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Discrete transect designs for estimating animal population Li, Leah Min 1994

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DISCRETE TRANSECT DESIGNS FOR ESTIMATING ANIMAL POPULATION by Leah Mm Li B.Sc. (Mathematics), Fudan University, 1987 A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE in THE FACULTY OF GRADUATE STUDIES Department of Statistics  We accept this thesis as conforming to the required standard  THE UNIVERSITY OF BRITISH COLUMBIA April 1994 ©Leah Mm Li, 1994  in partial fulfilment of the requirements for an advanced degree at the University of British Columbia, I agree that the Library shall make it freely available for reference and study. I further agree that permission for extensive copying of this thesis for scholarly purposes may be granted by the head of my In presenting this thesis  department  or  by  his  or  her  representatives.  It  is  understood  that  copying  or  publication of this thesis for financial gain shall not be allowed without my written permission.  Department  of  The University of British Columbia Vancouver, Canada Date  DE.6 (2188)  Abstract When studying the animal population in the forest, one sometimes collects data along a path which is called a transect. To estimate the size of an animal population, one travels along the transects and counts all animals sighted along the route. The goal of this project is to derive estimators of population size and to determine optimal transect designs.  11  Contents Abstract  ii  Table of Contents  iii  List of Figures  iv  Acknowledgements  v  1  Introduction  1  2  Mathematical Background  7  3  The Optimal Estimator for Fixed Design  10  4  Optimal Design  20  5  Examples  36  5.1  Disjoint Transects  36  5.2  Two Overlapping Transects  42  5.3  The Case with Startup Cost  48  References  71  111  List of Figures Figure 1.1  An area with 6 subregions  3  Figure 5.1.1 The optimal values of g  40  Figure 5.1.2 F 1 with b  1/4  41  1 with b—c.<1/4 Figure 5.1.3 F  42  Figure 5.3.1 Regions studied in Theorem 5.4  52  —  c>  Figure 5.3.2 The optimal values of g with a 2  =  0  62  Figure 5.3.3 The optimal values of g with a 2  =  0  63  Figure 5.3.4 The optimal values of g3 with u 2  =  0  63  Figure 5.3.5 The optimal values of g4 with u 2  =  0  64  Figure 5.3.6 The optimal values of g 2 1 with u  =  0.2  66  Figure 5.3.7 The optimal values of g with  =  0.2  66  Figure 5.3.8 The optimal values of g3 with  =  0.2  67  Figure 5.3.9 The optimal values of g4 with u 2  =  0.2  67  Figure 5.3.10 The optimal values of g with a 2  =  0.2  68  Figure 5.3.11 The optimal values of g with u  =  0.6  68  Figure 5.3.12 The optimal values of g with u 2  =  0.6  69  cr2 =  0.6  69  Figure 5.3.13 The optimal values of g3 with  Figure 5.3.14 The optimal values of g4 with u’ 2  =  0.6  70  Figure 5.3.15 The optimal values of g with u 2  =  0.6  70  iv  Acknowledgements I would like to thank my supervisor, Nancy Heckman, for her guidance, support, inspiration, and encouragement throughout the development of this thesis. I had very good fortune to study under her supervision. I would also like to thank Bent Jorgensen, for his careful reading and valuable com ments. I am grateful as well to all the graduate students, faculty, and staff who made my graduate study a rewarding experience.  v  1  Introduction  Many research programs for the management and conservation of wildlife will require, at some point, information on the number of animals in a region. Since counting all animals in a large area may be expensive and impracticable, the population size is usually estimated from the number of animals counted in sampled subregions. The transect method has become a popular technique for estimating animal popula tion size. It is being widely used either by individuals or groups. In transect sampling, the observer collects data while traveling along randomly located paths in the region. The transects are the strips running the length of the population area.  The width  and the length of the strip can be decided in advance according to the factors under consideration. The observer counts all animals sighted within the strip. We can travel in a few ways such as by light aircraft, by vehicles, or on foot. The method chosen depends on many factors like the cost of counting the sample and the size of the area involved. Another important factor is the nature of the area such as whether it is flat or mountainous and accessible or inaccessible. Light aircraft are  flOW  very widely used for wildlife censuses and surveys. They can  cover large areas quickly by taking photographs from the air and are the only means for censusing in areas where access on the ground is difficult. Their use only becomes limited when the vegetation is so thick that animals cannot be seen from the air. Counting from vehicles is practicable and economical. It gives excellent results in small to medium sized areas that can be traveled by vehicles. It is limited when ground access is difficult or the area to be covered is very large. Foot counts can be used for a small sized area which is flat and accessible. For a discussion of estimation of animal population size, see Seber [7] and [8]. We now consider a discretized version of the problem by dividing the whole area into a finite grid and defining each transect to be some collection of units of the grid. We 1  want to estimate the population size by counting the number of animals in a random selection of these discrete units of the grid. We denote the set of N(finite) subregions in the grid as X Z  =  (Z(xi),..  .  ,  =  {xi, x ,. 2  .  .  ,  XN}.  Let  Z(xj))’ E R”, with Z(x) being the number of animals in the ith  subregion. For a given weight function on X,  4  =  ((x ) 1 ,  .  .  .  ,  (xv))  e  RN,  we are  now interested in estimating N  I  =  q(x)Z(x)  based on observations from the randomly chosen subsets of X. (Here I includes both ob served and unobserved components of Z). For instance, if we are interested in estimating the total number of animals in the region, we take q(x)  =  1 for all x E X.  Definition 1.1 A set T(a) is called a Transect if it is a nonempty subset of X, for a in an index set A  =  {ai, a ,. 2  .  .  ,  ak}  Typically, A does not represent all subsets of X, but rather a collection of subsets chosen by the investigator. The researcher may choose the direction of the strips and the strip width, depending on physical features of the area and on the pattern of animal distribution. For example, if there is a river running east-west and we believe that animals are distributed along the river, then we choose transects running in a north south direction. Otherwise, if we choose east-west transects, our transects will contain either many or very few animals. We assume that A is fixed. Now we consider an example of an area with six subregions given in the following Figure:  2  Figure 1.1: An area with 6 subregions  If we wanted to use oniy east-west transects, we would let a 1 {4,5,6}, and A choose a 1 ) 2 T(a  =  =  2 , 1 {a } a . Then T(ai)  {l,4}, a 2  =  3 {2,5}, a  ,x 2 {x }, and T(a 5 ) 3  =  =  3 1 {x , 2 } x and T(a ) 2  =  {3,6}, and A  =  =  =  {1,2,3}, a 2  =  6 4 {x , 5 } x . If we  3 , 2 {ai,a } a , then T(ai)  4 , 1 {x } x ,  ,x 3 {x }. These transects are the north-south transects. 6  Let G be a probability measure on A, such that ZLjG{a}  =  1. We select transects  independently and randomly, according to the design measure G. It would be preferable to study simultaneous transect choice, that is, to study probability measures on A A x A x  ...  =  x A. However, this is a far more complicated problem and is not addressed  in this work. We assume that no animal is counted more than once and that animals tend to stay in their territories when observed. Thus we avoid the possibility of duplication in counting. Then the accuracy of the observations will depend on the visibility of animals, which depends on things such as animal size, vegetation covers available, and activity of the animal at the particular time of the year. Suppose the process is observed with error. Let  =  ) 2 (e(x ) 1 , E(x  e(xJ\r))t,  where e(x) is the observation error in the ith 3  subregion, so the observation from the ith subregion is (x)  =  Z and  1,2,.  are independent, and denote Z  =  Z+  Let aj, i  .  =  Z(x)+€(x). We assume .  .  ,  n be elements of an  independent random sample from A, each with distribution G. The a ’s are independent 1 of Z and of Z and  .  ,  We assume E() where  0, E(Z)  =  =  and  ,  and 6 are the covariance matrices  and 6 are strictly positive definite. We also assume  ji,  6 and  are  known based on prior information. We restrict ourselves to an affine estimator, I, which is design-unbiased, that is, with E(IZ)  =  I, where the conditional expectation is calculated by averaging over all  possible transect choices. In this sense, I is an unbiased estimator of I, no matter what the value of Z. Such an estimator is possible only if our design is random or if we census the entire population. Thus, it is essential that we choose values of a from A at random. We write the affine estimate of I based upon the randomly chosen transects T(a)’s in the form  1  =  ,. 1 l(a  =  1 n  ..  c(x)((x)  (x)) + 7(ai,... , a,) +  —  i1 XET(a:)  = fli where ZK  =  Z  —  =  Z+  CaBZ  €  —  .  vector c contains the weights  + 7(ai,.  .  .  ,  a)  +  Thus Z* has mean 0 and covariance matrix a(X),  + 6. The  used for observations along transect T(a). Bc  is a matrix of 0’s and l’s of dimension d by N, where d indicates the cardinality of T(a). BOZ* equals the vector of Z(x)  —  (x) values with x E T(a).  We denote the cost of sampling along the transect T(a) by S(a). Typically S(a) would be related to the length of the transect and the difficulty of traversing the asso ciated subregion.  4  Definition 1.2 We use the notation  S = S(a) to denote the total cost of sampling along the n randomly chosen transects. The expected value of the total cost is E(Sn) = n  S(a)G{a}. aEA  We are interested in finding an estimate and design measure that are in some sense optimal. This work extends that of Heckman and Rice [3], who did not include cost. Ideally we want estimates with high precision that are obtained at low cost. Since bias indicates systematic error, it is reasonable to restrict ourselves to estimators that are unbiased. We formulate our problem as follows.  Let c =  t, 2 (Cait, Ca  .  ,  Caj)  be the unknown vector of weights. For a given collection  of subsets, that is for a given A, find c and G to minimize E(I—I) + 2 E(S)  (1.1)  E(1IZ) = I.  (1.2)  subject to  The expected values are taken with respect to both the random transect choice and the random processes Z and e. We split this optimization problem into two steps by first fixing the probability measure G and minimizing (1.1) over c satisfying (1.2), then minimizing the resulting value of (1.1) over G.  5  In Chapter 2, some mathematical definitions, notation and results are provided. They will be frequently used in the following sections. In Chapter 3 we will show that for given design measure G, we can easily calculate the closed form of the optimal estimator. In Chapter 4 we will also show that an optimal design measure exists. In general, the optimal design may not be unique, and the explicit formula for the optimal design is often difficult to write out. In Chapter 5 we study several special cases for which we can find the closed form for optimal design, and also prove the uniqueness for the optimal design in some situatiolls.  6  2 Ill  Mathematical Background Chapter 3 and Chapter 4 we will show that the optimization problem (1.1) and (1.2)  is a convex programming problem. We now discuss the general solution for this type of problem. First we discuss the conditions for the existence of local minimizers for the optimiza tion problem (we call it Problem A): minimize f(x) over x E R subjecttol(x)Oandh(x)=O,fori=1,...,m,j=1,...,p. We define the Lagrangian function L by p  m  j=1  i=1  (Here the n is used for general R, it has nothing to do with the sample size n we use frequently throughout this paper). Definition 2.1 Linear Independence Assumption:  We say that the linear inde  pendence assumption condition holds at x E R if x satisfies the constraints of Prob lem(A), and the vectors Vl(x),i : l(x) = 0 and Vh (x),j = 1,2,... 3 are linearly independent. (The V denotes differentiation with respect to x). A general statement of the necessary conditions for x to be an optimal solution to Problem (A) is given by McCormick [5]. Theorem 2.1 Suppose that  f(.),  l(.), and h(.) are continuously differentiable func  tions. If x is a local solution to (A) and the Linear Independence Assumption holds at x, then there exists  =  (,i,  It2,.  .  ,  urn),  ) = (A 1  ttO,fori=1,2 7  ‘2,.  .  rn,  .  ,  ,) such that 7 A  ul(x)=O, fori=1,2,...,m, and 1 VL(x,pt,t)  =  0.  We now discuss the sufficient conditions for x to be an optimal solution to Problem (A). Definition 2.2 A set of points B  R is a convex set if and only if for 2 ,x e B and 1 x  we have 2 +(1—))x 1 \x B . Definition 2.3 A function  f is  convex on a convex set B if and only if for x ,x 1 2 e B  and 0 <) < 1, we have f(xi + (1  —  )X2)  ). 2 <f(xi) + (1— A)f(x  Definition 2.4 Problem (A) is called a convex programming problem if the functions  f(.)  and {l(.)} are convex and the functions {h(.)} are linear.  The following Theorem due to Kuhn aild Tucker [4] gives the sufficient conditions for x to be an optimal solution to the convex programming Problem (A). Theorem 2.2 Suppose that Problem (A) is a convex programming problem and  f(.),  l(.), and h(.) are continuously differentiable functions. Suppose at some point x satis fying the constraints of Problem (A), there exist multipliers 0, for i= 1, 2,  ...  ,  m,  tl(x)=0, fori=1,2,...,m, and ,A 1 VL(x, ) t  Then x is a solution to the problem.  8  =  0.  and A, such that  In practice, we can also solve the Problem (A) via computer programming. For example the FORTRAN subroutine NCONF is available in the IMSL MATH/LIBRARY which is a collection of FORTRAN subroutines and functions. We can write subroutines to calculate f(x), h (x) and l(x) for a given x and input an initial xo. Then NCONF 3 iterates to find a solution to Problem (A). Since NCONF uses a gradient search method,  f,  h, and i must be differentiable.  9  3  The Optimal Estimator for Fixed Design  The object of this chapter is to find the optimal estimator for a fixed design measure G. The explicit form for the optimal I is given in Theorem 3.3.  Notation and Preliminary Results The definition of I depends on the function  FG,  defined as follows:  Definition 3.1 Given G, we let  PG(x) = G{a : x E T(a)} denote the probability that x is in a transect chosen randomly according to the design measure G. We first define a matrix that appears in our optimal estimator. Lemma 3.1 describes its property. Theorem 3.1 gives (3.1), the equivalent formula of (1.2) and states that the optimal I has y = 0. Theorem 3.2 shows the set of ca’s satisfying the constraint (3.1) is nonempty, and provides important properties of a matrix MG which is used to calculate the optimal I. Theorem 3.3 proves the existence and the ulliqueness of the optimal I for the fixed design and gives its closed form. Definition 3.2 We define the notation  Aa  t Ba.(Z+5)Ba.  which is the covariance matrix of BajZ, the vector of Z(x) values for x E T(a). Lemma 3.1 Aaj is strictly positive definite for all a E A.  Proof: Recall that d is the cardinality of T(a). Since Z + 6 is strictly positive definite, for v E Rdi VtAaV  Ba( t V  10  + 5)BaitV > 0.  The equality holds if and only if  Batv = 0. If  Bav =  0, since Ba is full rank and BajBa V =  I, then  =  V = BaO = 0. t Ba.Ba.  So the equality holds if and only if v = 0. Hence Aa 2 is strictly positive definite. I Theorem 3.1 Fix G. Then we have the following results:  a. (1.2) is equivalent to the following two conditions:  BcG{a} =  (3.1)  aEA  ,. .,c)) 2 E(7(ci,c .  b. In minimizing (1.1) subject to (1.2), we have y  =  0.  0 a.e. ( Th GxGx... xG). G  Proof of (a): To prove (a), we write E(IIZ) = I as  E  (_‘  =  +7+  that is  t Ba(Z Ca  —  u)G{a} + E( 1 ) 7  =  (Z t 4  aEA  for all values of Z. So we have  BaCaG{a}  =  aEA  ,a 1 E(7(c , 2 .  .  11  =  0.  —  Proof of (b): We wish to minimize  E(l  1)2  —  + E(S)  =  E(I  =  E(I -  +p  —  —  )2  2E[(I  —  + E(I  -  1)2  t)2  —  -  + E(S)  4i)} + E(S).  However, the cross-product term is  E[(1  —  tIL)(j  —  =  E[E((I  =  E[(I  =  E(I  —  t)(I  qstp)E(I  —  p)Z)] t 4  —  —  Z)] t 4  #t)2  —  Thus, we write (1.1) as the following: E(I  —  1)2  + E(S)  =  E(1  —  2 t) —  E(I  —  t)2  + E(S).  So for fixed G, we only need to minimize E(1  —  2 t)  =  (c B Z* + E(n’ t  =  2 B t Z*) + E( E(n’ c ) 2 7 + 2E(n’7  7))2  BZ*). t c  Again, the cross-product term is zero, that is 7 E(n’  cB Z*)  =  CtBaZ*) E( 7 1 n  =  E 1 :  12  (3.2)  and 2 1 CtBciZ*) E(n  _2  2 +n i=1 ji  *) + n’(n B 2 t nE(c Z 1  —  *). tBa 2 1)E(CoitBciZ*Cc Z  We calculate k  2 G{az}E(caiBajZ*)  BiZ*) t E(ci 2 =  Ba(YZ + 8)BatCa. t 1 G{ai}ca  =  Aaca,, t G{aj}ca = and 12 E(CyitBy tBa Z*Cc Z *)  k  k  BatZ*cajiBajZ*) t G{aj}G{aj}E(ca  = i=1 j=1  k  k  =  3 }Caj G{ajG{a t Ca. Baj ( + li)Baj t i=1 j=1  By (3.1), we have k  o B t Z*Co Z *) Ba 2 t E(ci 1  =  k  }) 3 t Ca 3 G{a ( caitBaiG{ai})( + 5)( Ba j=1  i=1  = =  (Z t E(qS  =  E(I  —  2 p))  t)2 —  +  + q5ç.  Then we obtain k  E(1  —  2 t)  = n 1  Aaca +(1 t 1 G{aj}ca  )E(I_tp)2+(1  This is the sum of four parts; one is a function of ) does not depend on the 2 ). E(’y 2 7 other is E( 13  Ca’5.  