# Lesson 3: Unequal Probability Sampling

Lesson 3: Unequal Probability Sampling

## Overview

This lesson starts with the rationale for using unequal probability sampling in section 3.1. We then discuss in section 3.2 the Hansen-Hurwitz estimator which may be used when the sampling is with replacement. In section 3.3, we introduce the Horvitz-Thompson estimator which can be used when the sampling is with or without replacement. In section 4, a small population example is used to illustrate some properties of these two estimators. Through this example, one can see that both estimators are unbiased.

Lesson 3: Ch. 6.1, 6.2, 6.4 of Sampling by Steven Thompson, 3rd edition

## Objectives

Upon completion of this lesson, you should be able to:

• know why and when to use unequal probability sampling,
• how to perform unequal probability sampling,
• how to compute the Hansen-Hurwitz estimator and its estimated variance,
• how to compute the Horvitz-Thompson estimator and its estimated variance, and
• learn about the unbiasedness of these two estimators through an artificial small population example.

# 3.1 - Unequal Probability Sampling

3.1 - Unequal Probability Sampling

In simple random sampling, the probability that each unit will be sampled is the same. Sometimes, estimates can be improved by varying the probabilities with which units are sampled.

For example, we want to estimate the number of job openings in a city by sampling firms in that city. Many of the firms in the city are small firms. If one uses s.r.s, size of a firm is not taken into consideration and a typical sample will consist of mostly small firms. However, the number of job openings is heavily influenced by large firms.

Thus, we should be able to improve the estimate of number of job openings by giving the large firms a greater chance to appear in the sample, for example, with probability proportional to size or proportional to some other relevant aspects.

Selection probabilities

On each draw, the probability that a given population unit will be selected is denoted as: $$p_i$$, i = 1, 2, 3, ….., N.

Suppose that sampling is with replacement, the probability of selecting the ith unit in the population is $$p_i$$.

If the selection probabilities are unequal, the sample mean is not unbiased for population mean and sample total is not unbiased for population total.

Example: If larger firms are sampled with higher probability, the sample mean for job openings will be biased upward.

# 3.2 - The Hansen-Hurwitz Estimator

3.2 - The Hansen-Hurwitz Estimator

#### Try it!

Why do we use or talk about sampling with replacement?

When sampling with replacement, the variances tend to be larger. However, formula for replacement are simpler and easier to derive. When the sample size is small compared to N, with and without replacement are not too different. We often use the easier formula derived for the with replacement to approximate that for the without replacement.

Let $$p_i$$ , i = 1, ... , N denote the probability that a given population unit will be selected.

The Hansen-Hurwitz estimator is:

$$\hat{\tau}_p=\dfrac{1}{n} \sum\limits^n_{i=1} \dfrac {y_i}{p_i}$$

Since, $$E\left(\dfrac{y_i}{p_i}\right)=\tau$$

where $$\tau=\sum\limits^N_{i=1} y_i=\text{the population total}$$
thus, $$E(\hat{\tau}_p)=\tau$$ and $$\hat{\tau}_p$$ is an unbiased estimator for $$\tau$$.

Since $$Var\left(\dfrac{y_i}{p_i}\right)=\sum\limits^N_{i=1} p_i \left(\dfrac{y_i}{p_i}-\tau \right)^2$$, $$Var(\hat{\tau}_p)=\dfrac{1}{n}\sum\limits^N_{i=1} p_i \left(\dfrac{y_i}{p_i}-\tau \right)^2$$

An unbiased estimator for $$Var(\hat{\tau}_p)$$ is:

$$\hat{V}ar(\hat{\tau}_p)=\dfrac{1}{n}\cdot \dfrac{\sum\limits^n_{i=1} \left(\dfrac{y_i}{p_i}-\hat{\tau}_p \right)^2}{n-1}$$

and an approximate (1 - $$\alpha$$) 100% confidence interval for $$\tau$$ is:

$$\hat{\tau}_p \pm t \cdot \sqrt{\hat{V}ar(\hat{\tau}_p)}$$

For population mean $$\mu$$ = $$\frac{\tau}{N}$$ one uses:

$$\hat{\mu}_p=\dfrac{1}{N} \left(\dfrac{1}{n}\cdot \sum\limits^n_{i=1}\dfrac{y_i}{p_i}\right)=\dfrac{\hat{\tau}_p}{N}$$

$$E(\hat{\mu}_p)=\dfrac{\tau}{N}= \mu$$

$$\hat{V}ar(\hat{\mu}_p)=\dfrac{1}{N^2}\cdot \hat{V}ar(\hat{\tau}_p)$$

How do we perform unequal probability sampling according to given $$p_i$$ ?

