University of Groningen. Statistical inference via fiducial methods Salomé, Diemer

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1 University of Groningen Statistical inference via fiducial methods Salomé, Diemer IMPORTANT NOTE: You are advised to consult the publisher's version (publisher's PDF) if you wish to cite from it. Please check the document version below. Document Version Publisher's PDF, also known as Version of record Publication date: 998 Link to publication in University of Groningen/UMCG research database Citation for published version (APA): Salomé, D. (998). Statistical inference via fiducial methods s.n. Copyright Other than for strictly personal use, it is not permitted to download or to forward/distribute the text or part of it without the consent of the author(s) and/or copyright holder(s), unless the work is under an open content license (like Creative Commons). Take-down policy If you believe that this document breaches copyright please contact us providing details, and we will remove access to the work immediately and investigate your claim. Downloaded from the University of Groningen/UMCG research database (Pure): For technical reasons the number of authors shown on this cover page is limited to 0 maximum. Download date:

2 Appendix A. Proof of Lemma 5.3 Instead of proving the lemma directly, a slightly more general result is given, which will imply the assertion. After reduction by suciency, P = fp : 2 g is a 2{dimensional exponential family (the family of normal distributions), with R R, as the natural parameter space. The notation S will be used to denote the outcome space of the canonical sucient statistic ( P i X i; P i X2 i ). The objective isto make a distributional inference about the true value (t) of :7! R, where is of the form () =h a + b + c 2 d 2 ; (A.) with h continuous, strictly increasing, and such that h(z)!, as z!, and h(z)! as z!. Without loss of generality it can be assumed that d<0, but it has to be required that b>0. Under these restrictions on h, and, the hypothesis H z : (t) z, with z 2 R, is equivalent to the hypothesis H z : a t + a 2 (z) t 2 ; (A.2) with a = b, a 2 (z) = c, d h, (z) and =,a. Next, the parameter function : 7! R is chosen such that such that the pair ( ;) provides an equivalent parameterization. Theorem A. Let P = fp : 2 g be a 2{dimensional standard exponential family w.r.t. the measure, dened on (S;(S)), with as its natural parameter space. Assume that S = f(s ;s 2 ) : s 2 > f(s )g, where f is convex and bounded from below. If with (ds) = m(s)ds, and m is continuous and 37

3 38 Appendix A. nondecreasing in its second argument, then G s dened by G s (z) = R s R, expf rgm(r;l(r;z;s))dr ; expf rgm(r;l(r;z;s))dr (A.3) with l(r;z;s) = s 2 + a 2 (z)=a (r, s ), is a continuous probability distribution function. Proof. Notice that for xed z and s, the points (r;l(r;z;s)) 2 R 2 form a line through the point s, with slope a 2 (z)=a. The conditions of (A.) imply that the slope a 2 (z)=a is a continuous and strictly increasing function of z, such that a 2 (z)=a!, as z!,, and a 2 (z)=a! as z!. Dene (dr) = expf rgdr, and rewrite G s (z) =g (z)=(g (z)+g 2 (z)), where g (z) = g 2 (z) = Z s Z s, m(r;l(r;z;s)) (dr); m(r;l(r;z;s)) (dr): As m is continuous at the point s, it follows that, for all > 0, there exists a z such that z > z implies g 2 (z) <. This follows from the fact that, for all >0, there exists a z such that z>z implies Z s, m(r;l(r;z;s))(dr) < (m(s ;s 2 )+) Z s s, (dr): Now taking suciently small and z >z, provides the result that was required. Similarly, for all > 0, there exists a z such that z <z implies g (z) <. If z<z 0, then it follows that l(r;z 0 ;s) l(r;z;s) if r s ; l(r;z 0 ;s) l(r;z;s) if r s : As m is nondecreasing in its second argument it follows that g 2 (z) iscontinuous and nondecreasing, and that g (z) is continuous and nonincreasing. Taking these facts together, it can be concluded that (A.3) is a continuous distribution function on R. A.2 Proof of Lemma 6. A large part of the proof was given by M. van der Put. To complete his proof, however, the following result is needed.