Ca’5,  —  ). (3.3) )tS+E( 2 7  two are constants, and the  Therefore to minimize (1.1) we take  ) 2 7 E(  =  ) 7 0. From (a) we know that E(  =  0, so -y is a constant and must be 0 a.e. G’.  I  From (3.2) and (3.3), we know that minimizing E(I  1)2  —  + E(S) is equivalent to  minimizing k  H(c, G)  =  k  n  Aacaj + n t G{aj}ca  S(a)G{a}.  To minimize H for fixed G, we only need to minimize the first term. Theorem 3.2 Fix G. Then we have the following results:  a. There exist b. Let MG  Ca’5  = aEA  satisfying (3.1) if and only if PG(x)  =  0 implies (x)  =  0.  BaG{a}. Then MG is non-negative definite. MG is strictly 1 BatAa_  positive definite if and only if PG(x) > 0 for all x.  Proof of (a): To prove (a), first we suppose that there exist  x with PG(x) If x  0. We will show that q(xj)  =  e T(a),  since G{a : x  ith component of  BatCa  e  T(a)}  =  =  Ca’5  satisfying (3.1) and  0.  0, we have G{a}  =  0. If x  T(a), then the  is zero. Therefore  çb  =  çS(x)  Ca)jG{a} t (Ba  =  =  0.  aEA  To prove the converse, suppose that there exist  Ca’5  Let ca*(x)  PG(x) =  0 implies q(x)  =  0. We show that  satisfying (3.1).  =  q(x)/PG(x). Since PG(x)  we take 0/0 to be zero. For such  Ca*)iG{a} t (Ba  =  0 implies g(x)  =  0, ca*(x) is well-defined if  Ca*,  E T(a)}G{a}  = aEA  aEA  =  {(x)/PG(x)}G{a : x E T(a)}  =  14  So there exist  Ca*S  satisfying (3.1).  Proof of (b): Since Aa is strictly positive definite, then for v E RN  vtMGv  vtBatAa_lBavG{a}  = aEA  >0. Now we want to show that  VtMGV =  0 for some v  0 if and oniy if Po(x)  =  0 for  some x. For any v, VtMGV  =  0 if and only if  vtBatAa_lBavG{a} for all a E A. If G{a} MGV = t V  =  0, then BavG{a}  =  =  0  0. If G{a}  >  0, we have BaY  0. So  0 if and only if BavG{a}  =  0  for all a, that is if and only if vI{x E T(a)}G{a} for all i and a. Here v If  VMGV =  = (vi, v ,. 2  0 and v  .  ,  =  0  VN).  0, there exists v  0 and x such that  I{x E T(a)}G{a} = 0  for all a. So PG(X) =  G{a : x E T(a)}  Conversely suppose there exists x such that PG(x) component one and all the others zeros. Then vtMGv  15  0.  =  = =  0. Let v be a vector with ith  0. I  The Optimal Estimator Based on these Lemmas we have introduced, we now state the main result: Theorem 3.3 Fix G, and suppose FG(x) > 0 for all x. Then an optimal affine estima  tor I exists, is unique  1 =  almost everywhere and is given by:  n_ltMl  B(B(E +  +  I() with  E(1 The unique  —  Ca  1)2  + E(S) = n’ MG’# t  —  ( + 6) + 1 n  + E(S).  is given by Ca =  Aa’BaMG’cf’.  In addition, if G{a} and G{a } are non-zero, for some i andj, and if x e T(a)flT(a), 3 then we have  cov(Z(x),1(a)cr = a)) = cov((x),1(a) = as).  (3.4)  Comment: If the optimal estimator and design can be found using the method of La grangian multipliers, then the optimal estimator and design are characterized by Equa tion (3.4).  In Example 2, two overlapping transects case, we can find the optimal  estimator for the fixed design using Equation (3.4). Proof: From Theorem 3.1(b), to minimize (1.1) subject to (1.2), we take y = 0. So we  16  consider estimators of the form  CjtBajZ* + 4’p.  I=  Now we minimize H. By Theorem 3.2(a), the set of is nonempty. For a with G{a} = 0, the value of  Ca’S  satisfying the constraint (3.1)  does not contribute to the constraint  Ca  (3.1), nor to the function to be minimized. Therefore, we let A 0 and calculate the optimal 0 = {a A ,a 1 , 2  .  ,  a}, for r  Ca’S  e  for a  =  {a : G{a}  0}  . Without loss of generality we can write 0 A  k. Let C  *t_( t t iCai , 2 Ca , —  .  ,  Car  t  For fixed G, E(S) is fixed. We therefore wish to minimize the function f(c*) =  AacaG{a} t ca 0 aEA  subject to the linear constraint t caG{a} = Ba  4’.  0 aEA  First we will prove  f  f/vc* is positive 2 is a strictly convex function by showing 7  definite. Since Aai is positive definite, then the matrix 2Aai G{ai } G{a 2Aa } 2  —  2 \7C*  2Aar G{ar } is positive definite, so we know that mizer of  f  f  is a strictly convex function. Hence if the mini  exists, then it is uniqile. Furthermore, a local minimum of  f  will be a global  minimum. We find the minimizing c by using a Lagrangian multiplier vector ) differentiate the following: 17  e  RN, and we  AacaG{i} t Ca 0 aEA  with respect to  Ca  BA + t G{a}c  —  t\  aEAo  for each a in A , and set the results equal to zero. We then have the 0  following: 1  Ca =  Aa’BaA.  (3.5)  We use the constraint (3.1) to solve for A and obtain t CaG{a} Ba aEA  =  Aa’BaAG{a} t Ba aEA  =  MGA.  Since PG(x) > 0 for all x, by Theorem 3.2(b) MG is positive definite, so MG 1 exists. Therefore A = 2MG #. 1  (3.6)  Combining (3.5) and (3.6), we obtain Ca =  Aa’BaMG’çb.  We can easily calculate the optimal estimator which is the following: I  tMG_lBtA_lBZ* +  = =  fl_ltM_l  BjtAj_lBajZ* +  Substituting (3.7) into (3.3) and using (3.2) yields  E(I  —  1)2  + E(S)  cStM_BtA_lAA_BM_l#G{a}  =  aEA  qS + (1 t q 1 —n =  MG’qS t n’4 18  —  —  6 + E(S) t ) 1 n  1t + 6)# + #t64 + E(S). n (  (3.7)  To verify (3.4) we suppose that G{a} and G{a } are non-zero, for some i and 3 and xl  j,  ). For some 1’ and 1”, we have 3 T(a) fl T(a (xl) 2  . 111 3 Z) (Ba  , 1 (Bai)  Applying (3.5) we obtain BaA = 2AaCa.  So 1 = (BaA) A , = 2(Aa 1  ,. 1 Ca)  But , 1 (Aacaj  =  cov((Ba) i 1 , CajBajZ*)  =  cov((xi),1(a) = a).  By a similar approach, we may also obtain  )q  (3.4). I  19  =  2cov(Z(xi), I(a)c = a ). This proves 3  4  Optimal Design  Introduction We now proceed to find the optimal design measure G assuming that q(x) > 0 for all x. In Chapter 3, we showed that the unique optimal affine estimator for a given design measure exists if and only if PG(x) > 0 for all x. We therefore define the following: Definition 4.1 We denote 1’ to be the set of all probability measures on A such that PG(x) > 0 for all x.  Thus from Theorem 3.3, we want to minimize  F(G) over G  =  F. The following Lemmas and Theorem will show that an optimal design  exists. In Lemma 4.1, we show that F is a convex set. In Lemma 4.3 and Lemma 4.4 we show that F is continuous and convex on F. Unfortunately, it is easy to see that F is not a closed set, so more work is required to show that an optimal design exists. Lemma 4.5 provides us with the necessary result. Theorem 4.1 states that the optimal design exists. Theorem 4.2 gives a property of the optimal design and estimator. Now we discuss the property of the set F.  The Properties of the set F and the function F In this section we will show that F is a convex set, and function F is continuous and also convex on F. Therefore the problem of finding the optimal G is a convex programming problem which we have already defined in Chapter 2. Now we prove some important results. 20  Lemma 4.1 F is a convex set. Proof: Recall Definition 2.2 in Chapter 2. We must show  1 + (1 AG for allO  A  Let GA  =  —  2 A)G  e  F  land 2 ,G EF. 1 G 1 + (1 AG  —  measure on A. Assume GA  . Then it is easy to see that GA is still a probability 2 A)G F, then there exists some x E X, such that PG(x)  =  0.  This can be written as follows PGA(x)  =  (x)+(l—A)PG 1 APG ( 2 x)  =0. Since A  0 and 1  —  A  0, so we must have either (x) 1 PG  =  0  (x) 2 PG  =  0  or  which is a contradiction to our assumption of G ,G 1 2  F. I  To study F, it is convenient to view G as an element in Rk and it is useful to define a matrix norm. Let G be associated with the vector g G{aj norm  =  .  gj. We will sometimes write F(G)  as  =  (gi,g,.  g)t such that  F(g) and MG as Mg. Now we define the  for a matrix.  Let A be an rn x n matrix with entries,  lAW  = max  21  We define the norm AW of A by ajl.  The function  I II:  < m R  (a) lAW > 0 for all A  (d)  R satisfies  ’. m R  0 if and oniy if  (b) lAW (c)  —*  IIaAW  =  al lIAR  IA -i- BIl  The function  <  0 for all 1 <i < Tn, 1  =  IAII + IBII (This is called the triangular inequality). has the following property:  I I  For y  =  e  (yl,y2,.  n.  for all a E R, A E Rrnxn.  IIABII for A E Rm<fl, B  j  RnXT.  .  .  ,yk)  <  IIAII  IBII  (See Graham [2] for a discussion of these properties). E Rk, the matrix M is defined as k =  where R  =  tAai_lBa E 3 Ba  RNXN.  Now we prove the following lemma which will be  used in the proof of Lemma 4.3. Lemma 4.2 Suppose G  F, and G is associated with a vector g. For y  Rk with  IIII  sufficiently small, the matrix Mgy is positive definite. Proof: From Theorem 3.2, we know that Mg is positive definite. Let v E Rc with v  0 and  m  =  mm vtMgv>  IIvII=1  0.  Then we can write  vtMg+yv  fy(V)  = v (Mg+My)v t k  =  When  IlvIl  =  VtMGV  +  :: vtRjvyj.  1, we have RvI t jv  lRII 22  M  forM  maxWRII. So if we take y such that  =  BW  <  2Mk’  <  2Mk  then  Thus we obtain k  f(v) =  Now for vW  1>0.  0, then fy(j)  Therefore v(Mg + M)v  vW2  >  0.  