Population element: 1 2 N. 1.1 Sampling with Replacement: Hansen-Hurwitz Estimator(HH)

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1 Chapter 1 Samplng wth Unequal Probabltes Notaton: Populaton element: 1 2 N varable of nterest Y : y1 y2 y N Let s be a sample of elements drawn by a gven samplng method. In other words, s s a subset of f1; 2; ;Ng. 1.1 Samplng wth Replacement: Hansen-Hurwtz Estmator(HH) Suppose that a sample s of sze n s drawn wth replacement and that on each draw the probablty of selectng the th unt of the populaton s p. Then the Hansen-Hurwtz (1943) estmator of the populaton total, Y T = Propertes: y s dened as 1. ^YHH s unbased. 2. Var( ^YHH )= 1 n ^Y HH = 1 nx p ( y p, Y T ) 2. 2s 3. An unbased estmator for Var( ^YHH )s Remarks dvar( ^YHH )= 1 n(n, 1) y p X 2s ( y, ^YHH ) 2 : p 1. Because of the wth-replacement samplng, a unt may be selected on more than one draw. The HH estmator utlzes the unt's value as many tmes as the unt s selected. 2. Suppose x s an auxlary varable, and x1;x2; ;x N are known pror to samplng. We can try to ncorporate ths auxlary nformaton at the 1

2 samplng stage of probablty-proportonal-to-sze(pps) samplng. If we x select p / x,.e. p = P N x z,then ^YHH = 1 np2s y z. Ths s called the probablty-proportonal-to-sze-wth-replacement (PPSWR) estmator. EXAMPLE 1: Strp transect samplng Consder a study area of 100 (km) 2 parttoned nto strps 1 km wde but varyng n length. A sample of n = 4 strps are selected by draw-by-draw wth replacement. The number of anmals (Y )scounted n each of these 4 strps. The data are shown below: sampled length of strp strp (n km) p y Estmate the total number of anmals n ths area and ts standard error. 1.2 Any desgn: Horvtz-Thompson Estmator(HT) Suppose that wth any desgn, wth or wthout replacement, the probablty of ncludng unt n the sample s (> 0), for =1; 2; ;N. Then, the Horvtz-Thompson (1952) estmator for the populaton total Y T s dened as ^Y HT = X 2s where s contans the dstnct unts only and hence the sze of s could be less than the number n of unts drawn. If the samplng s wthout replacement, the sze of s must be n. Note that the Horvtz-Thompson (1952) estmator depends only on the dstnct unts n the sample, not on number of repeated selected unts. Revew of Incluson Probabltes - = P ( 2 s), the rst-order ncluson probablty, =1; 2; ;N. - j = P ( 2 s and j 2 s), the second-order ncluson probablty, 6= j. y 2

3 Then = E(D )and j = E(D D j ), where D = 1 f 2 s 0 f =2 s ;; 2; ;N: 1. Under a samplng wth replacement, t can be shown that for a xed sample sze n, =1, (1, p ) n and j = + j, [1, (1, p, p j ) n ]; where p = probablty of selectng the th unt of the populaton on each draw. 2. Under a smple random samplng wthout replacement, t can be shown that for a xed sample sze n, = n N and j = n(n, 1) N(N, 1) : Propertes of HT estmator: 1. ^YHT s unbased. 2. Two equvalent expressons for Var( ^YHT ): Var( ^YHT )= Var( ^YHT )= 1, y 2 + j=1 6=j ( j, j )( y, y j ) 2 j=1 j <j j, j j y y j (1) (f samplng s WOR): (2) 3. Based on (1) and (2), we can obtan twounbased estmators for Var( ^YHT ): ^V1 = X 2s 1, 2 ^V2 = XX ;j2s <j y 2 + XX j, j j ;j2s 6=j j, j j j y y j (3) ( y, y j j ) 2 : (4) 3

4 Remarks (a) ^V1 and ^V2 are unbased for Var( ^YHT )provded that all j > 0. (b) ^V2 s guaranteed to be postve for any samplng WOR method satsfyng j > j.also^v2 takes negatve values less often than ^V1. (c) Suppose we select / x,forall. If samplng s wthout replacement and the sample sze n s xed, t can be shown that = nx nz : P N x Then ^YHT = 1 np2s y z. Ths s called the probablty-proportonal-tosze-wthout-replacement (PPSWOR) estmator. EXAMPLE 2: Refer to EXAMPLE 1 Snce the sample s drawn wth replacement and only contans the dstnct unts, we have =1, (1, p ) n and j = + j, [1, (1, p, p j ) n ]. Based on these probabltes, nd the Horvtz-Thompson estmator of the total number of anmals and estmate ts standard error. EXAMPLE 3: Three-stage cluster samplng Suppose a regon s dvded nto L dstrcts. In dstrct, there are M households. Household j n dstrct has N j members. 1. In the rst stage, a SRS of ` dstrcts wll be drawn wthout replacement. 2. In the second stage, a SRS of m households wll be drawn wthout replacement from dstrct selected n the rst stage. 3. In the thrd stage, a SRS of n j household members wll be drawn wthout replacement from household j (n selected dstrct ) selected n the second stage. Then, (a) the probablty that dstrct wll be ncluded n the sample of dstrcts s `L, (b) the probablty that household j wll be ncluded n the sample of households from dstrct s m M, 4

