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% Title, subtitle, and other information
\title{Stat472/572 Sampling: Theory and Practice}
%\subtitle{Slides are intended for a course based on the book: \textit{Sampling: Design and Analysis, Third Edition} by Sharon L. Lohr}
\author[Yan Lu]{Instructor: Yan Lu}
\institute[UNM]{University of New Mexico}
\date{}

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% Title slide
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    \titlepage
    \vspace{-1cm} % Add space between the title page and chapter information
    \centering
    \textbf{\Large Chapter 6: Sampling with Unequal Probabilities} % Emphasize chapter title

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\section{Chapter 6}




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
{\bf Motivation}

Example 6.1: O'Brien et al. (1995)\\
A survey of nursing home residents in Philadelphia aimed to determine preferences regarding life-sustaining treatments.
\bi
{\large
\item The study involved 294 nursing homes with a total of 37,652 beds (the number of residents was not known at the planning stage).
\item Cluster sampling was used. Suppose an SRS of the 294 nursing homes is selected, followed by an SRS of 10 residents from each chosen home.\\
----- A nursing home with 20 beds has the same probability of being selected as a nursing home with 1,000 beds.\\
----- However, 10 residents from a 20-bed home represent fewer people than 10 residents from a 1,000-bed home.
}
\ei
\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf The above procedure results in a non-self-weighted sample.}

{\bf Possible design alternatives:}
\bi
\item A one-stage cluster sample.
\item A two-stage cluster design: an SRS of nursing homes followed by an equal proportion SRS of residents in each selected home.
\item If an SRS is used at the first stage, $t_i$ is expected to be proportional to the number of beds in nursing home $i$, leading to estimators with large variance.
\ei
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}

{\bf The Study}

\bi
\item A sample of $57$ nursing homes is drawn with probabilities proportional to the number of beds.
\item An SRS of 30 beds (and their occupants) is then taken from a list of all beds within each selected nursing home.
\item Each bed has an equal probability of being included in the sample (note the distinction between beds and occupants):

\begin{eqnarray*}
p &=& \ds\frac{\text{number of beds at the nursing home}}{\text{total beds in all nursing homes}} \\
&\times& \ds\frac{30}{\text{number of beds in the nursing home}} \\
&=& \ds\frac{30}{\text{total beds in all nursing homes}}
\end{eqnarray*}

\item The same number of interviews is conducted at each nursing home, ensuring the cost is known before selecting the sample.
\item This design likely results in estimators with smaller variance.
\ei

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
{\bf Unequal Probabilities}

\bi
\item $\pi_i$ represents the probability that unit $i$ is selected as part of the sample.
\item Most designs studied so far assume equal probabilities for $\pi_i$ across all units.
\item In general, $\pi_i$ can vary with $i$, allowing for more flexible sampling designs.
\item Sampling with unequal probabilities can yield significant advantages:
\bi
  \item Decreases variances without the need for explicit stratification.
  \item Allows deliberate variation in the selection probabilities of different primary sampling units (psus).
  \item Compensates for unequal probabilities by applying appropriate weights during estimation.
\ei
\ei

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
{\bf Sampling One PSU}

\bi
\item As a special case, we consider selecting just one ($n=1$) of the $N$ primary sampling units (psus) to be included in the sample.
\item Let the total for psu $i$ be denoted by $t_i$.
\item Our goal is to estimate the population total, $t$.
\item $\psi_i = p(\mbox{select unit }i \mbox{ on first draw}).$
\item $\pi_i= p(\mbox{unit }i\mbox{ in sample}).$
\item In the case of sampling one psu ($n=1$), $\pi_i = \psi_i$.
\ei

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

 {\bf Example:} A town has four supermarkets, ranging in size from
100 square meters ($m^2$) to 1000 $m^2$. We want to estimate the
total amount of sales in the four stores for last month by sampling
just one of the stores.
\bi
\item This is an illustration. We took a census, the total sales of
the four supermarkets are $300$ thousands.
\item Expect a larger store would have more sales than a smaller
store.
\item The variability in total sales among several 1000 $m^2$ stores
will be greater than the variability in total sales among several
$100 m^2$ stores.
\item The probability that a store is selected on the first draw
($\psi_i$) is the same as the probability that the store is included
in the sample ($\pi_i$).
\ei
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\begin{tabular}{cccccc}
Store & Size ($m^2$) & $\psi_i$ & $t_i$(in Thousands) & $\hat
t_{\psi} =\ds\frac{t_i}{\psi_i}$ & $(\hat t_{\psi} -t)^2$\\
\hline
A & 100 & 1/16 & 11 & 176 & 15,376\\
B & 200 & 2/16 & 20 & 160 & 19,600\\
C & 300 & 3/16 & 24 & 128 & 29,584\\
D & 1000 & 10/16 & 245 & 392 & 8,464\\
\hline
Total & 1600 & 1 & 300 & \\
\hline
\end{tabular}

\begin{eqnarray*}
E[\hat t_{\psi}] &=& \sum p(S) \hat t_{\psi S}\\
&=& \ds\frac 1 {16} (176) + \ds\frac 2 {16} (160) + \ds\frac 3 {16}
(128) + \ds\frac {10}{16} (392)\\
&=& 300
\end{eqnarray*}

