Chapter 4: Element sampling design: Part 2jkim.public.iastate.edu/teaching/chapter4.pdfIntroduction...

Chapter 4: Element sampling design: Part 2

Jae-Kwang Kim

Fall, 2014

Introduction

1 Introduction

2 Poisson sampling

3 PPS sampling

4 πps sampling

Kim Ch. 4: Element sampling design: Part 2 Fall, 2014 2 / 32

Introduction

Taxonomy

Equal probability sampling Unequal probability sampling

SRS (without replacement) πps sampling

SRS with replacement PPS sampling

Bernoulli sampling Poisson sampling

Systematic sampling Systematic πps sampling


Introduction

Why consider unequal probability sampling ?Example: N = 4 population of companies

Farm Size y (yield)

A 100 11B 200 20C 300 24D 1,000 245

Select n = 1 unit by

With equal probabilityWith probability proportional to size :

Compare the variances.


Poisson sampling

1 Introduction

2 Poisson sampling

3 PPS sampling

4 πps sampling


Poisson sampling

Definition:Ii

inde∼ Bernoulli (πi ) , i = 1, 2, · · · ,N.

If πi = π, it is called Bernoulli sampling.

Estimation (of Y =∑N

i=1 yi )

YHT =N∑i=1

Iiyi/πi

Variance

Var(YHT

)=

N∑i=1

(1

πi− 1

)y2i

Variance estimation

V(YHT

)=

N∑i=1

Iiπi

(1

πi− 1

)y2i


Poisson sampling

Optimal design: minimize Var(YHT

)subject to

∑Ni=1 πi = n

πi ∝ yi

To prove this, use Cauchy-Schwarz inequality(n∑

i=1

a2i

) n∑j=1

b2j

≥ ( n∑i=1

aibi

)2

with equality if and only if ai ∝ bi for all i = 1, · · · , n.


Poisson sampling

Disadvantage: sample size is random and it can decrease theefficiency of the HT estimator.

ExampleN=600 of students who took a test in a university. Want to estimatethe passing rate on the test. Use a Bernoulli sampling with π = 1/6.ns = 90 sample size is realized. Among the 90 sample students, 60students are found to have passed. What is a reasonable estimator ofthe total number of students who passed the test ?


Poisson sampling

Remedy1 Use an alternative estimator is

Y = N

∑Ni=1 Iiyi/πi∑Ni=1 Ii/πi

.

It is often called Hajek estimator. Its variance is

Var(Y).

=N∑i=1

(1

πi− 1

)(yi − y)2 .

2 Use rejective sampling:“conditional distribution of the Bernoulli sampling distribution givenn = n0 ⇐⇒ simple random sampling without replacement with sizen0”


PPS sampling

1 Introduction

2 Poisson sampling

3 PPS sampling

4 πps sampling


PPS sampling

Basic Setup

1 Let x1, · · · , xN be known characteristics of the population elementssuch that xk > 0 for all i . This xi is called the measure of size(MOS). Examples of MOS include the size of farm, the number ofemployees in a company, and the acreage of counties.

2 Wish to select a sample with the selection probability proportional toxi .

3 If the sample size is equal to one, then it is an easy job to select asample with probability proportional to xi .

4 Probability proportional to size (PPS) sampling idea: Use mindependent selection of a sample of size one with probabilityproportional to xi . Thus, it is a with-replacement sampling in thesense that, once drawn, an element is replaced into the population sothat all N elements participate in each draw.


PPS sampling

With-replacement sampling:

Pro: Easy to implement and to investigate its properties. Maybe agood approximation of without-replacement sampling if n/N isnegligible.Con: Sample elements can be duplicated. Inefficient.


PPS sampling

How to select a PPS sample with m = 1?

Method 1: Cumulative total method

[Step 1] Set T0 = 0 and compute Tk = Tk−1 + xk , k = 1, 2, · · · ,N.

[Step 2] Draw ε ∼ Unif (0, 1). If ε ∈ (Tk−1/TN ,Tk/TN), element k isselected.

Very popular. It needs a list of all xk in the population.


PPS sampling

Method 2: Lahiri’s method

[Step 0] Choose M > {x1, x2, · · · , xN}. Set r = 1.

