2σ 2 l 0 which simplifies to X σ2 µ ( µ 2 2σ 2 + l 0 ) = µ 2 + σ2 l 0 µ So our test is of the form {, for Xλ, φ(x) = 0, for X λ, where λ = µ 2 + σ2 l 0 µ

6 φ(x) = {, for Xλ, 0, for X λ, X N [0, σ 2 ] under α =E H0 [φ(x)] = λ f 0 (x) dx = Φ Solving for λ achieving a size α test: λ α = σφ ( α) ( ) λ σ X N [µ, σ 2 ] under β =E [φ(x)] = f (x) dx = λ α µ = ( ) µ. Example 2 The photon count N observed by a laser radar is a Poisson random variable: Under : N Poisson(λ 0 ) Under : N Poisson(λ ) We assume λ λ 0. We have probability mass functions (pmfs): p 0 (n) = λn 0 e λ 0 n! p (n) = λn e λ n!, n =0,, 2,...,, n =0,, 2,...,

7 The likelihood ratio is L(N) = p (N) p 0 (N) = λn e λ /N! λ N 0 e λ 0 /N! = ( λ λ 0 ) N e (λ λ 0 ) k The log-likelihood ratio is l(n) = ln L(N) =N ln ( λ λ 0 ) (λ λ 0 ) Hence we can express the test in the form l 0 = ln k. N l 0 +(λ λ 0 ) ln (λ /λ 0 ) 0 = η. N l 0 +(λ λ 0 ) ln (λ /λ 0 ) 0 = η. The most powerful test of size α is φ(n) = {, for Nη, γ, for N = η, 0, for Nη,

8 The power β of this size α test is β =E [φ(n)] = P ({N η α })+γ α P ({N = η α }) = P ({N η α })+γ α P ({N = η α }) = η α n=0 λ n e λ n! + γ α λ η α e λ η α! The Receiver Operating Characteristic (ROC) The performance of a binary test is characterized by the pair (α, β) For a likelihood ratio test, we achieve different pairs (α(k), β(k)) for different threshold values k The locus of points {(α(k), β(k)); k (0, )} specifies all achievable (α, β) that can be obtained by varying the threshold k. Such a curve is called a Receiver Operating Characteristic (ROC)

9 A typical ROC appears as follows: β = P D ROC Chance Line 0 α = P FA β = P D Properties of the ROCs of Likelihood Ratio Tests ROC Chance Line. All continuous likelihood ratio tests have ROCs that are convex downward. 0 α = P FA 2. Points on the chance line (β = α) can be achieved without observing any data by picking hypothesis at random with probability α. (i.e., flipping a biased coin with probability α of coming up heads ).

10 β = P D ROC Chance Line Properties of the ROCs of Likelihood Ratio Tests (Continued) 0 α = P FA 3. All continuous likelihood ratio tests have ROC s that are above the chance line. This is a consequence of property, because the points (α, β) =(0, 0) and (α, β) =(, ) are contained on all ROC s. 4. The slope of the ROC at any particular point is given by the threshold achieving that operating point, i.e., dβ dα = dβ/dk dα/dk = f L,θ (l) f L,θ0 (l) = k 0