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Mathematical Background of the P3r2 Test

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Mathematical Background of the P3r2 Test Mathematical background of the p3r2 test Niklas Hohmann∗ December 14, 2017 1 Motivation There are two pivotal principles for the construction of the test. The first one is • propability density functions (pdfs) are needed to construct certain statistical methods and to prove their optimality In the case of the p3r2 test, the statistical method used is a likelihood-ratio test. The notion of optimality for tests used here is a uniform most powerful (UMP) test. The second principle is that • a Poisson point process (PPP) can be modeled as a random measure At first sight, this complicates things. For example, the observation times y = (y1; y2; : : : ; yN ) (1) of a PPP are not taken as elements of RN , but are interpreted as a measure µy on the real numbers, which is the sum of Dirac measures: N X µy = δyj : (2) j=1 But this approach has got advantages that will become clear soon. The general goal in the next few pages is to establish a pdf for PPPs to construct a UMP test for PPPs. Ignoring more technical details, a random measure Xi is a random element with values in M(R), the set of all measures on R. Since a PPP can be ∗FAU Erlangen-Nuremberg, email: [email protected] 1 modeled as a random measure, it can be imagined as a procedure that is randomly choosing a measure similar to the one in equation (2) that can again be interpreted as observation times. Since Xi is a random element with values in M(R), the distribution Pi of Xi is a probability measure on M(R). So Pi assigns probabilities to sets of measures. Given a second random measure Xj with distribution Pj, the relationship of their distributions Pi and Pj can be examined by applying methods of measure theory. It is known from the Radon-Nikodym theorem that the existence of a pdf of a (probability) measure Pi in reference to a (probability) measure Pj is equal to absolute continuity of Pi in reference to Pj, which is written as Pi Pj. So to establish a pdf for PPPs, absolute continuity between the distributions of the relevant random measures has to be established. 2 Basic idea For simplification, look at the measures on the positive real numbers. Assume µr is a finite measure on [0; 1). Then it is uniquely determined by its Laplace transform Z Lr(t) = exp(−tx) µr(dx) : (3) If µr λM , there exists a density function fr of µr in reference to λM . So we get the alternative representation Z Lr(t) = fr(x) exp(−tx) λM (dx) (4) for the Laplace transform. Now, assume it is unknown whether or not µr λM , but it can be shown that Z Lr(t) = g(x) exp(−tx) λM (dx) (5) for all t and a given function g. So g behaves like a density in combination with the Laplace transform. If this property can be expanded and used to show that g behaves like a density in combination with all indicator functions of measurable sets, then g really is a density, which means µr λM and fr = g almost everywhwere. The same approach is used for random measures: first, a function is identified that behaves like a density in combination with the Laplace transform. This is done in section 4. Then, it is shown that this property expands and the function really is a density. This is done in section 5. 2 3 Modelling and Notation In terms of notation, I follow [2] and use the conventions ln(0) := −∞ and 1 1 exp(−∞) = 0. Choose fr 2 L (R; B(R); λ) := L (λ) satisfying 1. there is a compact, connected set M ⊂ R with supp(fr) ⊂ M for all r 2. fr(x) ≥ 0 for all x 2 R 3. R f(x)λ(dx) > 0 Here, r is an arbitrary index from an index set R and λ is the Lebesgue measure. Since everything outside the set M is irrelevant for modelling, I will use λM , the Lebesgue measure restricted to M instead of λ. The fr will serve as the rate functions of the PPPs. Define the measures Z µr(A) := fr(x) λM (dx) : (6) A 1 Then µr λM and fr is the density function of µr in reference to λM . Last, we denote the PPP with intensity measure µr with Xr and the PPP with intensity measure λM with XM . 