## Example 3-1: Total Number of Computer Help Requests

Estimate the total number of computer help requests for last year in a large firm.

The director of computer support department plans to sample 3 divisions of a large firm that has 10 divisions, with varying numbers of employees per division. Since number of computer support requests within each division should be highly correlated with the number of employees in that division, the director decides to use unequal probability sampling with replacement with $$p_i$$ proportional to number of employees in that division.

Division Number of Employees $$p_i$$ Assigned Numbers
1 1000
2 650
3 2100
4 860
5 2840
6 1910
7 390
8 3200
9 1500
10 1200

Total 15650

#### Questions

1. How do we practically implement unequal probability sampling according to the given $$p_i$$'s?
2. With the divisions selected by probability proportional to size, how do we construct the Hansen-Hurwitz estimator for $$\tau$$?

Division Number of Employees $$p_i$$ Assigned Numbers
1 1000 1000/15650 1-1000
2 650 650/15650 1001-1650
3 2100 2100/15650 1651-3750
4 860 860/15650 3751-4610
5 2840 2840/15650 4611-7450
6 1910 1910/15650 7451-9360
7 390 390/15650 9361-9750
8 3200 3200/15650 9751-12950
9 1500 1500/15650 12951-14450
10 1200 1500/15650 14451-15650

Total 15650 1

#### Using Minitab

We can perform probability proportional to size by using Minitab to calculate this for us:

1. Generate a column C1 that contains the value 1-15650

Calc > Make patterned data > Simple set of numbers

2. Sample 3 values with replacement from the column that contains 1-15650

Calc > Random Data > Sample from columns

The values generated by Minitab are given below:

The values given by Minitab are 1085, 6261, 9787. These numbers fall into division 2, division 5 and division 8.

So, we decide to sample division 2, division 5 and division 8.  We check the record to find the number of requests for these divisions.  The results are:

• For division 2, y1 = the number requests = 420
• For division 5, y2 = the number of requests = 1785
• For division 8, y3 = the number of requests = 2198

(For this random sample shown in the example, the division are distinct. For other random samples, it is possible that the same division may be selected more than once.)

The basic assumption is that number of requests is proportional to the size of the division.

##### Computing the Hansen-Hurwitz Estimator

We will need to compute the Hansen-Hurwitz estimator for Example 3-1.

The Hansen-Hurwitz estimator for $$\tau$$ is

\begin{align}
\hat{\tau}_p &=\dfrac{1}{3}\left(420 \cdot \dfrac{15650}{650}+1785 \cdot \dfrac{15650}{2840}+2198 \cdot \dfrac{15650}{3200}\right) \\  &=\dfrac{1}{3}(10112.31+9836.36+10749.59)\\
&=10232.75 \\
\end{align}

Each of the values, 10112.31, 9836.36, and 10749.59, look fairly stable so it looks like the variance will not be too large.

\begin{align}
\hat{V}ar(\hat{\tau}_p) &=\dfrac{1}{3}\cdot \dfrac{\sum\limits^3_{i=1} \left(\dfrac{y_i}{p_i}-\hat{\tau}_p \right)^2}{3-1}\\
&=\dfrac{1}{3}\cdot \dfrac{1}{2}((10112.31-10232.75)^2+(9836.36-10232.75)^2+(10749.59-10232.75)^2)\\
&=73125.74\\
\end{align}

$$\hat{S}D(\hat{\tau}_p)=270.418$$

(See  Example 1 on p. 68-69 in the text to see an example of when a unit is chosen more than once.)