\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\begin{eqnarray*}
V[\hat t_{\psi}] &=& E[(\hat t_{\psi} -t)^2]\\
&=& \sum p(S)(\hat t_{\psi S} -t)^2\\
&=& \sum\limits_{i=1}^N \psi_i \left(\ds\frac {t_i}{\psi_i} -t\right)^2\\
 &=& \ds\frac 1{16} (15,376) + \ds\frac 2{16} (19,600) +\ds\frac
 3{16} (29,584)\\
 &+&\ds\frac {10}{16} (8,464)\\
 &=& 14,248
\end{eqnarray*}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Consider an SRS of size 1}


{\normalsize
\begin{tabular}{cccccc}
Store & Size ($m^2$) & $\psi_i$ & $t_i$(in Thousands) & $\hat
t_{\psi} ={t_i}/{\psi_i}$ & $(\hat t_{\psi} -t)^2$\\
& & & & $= 4 t_i$ &\\
  \hline
A & 100 & 1/4 & 11 & 44 & 65,536\\
B & 200 & 1/4 & 20 & 80 & 48,400\\
C & 300 & 1/4 & 24 & 96 & 41,616\\
D & 1000 & 1/4 & 245 & 980 & 462,400\\
\hline
Total & 1600 & 1 & 300 & \\
\hline
\end{tabular}
}
\bi
\item Probability of selecting each unit is $\psi_i =1/4$
\item SRS estimator is unbiased
\item $V(\hat t_{srs}) = 154,488 \gg V[\hat t_{\psi}] = 14,248$\\
------The variance of SRS is significantly larger compared to that of unequal probability sampling.
\ei
\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
{\bf One-Stage Sampling with Replacement ($n > 1$)}
\bi
\item $\psi_i = p (\mbox{select unit (psu) } i \mbox{ on first draw})$\\
$=  p (\mbox{select unit } i \mbox{ on any given draw})$\\
---- The probability of selecting unit $i$ on the first draw is the same as its probability on any subsequent draw.
\item $\pi_i = p(\mbox{unit } i \mbox{ appears in the sample})$\\
---- This implies $\pi_i = 1 - (1-\psi_i)^n$, where $n$ is the number of draws.
\item $Q_i = \mbox{number of times unit (psu) } i \mbox{ is included in the sample}$\\
---- The total number of draws satisfies $\sum\limits_{i=1}^N Q_i = n$, and the expected number of times unit $i$ is selected is $E(Q_i) = n\psi_i$.
\ei
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\small
\bi
\item {\bf Estimator of the Total:}
\[
\hat{t}_{\psi} = \ds\frac{1}{n} \sum\limits_{i=1}^N Q_i \ds\frac{t_i}{\psi_i}
\]
\bi
\item In sampling with replacement, we obtain $n$ independent estimates of the population total, one for each unit included in the sample.  
\item The final estimate is the average of these $n$ independent estimates.  
\ei
\item Unbiasedness Proof: 
\begin{eqnarray*}
E(\hat{t}_{\psi}) &=& \ds\frac{1}{n} \sum\limits_{i=1}^N E(Q_i) \ds\frac{t_i}{\psi_i}\\
&=& \ds\frac{1}{n} \sum\limits_{i=1}^N n \psi_i \ds\frac{t_i}{\psi_i}\\
&=& \sum_{i=1}^N t_i\\
&=& t
\end{eqnarray*}
\ei
}
\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Variance:}
$$V(\hat t_{\psi}) = \ds\frac 1 n \sum\limits_{i=1}^N \psi_i
\left(\ds\frac {t_i}{\psi_i} -t\right)^2$$
\begin{eqnarray*}
\hat V(\hat t_{\psi}) &= & \ds\frac 1 n \cdot \ds\frac 1 {n-1}
\sum_{i\in \Re} \left(\ds\frac{t_i}{\psi_i} -\hat t_{\psi}\right)^2\\
&=& \ds\frac 1 n \sum\limits_{i=1}^N Q_i \ds\frac
{\left(\ds\frac{t_i}{\psi_i} -\hat t_{\psi}\right)^2}{n-1}
\end{eqnarray*}
where $\Re$ denote the set of $n$ units in the sample, including the
repeats.
$$E[\hat V(\hat t_{\psi})] = V(\hat t_{\psi}) (\mbox{ see proof on page
227})$$
\end{frame}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Estimator of the mean:}
$$\hat {\bar y}_{\psi} = \ds\frac{\hat t_{\psi}}{\hat M_{0\psi}},$$
where $\hat M_{0\psi} = \ds\frac 1 n \sum_{i \in \Re}
\ds\frac{M_i}{\psi_i}.$
$$\hat V(\hat {\bar y}_{\psi}) = \ds\frac 1 {(\hat
M_{0\psi})^2}\cdot \ds\frac 1 n \cdot \ds\frac 1 {n-1} \sum_{i \in
\Re}\left(\ds\frac {t_i}{\psi_i} - \ds\frac{\hat {\bar y}_{\psi}
M_i}{\psi_i}\right)^2$$
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
{\bf Probability proportional to size (pps) sampling}
\bi
\item Many totals in a psu are related to the number of elements in the
psu\\
---let $M_i$ be the size of psu $i$\\
---let $M_0 = \sum_{i=1}^N M_i$ be the size of the population
\item Take $\psi_i = M_i/M_0$\\
---a large psu has a greater chance of being in the sample than a
small psu.
\ei
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 {\bf One-Stage pps sampling}
\bi
\item $\ds\frac{t_i}{\psi_i} = t_i \ds\frac {M_0}{M_i} = M_0 \bar y_i$
\item Estimator of the total: $\hat t_{\psi} = \ds\frac 1 n \sum_{i\in \Re} M_0 \bar y_i$\\
\item Estimator of the population mean: $\hat {\bar y}_{\psi} = \ds\frac 1 n \sum_{i \in \Re} \bar y_i$,\\
-----this is the average of the sampled psu means
\item $\hat M_{0\psi} = M_0$ for every possible sample
\item Variance of the estimated mean: $\hat V(\hat {\bar y}_{\psi}) =\ds\frac 1 n \cdot \ds\frac 1
{n-1} \sum_{i \in \Re} (\bar y_i - \hat {\bar y}_{\psi})^2$
\item The pps estimators can be computed by treating the sampled psu means ($\bar{y}_i$) as individual observations, 
then calculating their mean and sample variance directly.  
\ei