[Step 1] Draw kr by SRS from {1, 2, · · · ,N}.[Step 2] Draw εr ∼ Unif (0, 1).

[Step 3] If εr ≤ xkr /M, then select element kr and stop. Otherwise,reject kr and goto Step 1 with r = r + 1.

The basic idea is based on the rejection algorithm due to Von Neymann.


PPS sampling

Justification

πk = Pr(k ∈ A)

=∞∑r=1

Pr

Kr = k, εr <xkrM,

r−1⋂j=1

(εj >xkjM

)

=

∞∑r=1

1

N

xkM×

r−1∏j=1

Pr{εj >

xkjM

}=

∞∑r=1

1

N

xkM

(1− xU

M

)r−1

=1

N

xkM

1

1− (1− xU/M)

=xk∑Ni=1 xi

.


PPS sampling

Let ai be the index of the element in the i-th with-replacementsampling.

Unequal probability with replacement: Consider p1, p2, · · · , pN > 0such that

∑Ni=1 pi = 1. We can construct pk from xk by

pk = xk/∑N

i=1 xi .

On the i-th draw, label k is selected with probability pk . That isPr (ai = k) = pk . Note that

πk = Pr (k ∈ A)

= 1− Pr (k /∈ A)

= 1− (1− pk)m

1 For m = 1, πk = pk .2 For m > 1 and pk ’s are small, πk

.= mpk .


PPS sampling

Estimator of Y =∑N

i=1 yi :1 First, define

Zi =yaipai

=N∑

k=1

ykpk

I (ai = k) .

Note that Z1, · · · ,Zm are independent random variables since the mdraws are independent.

2 Z1, · · · ,Zm are identically distributed since the same probabilities areused at each draw, where E (Zi ) = Y and

V (Zi ) =N∑

k=1

(ykpk− Y

)2

pk ≡ V1.

3 Thus, Z1, · · · ,Zm are IID with mean Y and variance V1. Usez =

∑mk=1 Zk/m to estimate Y .


PPS sampling

Hansen-Hurwitz estimator:

YHH ≡m∑

k=1

Zk/m

where Zi = yk/pk if ai = k .

Properties1 Unbiased estimator of Y =

∑Ni=1 yi

2 V(YHH

)= V1/m by standard result.

3 Unbiased estimator of V(tHH)

is

V(YHH

)=

1

m

1

m − 1

m∑i=1

(zi − z)2

4 For large m, YHH is AN (Y ,V1/m).


πps sampling

1 Introduction

2 Poisson sampling

3 PPS sampling

4 πps sampling


πps sampling

Ideally,1 The actual selection of the sample is relatively simple.2 The first-order inclusion probabilities πi are strictly proportional to xk .3 The second-order inclusion probabilities satisfy πkl > 0 for all k 6= l .

(measurable sampling design)4 The πkl can be computed without very heavy calculations.5 ∆kl = πkl − πkπl < 0 for all k 6= l to guarantee that the SYG variance

estimator is always nonnegative.


πps sampling

Motivation:

PPS sampling satisfies the above conditions but it can have duplicatedsample elements → inefficient.Want to find a fixed-size sampling design with πk ∝ xk where xk > 0and known. It is called πps design.

Remark: For fixed-size design, πk ∝ xk and∑N

i=1 πi = n leads to

πk =nxk∑Ni=1 xi

,

which can be contradictory to the fact that πk → 1 for n→ N. Also,πk can be grater than 1 if xk is extremely large.

Strict πps sampling may not be feasible from a particular population.We may need to truncate some of xi such that πi computed above isno greater than one.


πps sampling

Problem: Given π1, · · · , πN , where∑N

i=1 πi = n and πi ∈ (0, 1] for alli , how to select a nonreplacement sample of size n ?

Classification of the procedures for the selection of nonreplacementunequal probability samples

1 Draw-by-draw method2 Mass draw procedure (rejective sampling)3 Systematic procedure


πps sampling

Draw-by-draw method (n = 2)

Notationθi : the probability of selecting i in the first sample selection.θj|i : the probability of selecting j in the second sample selection giventhat i is selected in the first sample selection.