We will see that in ths example $$p_i$$ are chosen proportional to the values of a known positive auxiliary variable such as size, $$p_i=\dfrac{x_i}{\sum x_i}$$, the Hansen-Hurwitz estimator is also called p.p.s. (probability proportional to size).

Now, we need to ask ourselves, when and why would we need to use an unequal probability sampling? Let's think about the 'when' first.

When would we elect to use p.p.s.? What about if we were sampling from Penn State departments? They are of very different sizes, some are very large and others are very small. Would we automatically choose to use p.p.s.? The idea is that the thing that you are interested in has to be related to the size. If the thing that you are interested in is related to size, then you would want to use p.p.s. However, if what you are interested in has nothing to do with the size of the department, then there is no reason to use p.p.s.

Now, let us address the 'why'. By definition,

$$\tau=\sum\limits_{i=1}^N y_i$$ and  $$Var(\hat{\tau}_p)=\dfrac{1}{n}\sum\limits^N_{i=1} p_i \left(\dfrac{y_i}{p_i}-\tau \right)^2$$

For the special and unrealistic case $$\frac{y_i}{p_i}$$= constant, the constant will be$$\tau$$and the $$Var(\hat{\tau}_p)$$will be zero. Therefore, you want $$\frac{y_i}{p_i}$$ to be close to a constant. However, in reality, prior to sampling, the $$y_i$$ are unknown and we can not choose $$p_i$$ proportional to $$y_i$$. If we know $$y_i$$ is approximately proportional to a known variable such as $$x_i$$, then we can choose $$p_i$$ proportional to $$x_i$$. $$\hat{\tau}_p$$ will have low variances.

## Example 3-2: Total Number of Palm Trees

We want to estimate the total number of palm trees on 100 islands in a tropical paradise. The area of each island is known and it is reasonable to think that the number of palm trees on each island is approximately proportional to the size of the island.

We know that the sizes of the island are given (e.g., size of island 1 is 1 square mile, size of island 29 is 5 square mile and size of island 36 is 2 square miles. The total size of these 100 islands are 100 square miles. We find that $$p_i$$, ... , $$p_N$$ are:

How can we sample 4 islands by probabilities $$p_1$$, ... , $$p_{100}$$?

1. Assign an interval width of $$p_i$$ to ith unit
2. Generate 4 random numbers form a uniform distribution on (0,1)
3. Choose the units that correspond to the interval containing the random number.

In this example, we use Minitab >> Calc >> Random data >> Uniform and get: 0.335257, 0.0065551, 0.401869, 0.318977

The units selected are the islands 29, 1, 36, and 29, (since 0.335257 falls between 0.31 and 0.36, 0.0065551 falls between 0 and 0.01, 0.401869 falls between 0.40 and 0.42, and 0.318977 falls between 0.31 and 0.36.) The measurements ($$y_i$$) are:

 i size $$p_i$$ $$y_i$$ 1 1 0.01 14 29 5 0.05 50 29 5 0.05 50 36 2 0.02 25

Given these results we should now be able to estimate how many total palm trees are there on all of the islands put together:

\begin{align}
\hat{\tau}_p &=\dfrac{1}{4}\left(\dfrac{14}{0.01}+\dfrac{50}{0.05}+\dfrac{50}{0.05}+\dfrac{25}{0.02}\right) \\
&=\dfrac{1}{4}(1400+1000+1000+1250)\\
&=1162.5 \\
\end{align}

\begin{align}
\hat{V}ar(\hat{\tau}_p) &=\dfrac{1}{n(n-1)} \sum\limits^n_{i=1} \left(\dfrac{y_i}{p_i}-\hat{\tau}_p \right)^2\\
&=\dfrac{1}{4\cdot3}[ (1400-1162.5)^2+(1000-1162.5)^2+(1000-1162.5)^2+(1250-1162.5)^2]\\
&=9739.58\\
\end{align}

$$\hat{S}D(\hat{\tau}_p)=98.69$$

If we are interested in the mean number of trees per island in that population, then

$$\hat{\mu}_p=\dfrac{\hat{\tau}_p}{N}=\dfrac{1162.5}{100}=11.625$$

\begin{align}
\hat{V}ar(\hat{\mu}_p) &=\dfrac{1}{N^2} \cdot \hat{V}ar(\hat{\tau}_p)\\
&=\dfrac{1}{(100)^2}\cdot 9739.58\\
&=0.973958\\
\end{align}

$$\hat{S}D(\hat{\mu}_p)=0.987$$

# 3.3 - The Horvitz-Thompson Estimator

3.3 - The Horvitz-Thompson Estimator

Horvitz-Thompson (1952) introduced an unbiased estimator for $$\tau$$ for any design, with or without replacement.