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\small

{\bf Example: Estimating the Total Number of Physicians in the United States}

The file \texttt{statepop.dat} contains data from an unequal-probability sample of 100 counties in the United States. Our goal is to estimate the total number of physicians in the country.

\bi
\item Counties were selected using the cumulative-size method from the listings in the \textit{City and County Data Book} (1994), with probabilities proportional to their populations:  
\[
\psi_i = \frac{M_i}{M_0}
\]
where $M_i$ is the population size of county $i$ and $M_0$ is the total population of all counties.

\item Sampling was conducted with replacement.

\item Very large counties appear multiple times in the sample. For instance, Los Angeles County, the most populous in the United States, appears four times.

\item Since larger counties tend to have more physicians, probability-proportional-to-size (pps) sampling is expected to be effective for estimating the total number of physicians.
\ei
}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\begin{figure}[!htp]
       \centering
       \includegraphics[width=0.8\textwidth,
       height=1.0\textheight]{./plots/picture1.pdf}
\caption{Example 3.10}
       \label{archers}
     \end{figure}

\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\begin{figure}[!htp]
       \centering
       \includegraphics[width=0.90\textwidth,
       height=1.2\textheight]{./plots/p233.pdf}
\caption{Example 3.10}
       \label{archers}
     \end{figure}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\bi
\item The unusual observation is New York County, New York\\
{\footnotesize
\begin{tabular}{cccc}

& Landarea & Population Size & Number of Physicians\\
& square miles & &\\
 \hline
New York County & 28 & 1,489,066 &14,052\\
\hline
Mean for the 100 & 1188.04 & 1,048,753 & 2,979.87\\
selected counties & & &\\
\hline
\end{tabular}
}
\item The estimated total number of physicians in the United States is
$570,304$
\item The standard error of the estimate is $41401.23$
\item For comparison, the {\it City and Country Data Book} lists a
total of $532,638$ physicians in the United States, a value that is
less than $1$ SE away from the estimate
\ei

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
{\bf Two-Stage Sampling with Replacement}

\bi
\item The key difference between two-stage sampling with replacement and one-stage sampling with replacement is that in two-stage sampling, we must estimate $t_i$ (the total for primary sampling unit, PSU $i$).