So Mg+y is positive definite. I Now we prove that F has the following properties: Lemma 4.3 F is a continuous function on F. Proof: For any G  F, we want to show that F is continuous at G. It is enough  to show that for any (yl,  Y2,..  , .  yk)  e R’  >  0, there exists 6  satisfies  =  (g, 6  )  >  0, such that whenever y  =  6, we have F(g+y)—F(y)  <.  Since Mg is positive definite, from Lemma 4.2 we know that Mg+y is also positive definite for  small, and so Mg+y’ exists. We write  F(g + y)  —  F(g)  (M t n’  —  23  M’) + n S(a)y  =  ((Mg + M) t n#  =  )+n 1 (Mg + M)(—MM t n’  Recall that we have defined M  +n  —  S(a)y S(a)y.  By using the properties of matrix  =  norm we have the following F(g + y)  -  F(g)  fl’  j#My  -i 2  where C 1  =  (Mg + M)  IIRi yjj  M’  1+ n  S(a)y  (Mg + M)  S(a)y  +n <  1  Ciy (Mg + M)W + and C 2  n’ 2 WM’W  =  S(a). Now we need to show 1 n>  that I(Mg + M)W is bounded. For simplicity, we define A(y)  (aij(y)) E RNXN.  Mg + M  =  So we obtain A(O)  =  Mg  A(y)  =  detA(y)U  where Aii(y)  ...  A1N(y)  UA(y)=  AN1(y)  ...  ANN(Y)  where A(y) is the cofactor of the element ajj(y). A(y) is defined to be (—1) times the determinant obtained by omitting the ith row and jth column of the matrix A(y). 24  The detA(y) is a continuous function of y. So we have detA(y) For  WIJ  i  =  for some  det(A(O) + M)  detA(O)  —*  as  —*  we can write detA(y)  detA(O) < detA(O)L  —  So detA(y)  detA(y)  =  —  detA(O) detA(O)  >  detA(O) + detA(O)  —  —  detA(y)  —  detA(O)I  detA(O)  detA(O)L  =  We write =  1 UA(y) detA(y) detA(y)  UA(y)  —  +  UA0W  detA(y)  where 11 A (y) UA(y)  —  UA(o)  —  A(O)  —*  0 as y  —*  —  (O) 1 AN  0 for all i and  A(y) for  11 A (O)  N(y) 1 A  ...  —  N(O) 1 A  =  AN1(y) Since A(y)  —  —  A(O)I  ANN(Y)  ...  —  j, without loss of generality we have 1,  W3’W <5. This leads to the following UA(y)  —  UA(o)W  N.  Then we have detA(O) 25  ANN(O)  (N + UA(o)W)  =  . 3 C  Therefore F(g + y) if we take  62 =  —  F(g)  3+C C 1 e/(C ) and 2  3+C CiyC jy <e 2  < min{6 , 62}. I 1  Lemma 4.4 F is a convex function of G  F.  Proof: To prove this, we need to show  f(A)  =  ) + (1 0 AF(G  —  ) 1 A)F(G  0 + (1 F(AG  —  —  ) 1 A)G  0  for any G , G 0 1 E PandA e(0,1). Since we have f(o)=f(1)=0, then it is enough to show that  f  is a concave function for A  (0, 1),  that is f”(A) <0 for A e (0, 1). Suppose G E F, g is the associated vector in Rc, h E Rc, and 6 E R. To differentiate  f,  we use the fact that since Mg is positive definite, then Mg + 6Mh is also  positive definite for 6 small. Therefore M&h exists. So we can write M4sh  =  1 (Mg+Msh)  =  (Mg+6Mh)’  =  (I+6Mg_1Mh)_lMg_1  =  Mg’  —  6Mg’MhMg + 0(62).  So we have M’g+sh  Mg’  Mg’MhMg’  26  as S  0. We denote MAgo+(1_)gi by M. From Lemma 4.1, we know that F is a  —*  convex set. Then M is invertible. To calculate g  =  \go + (1  and h  — )gi  =  go  —  f’,  we use the above calculations with  g E RAT. Then as 6  —*  0, we obtain k  f))  =  F(go)  —  M_lcS 1 F(gi) + n4tM_lMgo_g  nS(a)(Go{a}  —  —  Gi{aj).  Furthermore, we can write MjshMhMsh  =  (Mg’SMg ) M 1 ) 2 )Mh(Mg’SMg hMg’+O(6 ).  We expand the right hand side of the above formula, and by a similar argument we can obtain the following: 1 MshMhMsh— Mg’MhMg S,  as S  —  MhMg’ 2Mg M 1 hMg  0. Therefore we have  f”)  =  #tM_lMgo_gi 1 1 _2n _g I[’. 0 M’Mg  Since Iit is positive definite, we obtain  f”(\)  i; 0.  Therefore F is convex function of G E F. I  The Main Results The task of this section is to show that the optimal design exists. First we give an important result which will be used later  ill  the proof of Theorem 4.1.  Lemma 4.5 Let {G } be a sequence in F such that 1  lim minG {a : x E T(a)} 1 x  =  0.  If g(x) >0 for all x, then  MG’qS t limsupqfi l—oo  27  =  Do.  (4.1)  Proof: To prove (4.1), it is enough to show that there exists a subsequence of {Gi},  ” }, 1 {G  such that lim tM  4, =  1-oo  Since Gj{a} is in [0, 1], and A is a finite set, there exists a subsequence {G } and G 1 such that i{a}—*G{a} for a11aA, 1 G where G is a probability measure on A. Therefore, since X is a finite set, ,(x) 1 PG  —*  PG(x) for all x E X.  Again, since X is a finite set, by assumption, there exists x E X such that PG(x) Then G{a}  =  0 whenever x  T(a). If x  =  0.  T(a) then (BatAa_lBa)ij = 0 for any  j.  Therefore ) 11 (IYIG  B >(BatAa_ , 1 {a})ij aG  =  aEA  Aa’Ba)jG{a} t (Ba  ‘S  aEA  =0. Since we can reorder the set X without loss of generality, then we can assume PG(x) for all 1  i  m and PG(x) > 0 for rn + 1 , 1 MG  M  —*  i  =  0  k. So for some matrix A(N_m)x(N_m),  °mxm  °mx(N—m)  °(N-m)xm  A(N_m)x(N_m)  as 1  —*  cc.  By an argument similar to that in Theorem 3.2 (b), we can show that A(N_m)x(N_m) is strictly positive definite. Therefore, M has exactly  in  eigenvalues equal to 0, with  corresponding eigenvectors spanning the rn-dimensional subspace of RN, ,x 1 (x , . {2 We now study MG . Let 11 ’j 1 jv  =  1 for i  =  1,2,  .  .  ).ij’s  .  , N.  .  .,Xm,O,.  )t 0 ., .  :  Xi  E R}.  and v ’s be the eigellvalues and eigenvectors of MG 1 , with 1 First we show that all the Asj’s lie in a bounded set. 28  Since MG , 1  —*  M,  Mv uniformly for v in compact set. Then we have  IV —* 1 MG  ,v 1 MG  sup V:IIVII=1  Mv  —  —*  0 as  — 00.  So sup  V1j 11 jILIG  —  ij 1 iWv  —*  0 as 1’  —*  cc.  1<i<N  But IV1I = 1 MG  so we have sup  \iivii  —  1<i<N  Mvi’W  —÷  0.  By the Cauchy- Schwarz inequality vl,t ()qivjij Since vi’lI  =  —  i 1 v  Mviii)  ’ 1 \jiv  —  ’ 1 Mv  1, then  sup  —  vi,ttMvi,,  0.  1<i<N  Since vl,tIV[vl,  iWv t v 1111 sup. =  where  ‘max  max  is the largest eigenvalue of M, then it follows that the \j’s are bounded.  Using the spectral decompositions for MG , and M, we obtain 1 , 1 MG  1 P , 1 D t 11 P 1  M=PDP  29  11 p , 1 where tp (viii,vii , 2 ..  .  =  j  ,ViIN)  PtV  I, and D 1 and D are diagonal matrices.  =  and Jvi’j  =  Since F ’ 1  1, all of the components of the matrix P 1 are in [—1,  =  11.  } of {P 1 So there exists a subsequence {P } and some F with 11  urn P 11  P.  =  l”—oo  Thus plp*  =  ,,tp 1 p , limji’ 11  =  1 D  I. Since we know , 1 A  0  0  0  o  112 0 A  0  o  o  Al’N  =  0  and A 11 are in a bounded set, there exists a subsequence {D } of {D 1111 } and D*, such 11 that D*.  111 D Therefore  p*D*p*t  111 = G 1 V 7 ,  But ,, 11 MG  —f  M  PDPt.  =  Suppose we have ordered the diagonal elements of D and D*. Because of the uniqueness of the eigenvalues of M, we have D=D*. From the discussion of M’s eigenvalues and eigenvectors, we know that the first m elements of D are 0 and the first m columns of P span {(x ,X 1 , 2 Therefore, the first  m  columns of F” span {(x ,X 1 , 2  .  .  Xm,  1,2,... #) t (F  ,.  (p*i#).  30  =  0  0,..  .  , .  .  Xm,  0,...  ,  O)t} and, for i  =  and —*  0.  Finally we show qSt ]41  —+ 00  as 1”’  o.  —÷  Now t  vl3’,,,  —1  (  ],,, t) 4  (p l 1 I,t)tD 11 ,,  =  —  —  1  N  nfl ti’, i 2 L’l” p)J  :  “l”i  m  nT-)  t) 1111 (P  2 tl\1  çi. II1-lll L\  i=1 00.  -+  I Theorem 4.1 Suppose q(x)  >  0 for all x. Then there exists G*  F that minimizes F.  Proof: By Lemma 4.3, F is a continuous function of the G{a}’s, i  1,2,.  .  .  ,  k.  However, I’ is not a closed set. For all x E X, we can find a sequence {G } E F, with 1 (x)—0asl-—*co. SowecannothaveGi—*GeF. 0 P  We define F infGj-’ F(G)  =  =  {G  e}. It is a closed bounded set in Rk. So  F : mint PG(x)  F(G) for some  E F.  0 Now we show that there exists an e inf F(G)  GEF  =  >  0 such that  inf F(G)  GEFEQ  =  F(G ) 0 .  If not, there exists a sequence {G ) decreasing and 1 } with F(G 1  urn mm PG (x) 1  l—+oo  31  =  0.  This contradicts Lemma 4.4  .  I  Thus we reach the important conclusion that the optimal design measure exists on the nonclosed set F. For a given G, the weights in the optimal estimator are functions of G, written as  CG.  So if we can find the design G* which minimizes F, then, by Theorem 3.3, the optimal weights,  CG*  can be calculated explicitly. Let d H(cG*,G*)  =  > d. Then we have  minminH(c,G)  =  H(CG,G) H(c,G) for any  e  c  Rd and G  e  F. Therefore,  (c*,  G*) minimizes H.  To minimize H subject to the constraints (3.1), we use the results in Chapter 2.  Here again G is associated with the vector g.  (gi,g,..  .  ,gk, c)  Rk*,  where k*  =  =  =  d + k. The constraint functions are:  (x) 1 h(x)  Now in our case we have x t  for i  [Btc —  hN+l(x)  =  =  1,2,...,k, for  —  j  =  1,2,..  1.  As noted before, to use the Lagrangian method we need to check the Linear Indepen dence Assumption at a point x. Now we define a matrix D transposes of the vectors Vl(x),i : l(x)  =  0, and Vh (x),j 3  =  =  D(x). Its rows are 1,2,. ..,N + 1. The  Linear Independence Assumption holds at x if and only if the rows of D are indepen— dent. Thus to check the linear independence we only need to check if the matrix is full rank.  32  For a given xt = (gt, Ct), let 1 be the number of zero  gj’s.  