Horvitz-Thompson estimatorample

$$\pi_i$$, i = 1, ... , N are given positive numbers that represent the probability that unit i is included in the sample under a given sampling scheme. The Horvitz-Thompson estimator is:

$$\hat{\tau}_\pi=\sum\limits_{i=1}^\nu \dfrac{y_i}{\pi_i}$$

Where $$\nu$$ is the distinct number of units in the sample. The Horvitz-Thompson estimator does not depend on the number of times a unit may be selected. Each distinct unit of the sample is utilized only once.

Read section 6.5 in the text. The section reviews the proofs for how the following two formula are derived.

Note that:

$$E(\hat{\tau}_\pi)=\tau$$

$$Var(\hat{\tau}_\pi)=\sum\limits_{i=1}^N \left( \dfrac{1-\pi_i}{\pi_i}\right)y^2_i + \sum\limits_{i=1}^N \sum\limits_{j\neq i} \left( \dfrac{\pi_{ij}-\pi_i \pi_j}{\pi_i \pi_j}\right) y_i y_j$$

where $$\pi_{ij}$$ > 0 denotes the probability that both unit i and unit j are included.

The estimated variance of the Horvitz-Thompson estimator is given by:

$$\hat{V}ar(\hat{\tau}_\pi)=\sum\limits_{i=1}^v \left( \dfrac{1-\pi_i}{\pi^2_i} \right) y^2_i + \sum\limits_{i=1}^v \sum\limits_{j\neq i} \left( \dfrac{\pi_{ij}-\pi_i \pi_j}{\pi_i \pi_j}\right)\dfrac{1}{\pi_{ij}} y_i y_j$$

Where $$\pi_{ij}$$ > 0 denotes the probability that both unit i and j are included.

An approximate (1-$$\alpha$$) 100% CI for $$\tau$$ is:

$$\hat{\tau}_\pi \pm t_{\alpha/2} \sqrt{\hat{V}ar(\hat{\tau}_\pi)}$$

where t has $$\nu$$ - 1 df

## Example 3-3: Estimating the Total Number of Palm Trees with Horvitz-Thompson Estimator

The Horvitz-Thompson estimator of the total number of palm trees. Since, for that example the sample is with replacement, the n draws are independent. It is relatively easy to compute the $$\pi_{i}$$'s .

For sample with replacement, we will compute:

\begin{align}
\pi_i &= \text{the probability of inclusion of the ith unit}\\
&= 1-P(\text{ith unit is not included})\\
&= 1-(1-p_i)^n \\
\end{align}

Recall: Samples 1, 29 and 36 are selected.

Since $$p_1=0.01$$, $$\pi_1=1-(1-0.01)^4=0.0394$$, and

$$p_2=0.05$$, $$\pi_2=1-(1-0.05)^4=0.1855$$
$$p_3=0.02$$, $$\pi_3=1-(1-0.02)^4=0.0776$$

Therefore,

\begin{align}
\hat{\tau}_\pi &= \sum\limits_{i=1}^\nu \dfrac{y_i}{\pi_i}\\
&= \dfrac{14}{0.0394}+\dfrac{50}{0.1855}+\dfrac{25}{0.0776}\\
&= 947.037\\
\end{align}

Next, we need to compute the estimated variance, $$\hat{V}ar(\hat{\tau}_\pi)$$. For this, we need to compute $$\pi_{ij}$$ .