\item If PSU $i$ appears in the sample more than once, there are $Q_i$ estimates of the total for PSU $i$: $\hat t_{i1}, \hat t_{i2}, \dots, \hat t_{iQ_i}$.

\item The estimator for the total is:
\[
\hat t_{\psi} = \frac{1}{n} \sum_{i=1}^N \sum_{j=1}^{Q_i} \frac{\hat t_{ij}}{\psi_i}
\]

\item The variance of the estimator is:
\[
\hat V(\hat t_{\psi}) = \frac{1}{n} \sum_{i=1}^N \sum_{j=1}^{Q_i} \frac{\left(\frac{\hat t_{ij}}{\psi_i} - \hat t_{\psi}\right)^2}{n-1}
\]
\ei

\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}

{\bf Estimator of the Population Mean}

\begin{itemize}
\item The estimator of the population mean is:
\[
\hat{\bar{y}}_{\psi} = \frac{\hat{t}_{\psi}}{\hat{M}_{0\psi}},
\]
where
\[
\hat{M}_{0\psi} = \frac{1}{n} \sum_{i \in \mathcal{W}} \frac{M_i}{\psi_i}.
\]

\item Variance of the estimated mean:
\[
\hat{V}(\hat{\bar{y}}_{\psi}) = \frac{1}{(\hat{M}_{0\psi})^2} \cdot \frac{1}{n} \cdot \frac{1}{n-1} \sum_{i=1}^N \sum_{j=1}^{Q_i} \left(\frac{\hat{t}_{ij}}{\psi_i} - \frac{\hat{\bar{y}}_{\psi} M_i}{\psi_i}\right)^2.
\]
\end{itemize}

{\bf pps sampling is a special case} when the probability of selecting each primary sampling unit (psu) is proportional to its size, i.e.,
    \[
    \psi_i = \frac{M_i}{M_0},
    \]
    where \( M_i \) is the size of psu \( i \), and \( M_0 \) is the total size of all psus in the population.

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}

{\bf Example 6.7: Average Number of Legs of Puppies}

\begin{itemize}
\item Goal: Estimate the average number of legs of healthy puppies in Sample City's puppy homes.

\item Sample City has two puppy homes:
    \begin{itemize}
        \item Puppy Palace: 30 puppies
        \item Dog's Life: 10 puppies
    \end{itemize}

\item Sampling Design: pps sampling
    \begin{itemize}
        \item Puppy homes are selected with probabilities proportional to the size of each puppy home:
        \[
        \text{P(Puppy Palace)} = \frac{30}{30 + 10} = \frac{3}{4}, \quad
        \text{P(Dog's Life)} = \frac{10}{30 + 10} = \frac{1}{4}.
        \]\\
        ----this ensures larger homes (with more puppies) have a higher chance of being selected\\
        ----pps sampling is effective because it accounts for the variation in home sizes, potentially reducing the variance of the estimator 
        \item After a home is selected, a simple random sample (SRS) of 2 puppies is taken from the chosen home.
    \end{itemize}

\end{itemize}

\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\begin{tabular}{c|cccccc}
\hline
& $M_i$ & $\psi_i$ & $m_i$ & $\hat t_i$ & $\hat
t_{\psi}=\ds\frac{\hat t_i}{\psi_i}$ & $\hat{\bar
y}_{\psi}=\ds\frac{\hat t_{\psi}}{M_0}$\\
\hline
 Puppy Palace & 30 & 3/4 & 2 & 30*4=120& 160 & 4\\
Dog's Life & 10 & 1/4 & 2 & 10*4=40& 160 & 4\\
\hline
\end{tabular}
\bi
\item Either possible sample results in an estimated average of
$\hat {\bar y}_{\psi} = 160/40 = 4$ legs per puppy
\item The variance of the estimator is zero
\ei
\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Notes: Sampling with Replacement}

\begin{itemize}
\item Sampling with replacement is advantageous because:
    \begin{itemize}
        \item It is simple and straightforward to select the sample.
        \item Estimators for both the population total. mean and their variance are easy to compute.
    \end{itemize}
    
\item However, if the population size (\( N \)) is small:
    \begin{itemize}
        \item Sampling with replacement becomes less efficient compared to many sampling designs without replacement.
        \item This is because repeated selections of the same unit reduce the effective sample size, increasing the variance of the estimators.
    \end{itemize}
\end{itemize}

\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
{\bf Weights in Unequal probability sampling with replacement}