Then we write the following  matrix °lxd  Q2  Q  Rlxc is a matrix obtained from  where Qi i if g  Qi  0.  Qe  ‘kxk  The remaining rows make up Qi.  in the following way: remove row The matrices Q2  R(N+l) and  are as follows: BaitCai Q2—  .  .  .  BaCa  1  and }Bai 1 G{a t  ...  t G{ak}Bak  Now we have the following result about the optimal design. Equation (4.2) shows a relationship between cost of sampling and variability of the estimator for certain transect. CBajZ*  Theorem 4.2 Suppose that I = n  +  gt  the problem of minimizing H subject to Equation (3.1),  and G is a local solution to G{a} = 1, and G{a}  0,  for i = 1,2,..., k. Suppose the matrix D is of full rank N + 1 + 1. (1 is the number of zero  G{a}’s). IfG{a}G{a} >0 Var(I(cc = a)  —  S(a) = Var(1(oo = a) 2 n  —  (4.2)  S(a). 2 n  Proof: The function to be minimized and the constraint functions are continuously differentiable at xt = (gt, Ct). By assumption, D is full rank N + 1 + 1. So the Linear Independence Assumption holds at x. By Theorem 2.1, x is a critical point of the Lagrangian equation k  L( , t 1 .  .  ,  k, 1  ,.. 1 A  .  ,  Ak,  /3) = n +2(t  k  k  AacaG{ai} + n t 1 ca  —  S(:)G{a}  2 B t G{aj}ca ) A a + /3(ZG{aj} 33  jG{a}  —  —  1).  Differentiating the Lagrangian equation with respect to G{a}, and equating to zero, we obtain flhCa.tAaCa  Differentiating with respect to  + nS(a)  Cat  — —  (4.3)  gives  G{aj}(2n’Aajca For G{a}  2CatBcL.) + /3 = 0.  —  2Ba.X) = 0.  0 the above can be written as Ba). = fl’AaCa.  By Theorem 2.1 we must have  jj  =  0. We substitute  j  = 0 and Baa). into Equation  (4.3) and obtain =  flhCatAaCa.  —  nS(a)  = n’Var(I(co = a)  So, when G{a}  —  nS(a).  0, Var(I(ca = a)  —  S(a) = n3. 2 n  I  It is important to note that if G{aj = 0, for some i, then the minimizing G is on the boundary of F. In this case, Equation (4.2) will not be valid anymore. In Chapter 5 we will have an example where the optimal design is on the boundary, that is the probability of choosing a particular transect is zero. In general, the optimal estimate and design must be calculated numerically. We have already proved that the optimal estimate of I and the optimal transect design exist. E(I  —  They may not be unique. 1)2  But all these solutions will minimize the sum of  and E(S).  There is no available formula for the optimal design in general, although it is simple to calculate numerically in specific cases by using computer programming. Recall we 34  have mentioned in Chapter 2, the FORTRAN subroutine NCONF can work quite well for this case. The FORTRAN subroutine can be used to minimize H with respect to x  =  (gt, Ct), or to minimize F with respect to gt and then calculate the optimal weights  by using Theorem 3.3. In the latter case if we input F(g), h(g)  gj  —  1, lj(g)  =  gj,  and the initial g, then by using the subroutine we will get the output: the optimal g. We prefer minimizing F because the space we work on has lower dimension than when we minimize H. In some cases, we can still use the equations containing conditional variances and covariances, given in Theorem 3.3 and Theorem 4.2, to solve for the optimal g and the optimal weights. In the next chapter we apply the results in some special cases.  35  5  Examples  As mentioned in Chapter 4, we cannot find general formulae for the optimal design. Also the optimal G may not be unique. So it is hard to understand how the cost function S(.) and variability measures  (  and (5)affect the optimal design.  In general, it is reasonable to choose transects which are cheap to sample, and which have high variability. A large prior variance on Z(x) indicates vague prior information. In this chapter we study several examples in order to quantify the balance between cost and variability. We study the case with k disjoint transects, the case with two overlapping transects, and a case with a very specific physical sampling plan. Sections 5.1 and 5.2 deal with the case of k disjoint transects and two overlapping transects. In these two cases we have the result that the optimal design is unique. In Section 5.3, we consider a sampling plan where there is an initial startup cost for each path. We consider the cost as two parts: the cost for setting up a sample location and the cost for traveling in the forest. In this case, we have explicit formulae for the optimal design for some values of the cost function and the variability measures.  5.1  Disjoint Transects  We consider the case that the set of possible transects is a partition of the whole region. Even in this simple case, the optimal G cannot be found explicitly. However we can still study some of the optimal G’s properties. Definition 5.1 The transects T(a), i  O when i  j, and  ui  T(a)  =  =  1,2,..., Ic are called disjoint if T(a)flT(a)  X.  For simplicity, we define Var( rT(a)  36  =  =  BaBaZ*) t Var(  =  cI;ItBatAaBaicb.  Theorem 5.1 Assume q(x) > 0 for all x. If the k transects are disjoint, then there is a unique optimal estimate and a unique optimal design G i = 1,2,.  .  .  ,  F, with G{a} > 0 for  k. The optimal estimator has Ca  =  Baqf 1 G{aj}  and the optimal G satisfies 2 nG{a}  —  nS(a) = /9  where 9 is a Lagrangian multiplier. Proof: We order the set X so that T(ai) = {xi,x ,... 2  }, 1 ,xL  ) = 2 T(a  +1,. 1 {XL  .  .  etc, with Lk = N. We assume L 0 = 0, so the cardinality of the ith transect is d = L  —  . For a given G, we have 1 L_ G{ai} 0 1 Aa MG=  o  G{a 1 Aa } 2  o  o  ...  Aak’G{ak}  Where Aag is a d x d matrix. For disjoint transects, PG(x) > 0 for all x implies G{a}>0fori= 1,2,••,k. ThenM ’ exists,and 0 Aai =  X  G{ai}’  0  0  }’ 2 2 x G{a Aa  0  0  MG’ Aak  By Theorem 3.3, we obtain Ca.  =  Aa.’Ba.MG’  37  x G{ak}  = Aaj —1 [OdXL , G{a} _ 1 —1  =  G{a}  =  G{aj}’Ba#.  —1  Aai, OdX(L_L)]çf  , ‘ct 1 [OdxL_ xd, OdX(L_L)]çb 2  To find the optimal design G, we minimize k  F(G)  k  G{aj}  =  nG{a}  =  B A t aBa a  +n  S(a)G{a}  +nS(a)G{a})  over F. We know from Theorem 4.1 that the minimizing G exists. We now show it is unique by showing that F is strictly coilvex. We have proved in Lemma 3.1 that Aaj is strictly positive definite. Since Bac equals  e  the vector of q(x) values with x have Ba.#  0. So b,  T(a), by the assumption that q(x)  >  0 for all x, we  0.  >  We calculate the following 1 b G{ai} 3 2  3 } 2 G{a  —1 702  bk  3 Since b/G{aj  >  0,  SO  2 2 7 F/7G is strictly positive definite. Thus F is a strictly ..  .  convex function. The minimum is therefore achieved uniquely for G E F. Using Lagrangian multipliers with the constraint  G{aj  =  1, the Lagrangian  function is k  L =  k  nG{a}  + nS(a)G{a}) +  G{a}  —  1),  where 3 is the Lagrangian multiplier. By Theorem 2.1, we know that if G is the optimal solutioll, then ãL/öG{a}  =  0, i  =  1, 2,... , k, that is  2 nG{aj  —  nS(a)  38  =  where 3 is the Lagrangian multiplier. I Remark: In this form it is easy to see that the optimal G{a} is a decreasing function of S(a) for given b. Also for fixed S(a), G{a} is an increasing function of b. This is a fairly obvious statement; the more expensive the sampling of transect T(a), the lower the probability that we choose the transect i; the more variable the observations along transect i, the more important it is to sample that transect.  Two Disjoint Transects To better understand the effects of cost and variability on the optimal design, we consider a simple case with only two disjoint transects. that T(ai) fl T(a ) 2 {XL+1,  ,  xjy}.  =  ) 2 0, T(ai) U T(a  Let G{ai}  =  =  } 2 g and G{a  For a given A  X, and T(a ) 1 1  =  —  =  (X2)  1 Ca  q(xL)  c(xL+1)  1 Ca21_g  and the optimal g satisfies b 1 — nS(ai) ng Let LS  =  S(ai)  —  ). If L\S 2 S(a  =  =  n(1—g) 2  —  ). 2 nS(a  0, then the optimal g is 1 =  1 +1 /12 2 (b 1 ) b 39  {ai, a }, suppose 2  ,... 1 {x , 2 x  ,XL},  ) 2 T(a  =  g. Here PG(x) > 0 for all x implies  g E (0, 1). From Theorem 5.1, the optimal weights are:  1  =  If zS  0, then c  b  —-i+ 2+1_0, g (1—g)  where b  =  2 / 1 b t (ri S)  c  =  /(n 2 b A S).  and  Thus for given b and S(a), we can find the unique optimal g fourth degree polynomial.  e  (0, 1) by solving a  Once we have the optimal g, it is easy to calculate the  optimal weights c and estimator I. The optimal values of g for different values of b and c are given in the following figure: Figure 5.1.1: The optimal values of g  1•  co 0  0 0  4’  Figure 5.1.1 also supports the remark: for fixed c, if the variability of observations along transect 1 is large, then the probability of sampling this transect is large. Also 40  if the cost of sampling transect 1 is much larger than sampling transect 2, then the probability of sampling transect 1 will be lower. Although an explicit formula for the optimal g would be very complicated, we can easily determine that g  <  1/2,  =  1/2, or  1/2, if b  >  —  c  <  1/4,  1/4, or  >  1/4  respectively. To show this we will graph a function F , which is related to F. Write 1 F(g)  =  +1b:3@18@2X1 —  rig  +  n(1—g)  ) 2 + nSg + nS(a  1 b  flS(n2Sg + 2 n S (1  —  g)  ) 2 + g) + nS(a  1 nS g +)+nS(a2 (+ )  =  nL.SF ( 1 g) ± nS(a ). 