5 (c) the probablty that member k wll be ncluded n the sample of members from household j s n j N j. It follows that member k wll be ncluded n the ultmate sample s: k = ` m n j : L M N j Suppose that the populaton has N =9persons: Household Dstrct Household members `L 1. North 1. Brown 1. John 1. North 1. Brown 2. Mary 1. North 2. Smth 1. Jane 2. South 1. Jones 1. Alce 2. South 1. Jones 2. Peter 2. South 2. Cox 1. Jacob 2. South 2. Cox 2. Sarah 2. South 2. Cox 3. Melnda 2. South 3. Elton 1. Davd m M n j N j k Assume that ` = 1 dstrct wll be selected n the rst stage, m = 2 households wll be selected from each dstrct selected n the rst stage, and n j = 1 member wll be selected from each household selected n the second stage. Based on ths samplng method, Peter Jones and Davd Elton are selected and ther ages are 25 and 60 respectvely. Fnd the HT estmate of the mean age of the populaton. EXERCISES Q1. Wrte down the Horvtz-Thompson estmator for the populaton mean, Y = 1 N y and obtan an unbased estmator for ts varance. Q2. Wrte down the Horvtz-Thompson estmator for the proporton of elements n the populaton that fall nto a gven category and obtan an unbased estmator for ts varance. 5

6 1.3 How to Draw a PPS Sample Wthout Replacement? It s dcult to draw a PPS sample wthout replacement. Over 50 methods have been proposed n the lterature but none s perfect. EXAMPLE 4 Suppose N =4andn = 2, and x1 =10;x2 =90;x3 =70; and x4 =30: Q. How to select a PPS sample of sze 2 wthout replacement? A1. Durbn (1967) method (for n =2) Step 1. Select 1 unt from the populaton wth probablty proportonal to x Assume unt s chosen at step 1. Step 2. Select a second unt from the remanng N, 1 unts wth probablty proportonal to 1 z j [ + 1, 2z 1 1, 2z j ]; for j 6= and z j = x j X T : For Durbn method, t can be shown that =2z and j = 2z z j (1, z, z j ) D(1, 2z )(1, 2z j ) ; where D = z (1, z ) 1, 2z : These j 's are needed n calculatng ^V1 and ^V2. It can also be shown that () all j > 0=) ^V1 and ^V2 are unbased for Var( ^Y HT ). () j, j > 0 for all 6= j =) ^V2 s always non-negatve. 6

7 A2. Systematc PPS method (Madow (1949) ndex x partal sum assgned nterval 1 x1 S1 = x1 (0;S1) 2 x2 S2 = x1 + x2 (S1;S2).... N x N S N = x1 + x2 + + x N = X T (S N,1;S N ) Dene d = X T n. Step 1. Select a unform random number u from the nterval (0;d). Step 2. Select the unts whose assgned ntervals contan u, u+d; u+2d; ;u+ (n, 1)d. For systematc P PPS method, t can be shown that = x d y ^Y sys = 2s s unbased for the populaton total Y T. = nz. Hence, Systematc PPS s popular because of ts smplcty but an unbased estmator for Var( ^Ysys ) s not avalable. Hartley and Rao (1962) examned ths method wth the unts rst arranged n random order and showed that when nz < 1 for all, Var( ^Ysys ) (1, n, 1 n )( y, Y T n )2 : whch can then be estmated by ^V3 X 2s (1, n, 1 n )( y, ^Y T n )2 : EXAMPLE 5 Suppose N =7andn = 3, and Unt, x S assgned nterval assgned nterval mod (0, 3) (0, 3) (3, 4) (3, 4) ( 4, 15) (4,10)or(0, 5) (15, 21) (0, 1) or ( 5, 10) (21, 25) ( 1, 5) (25, 27) ( 5, 7) (27, 30) ( 7, 10) 7

8 Suppose a unform random number u drawn from the nterval (0,10) s Fnd the PPSWOR sample. Soluton u ( 0, 1) ( 1, 3) ( 3, 4) ( 4, 5) ( 5, 7) ( 7, 10) PPS Sample Reasons for usng PPS samplng 1. An example of "mportance" samplng. For example n tax audtng: y = actual taxable ncome whch requres audtng x = reported taxable ncome 2. If y and x are perfectly correlated n the sense that y = ax ; = 1; 2; ;N, then = ny Y T and hence ^Y HT = X 2s y = X 2s Y T n = Y T : Thus, f y ax,forall, we can expect good performance of ^YHT under PPS samplng. Some observatons on the HT Estmator 1. Derent samplng methods may result n the same ncluson probabltes. In other words, the ncluson probabltes do not dentfy unquely a samplng method. 2. j and Var( ^Y HT ) are often hard to compute. 8

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