4 Identifying a possible pdf candidate In the case of a random measure Xr, the Laplace transform is given by Z Lr(f) = E exp − f(x) Xr(dx) : (7) + + Here, f 2 Cc , where Cc is the set of positive continuous functions with compact support. If Xr is a PPP with intensity measure µr, this simplifies to Z Lr(f) = exp exp(−f(x)) − 1 µr(dx) (8) Z = exp (exp(−f(x)) − 1)fr(x) λM (dx) : (9) Here, fr is the density of µr in reference to λM . Like in section 2, we need a function dr that behaves like a density in combination with the Laplace transform, meaning Z Lr(f) = E dr(XM ) exp − f(x) XM (dx) : (10) 1it is not a probability density function 3 I will show that the function Z dr(XM ) := cr · exp ln(fr(x))XM (dx) (11) with Z cr := exp 1 − fr(x)λM (dx) (12) has got that property. First, assume the rate function fr is a simple function, so n X fr(x) = αi1Ai (x) (13) i=1 with αi > 0 and measurable sets Ai. For technical reasons, add the set n A0 = M n [i=1Ai with α0 = 0, so n X fr(x) = αi1Ai (x) : (14) i=0 + Now, take any f 2 Cc . Then for every Ai, there is a series of simple functions (~sk;i)k2N withs ~k;i(x) % f(x)1Ai (x) almost everywhere for k ! 1. Then si;k(x) = (ln(αi) − s~i;k(x))1Ai (x) & (ln(αi) − f(x))1Ai (x) (15) for k ! 1. The function n X sk(x) = si;k(x) (16) i=0 is also a simple function and has got a representation n Xk sk(x) = bj;k1Bj;k (x) (17) j=1 with measurable Bj;k. Then sk(x) & ln(fr(x)) − f(x) (18) almost everywhere for k ! 1. Further, the sets Bj;k can be chosen to be com- patible with the sets Ai, meaning that for all i, there is a Ji ⊂ f1; 2; 3; : : : ; nkg satisfying [ Bj;k = Ai : (19) j2Ji 4 So we get an alternative representation of fr: n Xk fr(x) = α~j1Bj;k (20) j=1 with αi =α ~j for all j 2 Ji. Furthermore bj;k = ln(αi) − βj;k for j 2 Ji. So X s~i;k(x) = βj;k1Bj;k (x) (21) j2Ji and n n Xk Xk s~k(x) = s~i;k(x) = βj;k1Bj;k (x) % f(x) (22) i=1 j=1 almost everywhere for k ! 1. With fr being a simple function, the equation (12) reduces to n ! Z Xk cr = exp 1 − α~j1Bj;k λM (dx) (23) j=1 n ! Xk = exp (1 − α~j)λM (Bj;k) (24) j=1 n Yk = exp ((1 − α~j)λM (Bj;k)) : (25) j=1 + Now, we start showing equation (10). First, fix any f 2 Cc and an approxi- mating sequences ~k as discussed above. Then Z E dr(XM ) exp − s~k(x) XM (dx) (26) Z = E cr · exp ln(fr(x)) − s~k(x) XM (dx) (27) Z = E cr · exp sk(x) XM (dx) (28) " n # Yk = E cr · exp (bj;kXM (Bj;k)) : (29) j=1 Now, note that for a random variable Y having a Poisson distribution with parameter p > 0 and any a > 0, it is E [exp(aY )] = exp(p(exp(a) − 1)) : (30) 5 By using the independent increments of the PPP XM and the fact that XM (B) ∼ Poi(λM (B)), continue the series of equations with n Yk eq. (29) = cr · E [exp (bj;kXM (Bj;k))] (31) j=1 n Yk = cr · exp (λM (Bj;k)(exp(bj;k) − 1)) (32) j=1 n Yk = exp (λM (Bj;k) · (1 − α~j − 1 +α ~j exp(−βj))) (33) j=1 n ! Xk = exp α~jλM (Bj;k)(exp(−βj) − 1) (34) j=1 n ! Z Xk = exp α~j1Bj;k (x)(exp(−βj) − 1)λM (dx) (35) j=1 Z = exp (exp(−s~k(x)) − 1)fr(x)λM (dx) (36) = Lr(~sk): (37) By standard approximation arguments, this also holds for any fr as defined + in section 3 and all f 2 Cc . 5 Expanding the pdf property I will first show the expansion of the density behaviour for measures on the positive real numbers (see section 2). This is just to give an idea of the principles that will be used for random measures, so the fact that [0; 1) is not compact will be ignored. The underlying idea is to essentially reverse engineer theorem 15.5 from [2]. Once the expansion has been shown to work for measures on the positive real numbers, i will show that it also works for random measures. 5.1 Measures on the positive real numbers So assume we know that Z Lr(t) = g(x) exp(−tx) λM (dx)(t ≥ 0) (38) 6 with Lr being the Laplace transform of a finite measure µr on the positive real numbers and g the assumed density.
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