2 Minimizing F is equivalent to minimizing F . Since 1 1 F’ ( 1/2) always positive, if b  —  c  >  1/4, then F(1/2)  <  =  4(c  —  b) + 1 and F ’ is 1  0 and the optimal g is greatç than 1/2.  Figure 5.1.2: F 1 with b—c> 1/4  F1  ‘1  41  g  If b  —  c < 1/4, then 1 F ( 1/2) > 0 and the optimal g is less than 1/2.  Figure 5.1.3: F 1 with b  —  c < 1/4  1 F  0  0.5  g  1  Figure 5.1.2 and 5.1.3 show the shapes of F 1 for these two cases. Furthermore, for b sampling the  two  —  c  =  transects  variability measures satisfy b  1/4, the optimal g  are  —  c  =  1/2, that is the probabilities of  the same. This means when cost functions and the =  1/4, then the probability of sampling either transect  is the same.  5.2  Two Overlapping Transects  Now we deal with the case with two overlapping transects.  As in Section 5.1, the  optimal g cannot be found explicitly in this case. Again we discuss some properties of the optimal g. Theorem 5.2 shows that the optimal design is unique.  42  Definition 5.2 Two transects T(a) and T(a) are called overlapping if T(a)flT(a)  0. Suppose that A where 1 < r < L  =  ,a 1 {a }, T(a 2 ) 1  ,x 1 {x ,••• 2  =  ,XL},  and T(a ) 2  = {Xr+i, Xr+2,  ,XN},  N, so these two transects overlap. We partition X into three disjoint  sets as 1 X 2 X  3 x So T(ai)  =  Xi UX , and T(a 2 ) 2  =  2 , 1 {x , x .  =  ,Xr}  ,XL}  = {Xr+1,Xr+2,  = {XL+1,XL+2,... ,XN}.  2 UX X . Set X 3 2 is the overlapping part. Correspond  ingly, we partition all the vectors and matrices as follows: Ii,. t  —  Y  J. t  I t  kWi W2 ,Y3  —  *t  Z—I  1  *t 2  ,  *t 3  ,  t) 12 t,c 11 Calt = (c 2 Ca  t  =  t  I  23 = iC 22 C  t  11  12  13  21  >22  23  31  32  33  and 11  12  13  621  622  623  631 632 633 For example:  =  ),. 2 (() (x  .  .  ,  2 (xr)), Z 5 q  =  (Z(r+i), Z*(Zr+2),  is the covariance matrix of Z and Z . For simplicity, we write I’ 2  and =  +  jj. 6  So our optimal estimators are of the form  l(ai)  =  1 +Z Z 11 12 + q c 2  43  =  ,  Z*(XL)),  + 6 with  ) 2 l(a Suppose G{ai}  } 2 g and G{a  =  22 + 3 c * 2 tZ tZ 2 c * 3 + q.  =  1  =  —  g. Here PG(x) > 0 for all x implies g E (0, 1).  First we define TV =  for i  =  —  1,3. The optimal design will depend on V , V 1 , and S(ai) 3  —  ). Next we 3 5(a  ’s are Bayes risks. 1 show that the V Lemma 5.1 Suppose that Z is multivariate normal. Then under squared error loss, tZ and * 1 tZ based on Z 3 1 and V V 3 are Bayes risks of estimating, respectively, * . 2 Proof: First we show that I  —1 jtT T = Pi ‘112’22  *  2  is the Bayes estimate of qitZl* based on Z ’. 2 To show this, we must show that w f(w)  Differentiating  f  =  Z t 1 E(qf  =  tZ) 1 E( 2  =  #tlIEc  —  IJ22_1I12141 minimizes the following:  =  2 wtZ) —  —  tZwtZ) + E(wtZ) 1 2E(qS 2  II 1 2ç w 12  + wlPw.  with respect to w and equating to 0 yields =  1 + 2P q 21 —2W w =0. 22  So the minimizing w is W = I’221’21#1.  tZ is 1 So the Bayes estimate for 4 Il  =  . 2 wZ  *. 2 t# Zi* from Z Now we can calculate the Bayes risk of estimating 1 44  Z t 1 E(qf  —  l)2  2 E(tZ)  =  tZ) 1 E(# 2  =  E(I)  —  —  =  ‘1’12’112221)4i  —  *. 2 tZ from Z # * Therefore, V 1 is the Bayes risk of estimating 1 * from Z 3 *. 2 Similarly, we can show that V 3 is the Bayes risk of estimating qfitZ Theorem 5.2 Suppose that A  •  {ai,a } 2 , and 1 T(a and T(a ) ) are two overlapping 2  =  transects. Then, for a fixed g E (0, 1), there is a unique optimal estimate with 11 c  =  12 C  =  —  #2  + +  22 C  =  #2  23 C  =  #37(1  22  1#  —  ‘I’21#i  I22  —  (1  ;  g 11221123#3 (1—g)  g).  Furthermore, a unique optimal g E (0, 1) exists. If /S g= If S  =  0, then the optimal g is  1 )’ 1 V / 12 3 1+(V  0, then the optimal g satisfies v +1=0 -—+ (1—g) 2 2 g  where u  =  Vi/(n S 2 ) and v  =  2 / 3 V S ). (n  Proof: First we find the optimal weights cj’s for a given g. For g E (0, 1), the opti mal estimator exists and is unique by Theorem 3.3. Since the optimal weights satisfy constraint 3.1, (c,,t, , 12 c  Ot)g  (ot  22 c ,  23 c t)(l —  45  g)  =  (#it,  #2,  t). 3 #  Then we obtain 11 = c 12 + (1 gc C23  =  (5.1) 22 = g)c  —  13/(1  —  (5.2)  4’2  (5.3)  g)  Since x E T(ai) fl T(a) if and only if x E X , then by Equation (3.4), we have 2 *,l(cc 2 cov(Z  ai)  =  *,l(a)c 2 cov(Z  *, Zi 2 cov(Z 12 c * 2 ) 11 + tZ c  =  *, Z 2 cov(Z 2). c * 3 3 22 + tZ c 2  =  =  ), 2 a  that is  Calculating the above yields 11 + I’ C 21 k11 12 = c 22  22 c 22 q1  23 c 23 ‘P +.  (5.4)  We combine the formulae (5.1)-(5.4), and obtain the expressions for c 12 and . 22 c To find the optimal g, we suppose q(x) > 0 for all x. From Theorem 4.1, there exists an optimal g F(g)  (0, 1). Now we want to minimize =  =  tA _l[c g c 1  tAa Ca C ( 1 a +2  —  )(1 2 g)] + nS(ai)g + nS(a  —  g)  V 3 + 2332#2 + 2222 + 22t211 + 1—g g t ’Ji 1 t 3 24 i 2 hlP 22 IJ 32 cS W 22 + ] P 32 3 4 23 1 + 3 # _1lJ, 22 ti 21 ltJ +  )(1 2 +nS(ai)g + nS(a +  1 —g  —  g)  )(1 2 + C] + nS(ai)g + nS(a  —  g).  tZ from 3 4, Since V 1 and V 3 are the Bayes risks of estimating, respectively, qSitZl* and * *, they are positive. Now we write F as 2 Z F(g)  =  _1[  =  flZS[n2g/S + n (1 2  +  ) + n’C 2 ] + nSg + nS(a  —  46  g)/S  )+n 2 + g] + nS(a  C.  So minimizing F is equivalent to minimizing V  1—g  g  For given V , V 1 , S(ai) and S(a 3 ), we can find the unique g E (0, 1) by solving a 2 fourth degree polynomial, which is obtained by differentiating F 2 and equating the result to zero. Thus, we solve U  V  g  (1—g)  2+1  Since 2u 2v 3>0, F ” 2 (g)=—-+ g (1—g) there is a unique optimal g in (0, 1). If /S  =  0, then the optimal g solves (1 1 V  =  —  . 2 g 3 V  Then the optimal g is 1 =  1+ 1 /V 3 (V 2 )  I Recall F 1 in Section 5.1. F 2 has same form as F . So the graph of the optimal values 1 of g for given values of u and v will be exactly the same as in Figure 5.1.2 except we use u and v instead of b and c for our overlapping transects case. Although in this case an explicit formula for the optimal g would also be very com plicated, as in the previous example, we can still determine that g < 1/2, >  1/2 if u  —  v < 1/4,  =  =  1/2, or  1/4 or > 1/4 respectively. The calculations are identical to  those in Section 5.1. Thus, when u is much larger than v, that is, when V 1 is much greater than V , we 3 tZi* is 1 are more likely to sample transect T(ai). A large value of V 1 indicates that # Z * . So if V *. 1 difficult to estimate from 2 1 is large, it is important to observe Z 47  5.3  The Case with Startup Cost  Many sampling designs involve collecting data along transects with starting points at very different locations. In this case, there are two components to sampling cost. The first component, which we call the startup cost, is the cost incurred by beginning sam pling at a particular location. This cost usually involves travel expenses and sometimes, the hirillg and training of field assistants. The second component of cost reflects the direct cost of sampling along the path. We therefore consider that the cost contains two parts: c represents the price of setting up a sampling location, and y represents the price of traveling one unit distance at that location. We assume c > 0 and y > 0, and that neither c nor y depends on the transect. We now consider a simple case. Suppose we can start at two separate locations A and B. From location A we can either travel one or two units distance, and from B we can only travel one unit distance, counting all the animals sighted along the route. The physical description of the sampling region thus determilles our choices of tran sects. We can start at either place to travel one unit of distance, start at location A to travel two units of distance, start at both locations to travel one unit of distance from each location, or travel the whole area. Our optimal strategy will depend on the costs involved and the accuracy of our observations. It is reasonable to consider that setting up a samplillg location is more expensive than traveling a unit distance in the forest although we do not assume that in our calculation. The question arises if we should spend more money to visit the whole area, or if we should just travel in one or two subregions. The region is divided into three subregions, x , x 1 , and x 2 . 3  Let x 1 and x 2 be,  respectively, the regions for the first unit of distance and the second unit of distance when we start at A. Let x 3 be the area we cover when we start at B and travel through the one unit of distance. Then there are five possible transects. 48  Suppose A  The transects and the corresponding costs are given  =  as follows:  We write gj  =  T{ai}  =  }, 1 {x  8(ai)  } 2 T{a  =  }, 3 {x  )=c+y 2 S(a  } 3 T{a  =  2 , 1 {x } x ,  ) =c+2y 3 8(a  ) 4 T{a  =  3 , 2 {xi,x } x ,  ) =2c+3y 4 S(a  T{as}  =  {xi,x } 3 ,  ) 5 S(a  G{a} for i  1, 2, 3,4, 5, and g  =  =  =  =  c+y  2c + 2 y.  . t (gl,g2,g3,g4,gs)  Since A and  1 and x B are separate locations, we assume the numbers of animals in regions x 2 are uncorrelated with the number of animals in region x . Also we assume the measurement 3 errors from three subregions are uncorrelated. To simplify calculations, we also assume that the variances of Z(x) and  j  do not depend on i. Therefore, we suppose 0  12  0  °12  o  o  2  and 6=  72  0  0  0  2  0  o  o  72  I  Now we let r  = 2  +  72  Then straightforward calculation shows that the matrix MG may be written as follows MO MG=  O  (g2+g4+g5)  49  where M Suppose that n  =  122 _ 2 ) 4 (gi +gs) + (g3+g (g + g4) r2_c7122  =  r2_ui2(93  +94)  (93  + g4)  r2_j122  1 and (1,  =  )t 11  Then F can be written F(g)  =  MG’#+ZS(aj)th  91  + g3 + g4 1 g3 +  +95  12 U 21  +c(gi  +g2  94  +93  —  1 + g3 + g4  1 gi +g + g4 +  1 g2  +  94+95  +  12 2U  95)  +2g4 + 2 gs)+y(gi  +92  + 293  +  93  +394  Denote 12 U  *  =  r  which is the correlation between 1 (x and Z(x 2 ) ), 2 C S  =  —,  r  and t=.  It is obvious that s and t are both positive. For simplicity, we write  =  91+93+94+95  2 P  =  93+94  3 P  =  g2+g4+g5.  50  +  g4  +95  + 2g5).  Then we can rewrite the problem as minimizing F*(g)=  subject to and gO for i=1,2,...,5. For specific values of , 12 s, and t, the optimal gj’s can be calculated easily using u the FORTRAN subroutine NCONF. Here we write a subroutine to calculate F*(g), h(g)  =  gj  —  1 and l(g)  =  gj for a given g, then input an initial value go. Then we  can get the solution for this problem.  The case with u 2  =  0  The following Theorems apply to the case where  =  =  0, that is, the numbers of  animals in these three subregions are uncorrelated. This simplification of the problem allows us to study how the costs and variabilities affect the optimal design. In Theorem 5.3, we show that the optimal design has g5 discuss the values of g,  g2,  =  ,x 1 }. To 3 0, that is we never sample {x  g3 and g4, we denote ,‘__1__1/2  \s + t / 2 N 112 s+2tJ 12 tiN’  In Theorem 5.4 (a), (b) and (c), we give explicit formulae for the optimal gj’s. These formulae depend on whether ci and /3 lie in regions (a), (b) or (c) of Figure 5.3.1. We cannot calculate formulae for the optimal gj’s for all points in region (d). However, we do find formulae for the subregion  <  1/2 (see Theorem 5.4(d)).  51  Figure 5.3.1: Regions studied in Theorem 5.4  C  a  For the case 12  =  F(g)  0, the problem becomes minimizing  =  1 1 —+—+ 2 P 1 P  1 3 +P 2 +P ) ) +t(P 3 +P  subject to and 9iO for i=1,2,...,5. We call this constrained minimization problem (B). We write the set 4,5, and g=1} 0,P 0,g0,i=1,2,3, 3 > 2 C={g:Pi>0,P >  =  {g : g3 +g4  >0,92  +g4  +95  >0, g  0, i  =  1,2,3,4,5,  and  =  1}.  Recall in Chapter 2, Theorem 2.1 requires that the Linear Independence Assumption holds at the optimal g. This is a simple case with lj(g)  =  g,  0 and h(g)  =  g—1  =  0. Thus we know that Vl(g) is a vector with ith component 1 and the others 0’s: Vh(g) is a vector with all components equal to 1. So for any g corresponding to a probability measure, the gj’s cannot be all 0, so the vectors, Vl(g),i g, 52  =  0, Vh(g), are linearly  independent.  Therefore the Linear Independence Assumption always holds for this  problem. On the other hand, Theorem 2.2 requires the convexity of C and F*. This is easy to see since C is just a simple case of F, and from Lemma 4.4 F* is also a convex function on C. Now we can apply Theorems 2.1 and 2.2 to our problem. The Lagrangian function is  Suppose g is an optimal design. From Theorem 2.1, we know that there exist \ and ‘s such that, for this g; 0 for i  ui  =  0 for i  =  1,2, 3, 4,5  (5.5)  1,2, 3, 4,5  (5.6)  1,  (5.7)  1 ——+s+t+X—j=0  (5.8)  gj 8L  =  =  (5.9) (5.10)  8F2F2]32  (5.11)  (5.12)  2 app  Conversely suppose there exists g E C and A and tj’s satisfying Equations (5.5)—(5.12). Then g is an optimal solution to the problem. This is the result of Theorem 2.2. We now use these equations to calculate the optimal  gj’s.  Theorem 5.3 If g is a solution of(B), then we must have  0.  g5  53  Proof: Suppose there is an optimal g with g 5  Substitution of  [L5 =  0. Equation (5.6) implies that  =  0 into Equation (5.12) yields +  =  —  2.s  —  2t,  and substitution into Equations (5.8)—(5.11) yields ui  1 =  1 =  11  1 3 [L  =  1  —————S  n 2 r3  2 r  F  1 4 [t  =  By Equation (5.5) we must have ——s—t 2  >  —s—t  >  0  ————S 2 2  >  0  0  —  —  —+t  0,  or equivalently 5+1;  1 > P , we have 1/P 2 2 1 Since it is obvious that P s+t >t  54  . However, from the above, 2 1/P  0.  which leads to a contradiction. I The optimal designs are given in the following: Theorem 5.4 (a) g  (b) If g  =  = ( )t 1 , 0  , 3 (O,O,g , 4 O)t, g g3  0, g4  is a solution of(B) if and only ifa  1.  0, is an optimal design, then a<1  and g3  =  g4 Conversely, if a < 1, and 3 (c) If g  = 4 2 (O,g , 3 , 0)t, g  2 g  1 =  a a.  1, then g as given is an optimal design. 0, g3  0, g4  0 is an optimal design, then  13<1 a+i3>  1  and g2  =  1—/3  g3  =  1—a  g4  =  a+/3—1.  Conversely, if a < 1, /3 < 1, and a + 3> 1, then g as given is an optimal design. (d) If g  =  , 2 (gi,g , 3 0,0)t, g g 1  0, g #0, g3 < 1/2  55  0 is an optimal design, then  and 1  gi  =  g2  =  g3  =  1  Conversely, if  <  Proof of (a): We  77.  1/2, then g as given is an optimal design. note  that for g  (0,0,0,1,0),t P 7  =  2 P  =  3 P  =  1. Suppose that g  is an optimal design. Then by Theorem 2.1, there exist ) and jj’s satisfying Equations (5.5)—(5.12). Since g4 1 P  =  2 P  =  3 F  =  =  1, Equation (5.6) implies that  4 [L  0. Substituting  1 into Equation (5.11) yields =  3  —  2s  —  3t,  and substitution into Equations (5.8)—(5.10) and (5.12) yields =  2—s---2t  2 It  =  2—s—2t  3 It  =  1—s—t  IL5  =  1_t.  By Equation (5.5), we must have 2  —  s  —  2t  0  1 —s—t>0  1  —  t  0,  or equivalently s+t<1. 56  4 = L  0 and  Now we suppose that s+t  1. By the calculations above, there exist A and [Li’s satisfying  Equations (5.5)—(5.12) with g  (0,0,0, 1, 0). Thus, by Theorem 2.2, g  =  is an optimal design. Therefore g  )t 0  (0,0, 0, 1  =  =  (0,0,0, 1,  )t 0  is a solution of (B) if and only if  a>1. Proof of (b): We consider g 1 0. Then P  2 P  =  )t (0, 0, g3, g4, 0 where g3 and g4 satisfy 93  =  1, and P 3  =  0, and  g4. Suppose that g is an optimal design. Then  by Theorem 2.1 there exist A and t’s satisfying Equations (5.5)—(5.12). Since g Equation (5.6) implies  j  =  0. Substituting u 4  1 0, P  =  =  2 P  =  1 and P 3  =  =  1  0, —  into Equation (5.11) yields A Since g3  =  1 (1—g3) 2  2+  0, Equation (5.6) implies ,u  =  —  2s  —  3t.  1 0. Substituting A, P  (5.13) =  2 P  =  1 and u 3  =  0  into Equation (5.10), and also using Equation (5.13) yields g3  1/2  1 =  =  1—a,  and therefore g4  for 0 < a < 1. It is obvious that  93,  =  1—g3  =  a,  g4  e  =  2  (0, 1). Using these results we can simplify A  in Equation (5.13) to A  —  s  —  2t,  and substitution into Equations (5.8), (5.9) and (5.12) yields [Li  =  1—t  2 It  =  2—s—2t  5 [L  =  1_i. 57  By Equation (5.5), we must have 2—s—-2t  >  1—t  0 0.  This is equivalent to 8>1. Now we suppose that c < 1 and  >  1. By the argument above, there exist A and nj’s  satisfying Equations (5.5)—(5.12) with g g  =  (0, 0, 1  ,  —  =  1  =  —  c, c,  Thus by Theorem 2.2  .  =  4 2 (0,g , 3 , 0)t, g  in which case Pi  + g4. Suppose g is a solution of (B) and satisfies  g2  g  (0, 0, 1  )t a, 0 is an optimal design.  Proof of (c): We consider g and F 3  =  —  9  g  =  2 P  =  L 0, g3  3 g  4 +g  0, and  0. Then by Theorem 2.1 there exist A and tj’s satisfying Equations  (5.5)—(5.12). Since g 4  0, from Equation (5.6) we have  4 = [t  0. Substituting  =  0  into Equation (5.11) yields A  =  =  Since g  0  ,  Equation (5.6) implies i’2  =  0. Substituting A, and  (5.9) yields (93+94)  2  2 s+2t’  and so  g2  =  l—(g3+g4)  =  i—(_2_h/2  =  1-a.  + 2t1  58  /22 =  0 into Equation  Since g3  0, Equation (5.6) implies  0. Substituting \, P 1  3 = t  2 and ,u P 3  =  =  0 into  Equation (5.10) yields 1  2 (g2+g4) =  and thus /__1__1/2  g4=(  \S  )  +  —g2  tj  =  From Equation (5.7) we have g3  =  4 1—g2—g  =  1—o  Substituting these results into Equations (5.8) and (5.12) we obtain 1  1  S  =  s  1 5 [L  = L  Since we must have g,  >  0, for i  =  2, 3,4, we require c<1 9<1 1.  a+i3> Now we suppose that c  <  1,  <  1 and a+/ 3  >  1. By the arguments above, there exist \  and j’s satisfying Equations (5.5)—(5.12) with g by Theorem 2.2, g  =  (0, 1  —  Proof of (d): We consider g 3 P  =  g2. Suppose g is  /3, 1 =  —  a, /3 + a  —  =  (0, 1  —  /3, 1  —  —  )t• 1, 0 Thus  )t 1, 0 is an optimal design.  