Inclusion probabilitiesSecond order inclusion probability (for i 6= j)

πij = θiθj|i + θjθi|j

First order inclusion probability: Since∑

j 6=i πij = πi ,

πi = θi +∑j 6=i

θjθi|j

Restrictions on θi and θj|i :

θi +∑j 6=i

θjθi|j = πi .


πps sampling

Example

Unequal Nonreplacement sampling of size n = 2 from the population ofsize N = 5.

ID pi = πi/n πi1 0.20 0.42 0.10 0.23 0.25 0.54 0.25 0.55 0.20 0.4

How to choose θi and θj |i ?


πps sampling

Example

If we choose θi = pi and θj |i ∝ pj , then

ID pi θi θj |1 θj |2 θj |3 θj |4 θj |51 0.2 0.20 0 4/18 4/15 4/15 4/162 0.1 0.10 2/16 0 2/15 2/15 2/163 0.25 0.25 5/16 5/18 0 5/15 5/164 0.25 0.25 5/16 5/18 5/15 0 5/165 0.2 0.20 4/16 4/18 4/15 4/15 0

π1 = 0.2 + 0.1(4/18) + 0.25(4/15) + 0.25(4/15) + 0.2(4/16)

= 0.4056 6= 0.4

π2 = 0.1 + 0.2(2/16) + 0.25(2/15) + 0.25(2/15) + 0.2(2/16)

= 0.2167 6= 0.2


πps sampling

Draw-by-draw method (n = 2)

Brewer (1963) method: Use

θi ∝pi (1− pi )

1− 2pi

andθj |i ∝ pj

Durbin (1967) method: Use

θi ∝ pi

and

θj |i = π−1i πij ∝ pj

(1

1− 2pi+

1

1− 2pj

)


πps sampling

Motivation for Brewer’s method

If we choose θj |i = pj/(1− pi ) for i 6= j , using

πi = θi +∑j 6=i

θjθi |j ,

we have

2pi = θi +∑j 6=i

θjpi

1− pj

= θi + pi∑j 6=i

θj1− pj

= θi + pi

(B − θi

1− pi

)Therefore,

θi = (2pi − piB)/{1− pi/(1− pi )} ∝ pi (1− pi )/(1− 2pi )


πps sampling

Motivation for Durbin’s method

For θi = pi = πi/2, the choice of θj |i = πij/πi satisfies

πi = θi +∑j 6=i

θjθi |j ,

πij = θiθj |i + θjθi |j

Durbin used

πij =πiπj

2(1 + A)

(1

1− πi+

1

1− πj

)where A = 0.5

∑Ni=1 πi/(1− πi ) =

∑Ni=1 pi/(1− 2pi ).


πps sampling

Remark

The joint inclusion probabilities from the two methods are identical.

Satisfies πij < πiπj : nonnegative variance estimation forSen-Yates-Grundy formula.

Extension to n > 2 is possible, but involves heavy computation for theconditional selection probabilities. Discussed by Rao (1965),Sampford (1967), and Fuller (1971).


πps sampling

Systematic πps sampling

1 Choose R ∼ Unif (0, a]

2 Unit i is selected if

i−1∑j=1

xj < R + ka ≤i∑

j=1

xj

for some k = 0, 1, · · · , n − 1, where a =∑N

i=1 xi/n is the samplinginterval for the systematic sampling.


πps sampling

Example: Systematic πps with n = 2

ID πi CL CU

1 0.4 0 0.42 0.2 0.4 0.63 0.5 0.6 1.14 0.5 1.1 1.65 0.4 1.6 2.0

A = {1, 3} for 0 < R ≤ 0.1

= {1, 4} for 0.1 < R ≤ 0.4

= {2, 4} for 0.4 < R ≤ 0.6

= {3, 5} for 0.6 < R ≤ 1


πps sampling

Systematic πps sampling

First order inclusion probability: let l be the integer satisfyingl · a ≤ CLk < CUk ≤ (l + 1)a.

Pr (k ∈ A) = Pr {CLk < R + l · a ≤ CUk}

=

∫ CUk−l ·a

CLk−l ·a

1

adt =

nxk∑k∈U xk

.

Easy to compute. Very popular

πij = 0 for some i , j .

Efficiency depends on the sorting variable.


Chapter 4: Element sampling design: Part 2jkim.public.iastate.edu/teaching/chapter4.pdfIntroduction...

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