4 A.2. Proof of Lemma Lemma A. Let a; b be two integers, then (a +;b+) a a b b (a + b) a+b ( + 2 (a + b)): Proof. Notice that (a +;b+)> establish that,(a + ),(b +) 2,(a + b +),(a+),(b+) (a+b+2),(a+b+). Hence, it is sucient to a a b b : (A.4) (a + b) a+b For a = b = equality holds. As the inequality is symmetric in a and b it is sucient to verify that it holds for a + and b. Replacing a by a + on the r.h.s. of (A.4) we obtain,((a +)+),(b +) 2,((a +)+b +) and on the l.h.s. of (A.4) = a + a + b +,(a + ),(b +) ; 2,(a + b +) (a +) (a+) b b a + (a +) a b b = ((a +)+b) (a+)+b a + b +(a + b +) ; a+b so that (A.4) can be rewritten as a a+b a + a + b + : a a + b, The fact that a + b >acompletes the proof, because x+ x increasing for x>0. x is monotonically Proof of Lemma 6.. Notice that e G(z) is continuous and dierentiable on (0; ), and hence it has to be veried that 0 e G(z), and e G 0 (z) > 0. In the case a = b =, it is easy to see that e G(z) = 2 on [0; ), and hence it is a distribution function. Using l'h^opital's rule, it can be obtained that eg(0) = 0 if a =; b b+ if a>; and e G(,) = if b =; a+ if b>: To establish that e G 0 (z) 0, in the case a + b 2, it suces to show that B(z) decreases monotonically or, equivalently, that G 0 (z) G(z)(, G(z)) z(, z) ; (A.5)

5 40 Appendix A. holds for z 2 (0; ). Suppose that b =and a >, then G(z) =z a, and (A.5) can be reformulated as a(, z), z a,which is easily veried. By symmetry, it can be obtained that (A.5) holds for a = and b >, and hence it remains to be veried that it holds in the case that a; b >. Now, reformulate (A.5) as H(z) =z(, z)g 0 (z)+g 2 (z), G(z) 0; and notice that in both end points H(z) = 0. Hence, sucient conditions for H(z) 0 are: (i) H 0 (0) > 0, (ii) H 0 () < 0, and (iii) H 0 (z) = 0 has only one solution on (0; ). As G 0 (z) > 0 and (G 00 (z))=(g 0 (z)) = (a,)=z,(b,)=(,z), it is possible to dene eh(z) = H0 (z) G 0 (z) =(, z)(a, ), z(b, ), 2z +2G(z); and the conditions (i){(iii) can be translated into: (i) e H(0) > 0, (ii) e H() < 0, and (iii) e H(z) = 0 has only one solution on (0; ). Notice that e H(0) = a, > 0, eh() =,(b, ) < 0. Hence, it remains to check (iii). A sucient condition is that e H(z) is monotonically decreasing on (0; ) or, equivalently, eh 0 (z) =2G 0 (z), a, b<0; (A.6) for z 2 (0; ). In both end points this inequality holds, hence it is sucient to check the local extrema of H e 0 (z) in(0; ). G 0 (z) has only one local extremum at z = a, (a,)+(b,). Inequality (A.6) in this local extremum can be reformulated as (a, ) a, (b, ) b, (a; b) > (a, +b, ) a,+b, 2 (a + b) : Replacing a by a +, b by b +, and applying Lemma A. will then complete the proof. A.3 Partial proof of Conjecture 6. Recall that B(0;Q semi,bayes )=,B(;Q semi,bayes )=n=(2(n +)). By numerical investigation it can be obtained that B(; Q semi,bayes ) is strictly decreasing. Now, dene the function f(; c) =E +c, (, c)b X () ;

6 A.3. Partial proof of Conjecture 6. 4 where B x () =,, G Bayes;x () : G Bayes;x () For xed 2 (0; ), the function f(; c) is continuous and strictly increasing in c, and satises: f(; 0) = 0 and f(; ) =. Hence, the equation f(; c) = 2 has a unique solution c(), for every 2 (0; ). If it can be shown that, for every xed c 2 (0; ), the function f(; c) is continuous and strictly decreasing in, then it follows that the solution c() of f(; c()) = 2 is continuous and strictly increasing. Clearly, if this can be shown for one particular c, then it automatically holds for all c 2 (0; ). Now, take c = 2 such that f(; c) = E G semi,bayes;x (). Hence, f(; c), 2 = B(; Q semi,bayes), which is strictly decreasing.