Since \begin{align}
P(A\cap B) &= P(A)+P(B)-P(A\cup B)\\
&= P(A)+P(B)-[1-P(A^c \cap B^c)]\\
\end{align}

Then we get:

$$\pi_{ij}=\pi_i+\pi_j-[1-(1-p_i-p_j)^n]$$

This means that we have to run through each of the unit pairs such as:

$$\pi_{12}=0.0394+0.1855-[1-(1-0.01-0.05)^4]=0.00565$$

$$\pi_{13}=0.0394+0.0776-[1-(1-0.01-0.02)^4]=0.00229$$

$$\pi_{23}=0.1855+0.0776-[1-(1-0.05-0.02)^4]=0.01115$$

plugging in the values, we obtain:

$$\hat{V}ar(\hat{\tau}_\pi)=92692.9$$

Thus, $$\hat{S}D(\hat{\tau}_\pi)=\sqrt{92692.9}=304.455$$

where $$\nu$$ = the # of distinct units, $$\nu$$ = 3, therefore the df = $$\nu$$ - 1 = 2

## Try it!

Is there some popular estimator that can be derived as a Horvitz-Thompson estimator?

Yes, under simple random sampling, the inclusion probability of the ith unit is:

$$\pi_i=n/N$$

Thus,

\begin{align}
\hat{\tau}_\pi &= \sum\limits_{i=1}^n \dfrac{y_i}{\pi_i}\\
&= \sum\limits_{i=1}^n \dfrac{y_i}{n} \cdot N\\
&= N \bar{y}\\
\end{align}

# 3.4 - Small Population Example

3.4 - Small Population Example

## Example 3-4: Wheat Production

( Reference Section 6.4 of the text)

 unit (Farm) i 1 2 3 Selection Prob, $$p_i$$ 0.3 0.2 0.5 Wheat produced 11 6 25

N = 3 farms, n = 2 sample with replacement.

 s p(s) ys 1, 1 0.3(0.3) = 0.09 (11, 11) 2, 2 0.2(0.2) = 0.04 (6, 6) 3, 3 0.5(0.5) = 0.25 (25, 25) 1, 2 0.3(0.2) = 0.06 (11, 6) 2, 1 0.2(0.3) = 0.06 (6, 11) 1, 3 0.3(0.5) = 0.15 (11, 25) 3, 1 0.5(0.3) = 0.15 (25, 11) 2, 3 0.2(0.5) = 0.10 (6, 25) 3, 2 0.5(0.2) = 0.10 (25, 6)

Question: Compute the Hansen-Hurwitz estimator.

When (1,1) is sampled, the Hansen-Hurwitz estimator is:

$$\hat{\tau}_p=\dfrac{1}{2}\left(\dfrac{y_1}{p_1}+\dfrac{y_1}{p_1}\right)=\dfrac{1}{2}\left(\dfrac{11}{0.3}+\dfrac{11}{0.3}\right)=36.67$$

When (1,2) is sampled, the Hansen-Hurwitz estimator is:

$$\hat{\tau}_p=\dfrac{1}{2}\left(\dfrac{y_1}{p_1}+\dfrac{y_2}{p_2}\right)=\dfrac{1}{2}\left(\dfrac{11}{0.3}+\dfrac{6}{0.2}\right)=33.33$$

Similarly, we can fill out the table and get the Hansen-Hurwitz estimators as shown below:

 s p(s) ys $$\hat{\tau}_p$$ 1, 1 0.3(0.3) = 0.09 (11, 11) 36.67 2, 2 0.2(0.2) = 0.04 (6, 6) 30.00 3, 3 0.5(0.5) = 0.25 (25, 25) 50.00 1, 2 0.3(0.2) = 0.06 (11, 6) 33.33 2, 1 0.2(0.3) = 0.06 (6, 11) 33.33 1, 3 0.3(0.5) = 0.15 (11, 25) 43.33 3, 1 0.5(0.3) = 0.15 (25, 11) 43.33 2, 3 0.2(0.5) = 0.10 (6, 25) 40.00 3, 2 0.5(0.2) = 0.10 (25, 6) 40.00

Question: Compute the Horvitz-Thompson estimator.

$$\pi_1=0.09+0.06+0.06+0.15+0.15=0.51$$
$$\pi_2=0.04+0.06+0.06+0.10+0.10=0.36$$
$$\pi_3=0.25+0.15+0.15+0.10+0.10=0.75$$