3 , 2 (gi,g , 0,0)t, g in which case P 1  an optimal design and satisfies g  59  a, /3 + a  0, g  =  gl+g3,  2 P  0 and g3  =  g3, and  0. Then  by Theorem 2.1 there exist ) and ij’s satisfying Equations (5.5)—(5.12). Since g3 Equation (5.6) implies  Since g  /13 =  0. Substituting p  0, Equation (5.6) implies  2 (5.8) and using the fact that P  =  0,  0 into Equation (5.10) yields  0. Substituting  =  0 and \ into Equation  g3, we obtain  =  /1hI2  =1,  g3=J)  which we require to be less than 1. Substituting 2 P  =  /12 =  3 g3, and P  =  Since g  0, Equation (5.6) implies  0 and \ into Equation (5.9) and using the fact that P 1  /12  =  1  i  <  =  0.  —  g, we obtain 1 g2 =  Thus, g  =  1  — —  g3 =  1/2  —  .  Since we assume g > 0, we require  1/2.  Substituting the g’s and ) into Equations (5.11) and (5.12) yields /14  By Equation (5.5), we require s + t g  =  (1/2  71 2 / 1 _, ) , 0 t  Now suppose that  g  =  /15  =  —4+s+t.  4. But this follows since t  =  (1/)2 >  4. Thus  is an optimal design.  <  fying (5.5)—(5.12) with g  =  1/2. By the calculations above, there exist .\ aild jj’s satis  =  (1/2  —  i,  1/2, i, 0,  )t• 0  Thus by Theorem 2.2 we know that  (1/2 —i,1/2,i,0,0) t is an optimal design. I  The conclusions that can be drawn from the above theorems are that, if the ob servations in the three subregions are uncorrelated, we will never sample the transect ,x 1 }. 3 {x  Sampling the transect {x ,x 1 } yields the same amount of information as sam 2  pling {x ,3 1 x } , but at a smaller cost (c + 2 y compared to 2c + 2y). We should look 60  at all subregions when both .s and t are small. We will never sample the whole region whenever (s, t) is in the set K, which is defined as follows K  =  {(s, t) : 1/(s + t) 2 + (2/(s + 2t))’ 72 < 1}.  The following figures show the five components of an optimal design as a function of s and t. For values of s and t between 0 and 8, we calculate the optimal values of gj using FORTRAN subroutine NCONF. We omit the plot of g5, since we have shown in Theorem 5.3 that g  0.  Recall that s is the startup cost/the variance of Z(x) and t is the “per unit of travel” cost/ the variance of (x). Figure 5.3.2 shows the optimal gi, the probability of choosing }. It appears that the probability of sampling {x 1 {x } is always 0 when t <4, that is 1 when r > 1/2. Fort > 4 from Theorem 5.4 (d) we know that  91 =  1/2—17  =  . 2 1/2—l/t’/  Figure 5.3.2 illustrates this fact. From Theorem 5.4 (a)—(c), we know that g’  =  0 when  s and t are small. This, too can be seen in the figure. Figure 5.3.3 shows the the optimal g2, the probability of choosing {x }. It seems 3 that the probability of choosing 3 {x is always 0 when values of s and t are both small. } From Theorem 5.4 (a) and (b) we know that g s + 2t  =  0 when a  1 or /3  1, that is, when  2. Also from Figure 5.3.3 and Theorem 5.4, we see that g is non-decreasing in  1/2fort>4. 2 salldt,andg Figure 5.3.4 shows the optimal 93, the probability of choosing {x ,x 1 }. It seems 2 that the probability of choosing 2 ,x is always 0 when values of s and t are both 1 {x } small. Furthermore, from Theorem 5.4 (b) and (c), we see that g3 function of s and t, for a +  >  1  —  a, an increasing  1. This corresponds to moderate values of s and t. By  2 for t > 4. This is clearly seen in Figure 5.3.4. Theorem 5.4 ’ 1 t  Theorem 5.4 (d), g3  =  (a) states that g3  0 if a  =  /3  =  1, that is, if .s + t  1.  Figure 5.3.5 shows the optimal g4, the probability of choosing 3 1 {x , 2 } x , that is, the whole area. It appears that probability of choosing the whole area is 1 when values 61  of s and t are both small. From Theorem 5.4 (a) we know g4  =  1 when  1, that is  when s + t < 1. Then when s and t increase, the probability of choosing the whole area goes to 0. This can been seen in the figure. Both the theorem and the figure indicate that we should visit the whole region when the costs are low. If the costs are high, we should not visit the whole area. In particular, by Theorem 5.4 (d), g4  Figure 5.3.2: The optimal values of g with a 2  0  C  C%4 C 0  00  62  =  0  0 if t  4.  0  0 O.20.40.6°.8 1  0  0  *  q  I:  t.)  CD  C,)  CD  0  0  0  CD  CD  1  CD C.;’  00.20,40  CD  q  Figure 5.3.5: The optimal values of g4 with  °2 =  OD• 0 (0•  0 0  c9  00  The case with u  0  1 and x 2 are correlated. (The correlation In this case, the observations in regions x between (x ) and 2 1 ) is equal to x  °2)  Recall when a 12  =  0, then we will never  sample {x ,x 1 } because it costs more but we get the same amount of information as 3 ,x 1 {x } . But for sampling 2  ,x 1 {x } , 0, the situation is quite different. Sampling 3  2 when the observations in x 1 and x 2 are although costly, yields information about x correlated. So the transect T(as)  =  ,x 1 {x } will be chosen sometimes because sampling 3  ) 3 ) will yield more information than sampling T(a 5 transect T(a  =  ,x 1 {x }. 2  Although we do not have explicit expressions for the optimal designs for this case, we can always calculate the results using numerical methods. The following figures show the five components gi through g5, of an optimal design as a function of s and t for different o ’s. These values were calculated using NCONF. 2 Figures 5.3.6—5.3.10 display the  five  components  64  of  an  optimal design as a function  of s and t when u’ 2  =  0.2. Here g5 is included since it is no longer 0 for this case.  Figures 5.3.11—5.3.15 are similar to Figures 5.3.6—5.3.10 except  12 =  0.6.  Comparing Figures 5.3.6 and 5.3.11 with Figure 5.3.2, we can see the probability of choosing {x } increases with u 1 . That is because, for fixed cost, when the correla 2 tion between the observation in regions x and x 2 increases, sampling x 1 yields more information about x . It is therefore reasonable to have a larger probability of sampling 2 {Xi}.  Comparing Figures 5.3.7 and 5.3.12 with Figure 5.3.3, we can see when  ‘2  increases  from 0 to 0.2 to 0.6, the probability of choosing {x } decreases. This might be because, 3 for fixed costs, when the correlation of the observations in regions x 1 and x 2 increases, sampling x 1 yields more information than sampling x . 3 Comparing Figure 5.3.10 to Figure 5.3.15, and also noting the fact g5 =  0 when  0, we can conclude that the probability of choosing {x } increases as 3 ,x 1  increases, at least for small .s and some t’s. Recall small values of s corresponding to low startup costs, in which case, the cost of sampling {x ,x 1 } is comparable to that of 3 sampling {x, 2 x } . Similar to the previous discussion, because of the correlation of the observations in regions x and x , sampling the transect {x 2 ,x 1 } yields more information 3 ,x 1 than sampling {x }. Thus it is worthwhile to sample {x 2 ,x 1 } even though it is more 3 costly than sampling 2 ,x 1 {x } . From these figures, we cannot find any obvious relationship for g3 and g4 when u’ 2 changes from 0 to 0.2 to 0.6. The way that .s and t affect the optimal gj is quite similar to the case when o 2 except for g5. In the uncorrelated case  (°2 =  65  0  0), the optimal g 5 is equal to 0, and thus  does not change with s and t. But for the dependent case (a 2 depends on the values s and t.  =  0), the optimal g5  0  cP  00.20A0.60.8 1  cP  C  0  CD  CD C  C  C  I:  — *  q  CD  (1)  CD  —  CR  CD  CD CR  ,  00.20.40.60.8 1  0  00.20.40.60.8 1  0  .3 q *  0,  Ct,  0  0  (D 0  Cl’  aq  (l  ct  0  C.T  ‘1  c,q  00.20.40.60.8 1  00  0  O0.20.40.6°.8 1  *  q  cP  Cl)  0  0  ,-.  *  q  C”  CD  IJ  0  0  CD  CD  CD  0  Ct’ CR  CD CR  cjq  00.20.40.60.8 1  0  00.200.60.8 1  (3  0  C’)  CD  I:  0  0  0  —  CD  CD  aq CD  00.20.40.60.8 1  CD  —  q,  0  cP  00.20.40.60.8 1  0  —  C,’  *  0  C’)  CD  0  —‘  *  0  CD  e÷.  CD  0  3  CD  c,q  CD  cy’  CD c;1 3  ci  00.20.40.60.8 1  References [1] Bertsekas, D.P. Constrained Optimization and Lagrange Multiplier Methods, Aca demic Press, New York, 1982. [2] Graham, A. Nonnegative Matrices and Application Topics in Linear Algebra. Ellis Horwood Ltd. 1987. [3] Heckman, N. and Rice, J. “Linear Transects of Two-dimensional Random Fields: Estimation and Design”, Technical Report, Department of Statistics, UBC, No. 132, 1993. [4] Kuhn, H.W. and Tucker, A.W. “Nonlinear Programming”, Proceedings of the Sec ond Berkeley Symposium on Mathematical Statistics and Probability, J. Neyman (ed.), 481—493, University of California Press, Berkeley, 1951. [5] McCormick, G.P. “Optimality Criteria in Nonlinear Programming”, SIAM—AMS Proceedings, Vol. 9, 27—38, 1982. [6] Rao, C.R. Linear statistical inference and its applications, Wiley, New York, 1973. [7] Seber, G.A.F. The estimation of animal abundance and related parameters, C.Griffin, London, 1982. [8] Seber, G.A.F. “A Review of Estimating Animal Abundance”, Biometrics, Vol. 42, 267—292, 1986.  71  

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