When (1,1) is sampled, the Horvitz-Thompson estimator is:

$$\hat{\tau}_\pi=\left(\dfrac{11}{0.51}\right)=21.57$$

When (1,2) is sampled, the Horvitz-Thompson estimator is:

$$\hat{\tau}_\pi=\left(\dfrac{11}{0.51}+\dfrac{6}{0.36}\right)=38.24$$

Similarly, we can fill out the table and get the Horvitz-Thompson estimators as shown below:

 s p(s) ys $$\hat{\tau}_p$$ $$\hat{\tau}_\pi$$ 1, 1 0.3(0.3) = 0.09 (11, 11) 36.67 21.57 2, 2 0.2(0.2) = 0.04 (6, 6) 30.00 16.67 3, 3 0.5(0.5) = 0.25 (25, 25) 50.00 33.33 1, 2 0.3(0.2) = 0.06 (11, 6) 33.33 38.24 2, 1 0.2(0.3) = 0.06 (6, 11) 33.33 38.24 1, 3 0.3(0.5) = 0.15 (11, 25) 43.33 54.90 3, 1 0.5(0.3) = 0.15 (25, 11) 43.33 54.90 2, 3 0.2(0.5) = 0.10 (6, 25) 40.00 50.00 3, 2 0.5(0.2) = 0.10 (25, 6) 40.00 50.00 Mean 42 42 Variance 34.67 146.46

$$\text{Mean of }\hat{\tau}_p = E(\hat{\tau}_p)=\sum{p(s) \cdot \hat{\tau}_p (s)}$$
$$= 0.09\times36.67+0.04\times30.00+0.25\times50.00+0.06\times33.33+0.06\times33.33$$
$$+0.15\times43.33+0.15\times43.33+0.10\times40.00+0.10\times40.00$$
$$= 42$$

From the table above we can see that both $$\hat{\tau}_p$$ and $$\hat{\tau}_\pi$$are unbiased. This example is a small population example to illustrate conceptually the properties of these estimators. We can compute the variance for $$\hat{\tau}_p$$and the variance for $$\hat{\tau}_\pi$$ directly from the definition of variance.

Since mean of $$\hat{\tau}_p$$ is 42 and E(g) = $$\sum p(s)*g(s)$$, we can compute the variance of $$\hat{\tau}_p$$ as:

$$Var(\hat{\tau}_p) = E[\hat{\tau}_p-\text{mean of }\hat{\tau}_p]^2$$
$$= 0.09\times(36.67-42)^2+0.04\times(30.00-42)^2+0.25\times(50.00-42)^2+0.06\times(33.33-42)^2$$
$$+0.06\times(33.33-42)^2+0.15\times(43.33-42)^2+0.15\times(43.33-42)^2+0.10\times(40.00-42)^2$$  $$+0.10\times(40.00-42)^2$$
$$=34.67$$

Similarly, we can compute that the variance of $$\hat{\tau}_\pi$$is 146.46.

Now, how do we compute the MSE of $$\hat{\tau}_p$$?

By definition, $$MSE (\hat{\tau}_p) = E(\hat{\tau}_p-\tau)^2$$, in this case, since $$\hat{\tau}_p$$is unbiased and $$E(\hat{\tau}_p) =\tau$$ ,$$MSE (\hat{\tau}_p)$$is the same as $$Var(\hat{\tau}_p)$$.

Remark 1. The above demonstration is just a teaching tool. In reality we will not know the population and will not come across small population problems like this other than in exams and homeworks. What we know are:

 Unit 1 2 3 Selection probability 0.3 0.2 0.5

And, we draw a sample. If the sample we draw is (1,2) then $$\hat{\tau}_p$$ = 33.33 and$$\hat{\tau}_\pi=38.24$$ .

We will not be able to find the real population total nor the real variance of the estimator. However, we will be able to estimate them.

Remark 2. Now, should we use $$\hat{\tau}_p$$or should we use $$\hat{\tau}_\pi$$?

There are no clear answers. Both estimators ar acceptable when $$y_i$$ and $$p_i$$ are proportional.

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