{\bf One stage sampling with replacement with {\bf only one psu}}
\bi
\item $\pi_i = \psi_i$
\item $w_{ij}= w_i = 1/\psi_i$
\item $\hat t_{\psi} = \sum\limits_{i \in \cal W} \sum\limits_{j=1}^{M_i}
w_{ij} y_{ij}$
\item $\hat{\bar y}_{\psi} = \ds\frac{\sum\limits_{i\in \cal W}
\sum\limits_{j=1}^{M_i} w_{ij} y_{ij}}{\sum\limits_{i\in \cal W}
\sum\limits_{j=1}^{M_i} w_{ij}}$
\ei
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf One Stage Sampling with replacement: general Case ({\bf{more than one psu}})}\\
$\pi_i \approx n \psi_i$, the larger the population, the closer of
$\pi_i$ to $n\psi_i$
\begin{eqnarray*}
w_i &=& \ds\frac{1}{\mbox{expected number of hits}}\\
&=& \ds\frac 1 {E[Q_i]}\\
&=& \ds\frac 1 {n\psi_i}
\end{eqnarray*}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\bi
\item $w_{ij} = w_i = \ds\frac 1{n \psi_i}$
\item $\hat t_{\psi} = \sum\limits_{i \in \cal W} \sum\limits_{j=1}^{M_i}
w_{ij} y_{ij}$
\item $\hat{\bar y}_{\psi} = \ds\frac{\sum\limits_{i\in \cal W}
\sum\limits_{j=1}^{M_i} w_{ij} y_{ij}}{\sum\limits_{i\in \cal W}
\sum\limits_{j=1}^{M_i} w_{ij}}$
\item $\psi_i$s are unequal, the sample is not self-weighting
\item In one-stage pps sampling, elements in large psus have smaller
weights than elements in small psus 
\ei
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Two stage cluster sampling with replacement}

Sampling weight for element sampled from psu $i$
$$w_i = \ds\frac 1 {n \psi_i}\ds\frac{M_i}{m_i}$$
pps sampling
\bi
\item $\psi_i = \ds\frac{M_i}{M_0}$
\item $w_i = \ds\frac{M_0}{n M_i}\ds\frac{M_i}{m_i} =
\ds\frac{M_0}{n m_i}$
\item If $m_i$s are the same within each psu, sample is
self-weighting\\
----equal interviewer works loads\\
----sample size of ssu's known in advance\\
----if $t_i \propto \psi_i$, more efficient
\ei
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Review: SRS without replacement}
\bi
\item $\pi_i = p(\mbox{unit }i \mbox{ in the sample}) =
n/N$
\item $\sum\limits_{i=1}^N \pi_i = n$
\item $\pi_{ij} =p(\mbox{unit i and j in the sample}) =\ds\frac
{n(n-1)}{N(N-1)}$
\item $\sum\limits_{i=1, j\neq i}^N \pi_{ij} = n(n-1)$\\
 Proof: Let
$$Z_i = \left\{
\begin{array}{cc}
1 & \mbox{if unit}\quad i\in S \\
0& \mbox{otherwise}
\end{array}\right.$$
$p(z_i =1) =\pi_i$\\
$E(z_i) = 1* p(z_i =1) = \pi_i$\\
$\sum\limits_{i=1}^N \pi_i = \sum\limits_{i=1}^N E(z_i) =
E\left(\sum\limits_{i=1}^N z_i\right) =n$
\ei
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\begin{eqnarray*}
\sum\limits_{i=1, j\neq i}^N \pi_{ij} &=& \sum\limits_{i=1, j\neq
i}^N p(Z_i =1, Z_j=1)\\
&=& \sum\limits_{i=1, j\neq i}^N E(Z_i Z_j)\\
&=& \sum\limits_{i=1}^N E[Z_i(n-Z_i)]\\
&=& \sum\limits_{i=1}^N (E(n Z_i) - E(Z_i^2))\\
&=& \sum\limits_{i=1}^N \left(n \cdot\ds\frac n N - \ds\frac n N\right)\\
&=& n(n-1)
\end{eqnarray*}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Example 6.8: Supermarket Example}

\begin{tabular}{cccc}
Store & Size ($m^2$) & $\psi_i$ & $t_i$(in Thousands) \\
\hline
A & 100 & 1/16 & 11 \\
B & 200 & 2/16 & 20 \\
C & 300 & 3/16 & 24 \\
D & 1000 & 10/16 & 245 \\
\hline
Total & 1600 & 1 & 300  \\
\hline
\end{tabular}
\bi
\item Want to select two psus without replacement and with unequal
probabilities.\\
---Recall $\psi_i = p(\mbox{select unit }i \mbox{ on first draw})$
\ei

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 $$p(\mbox{store } A \mbox{ chosen on first draw}) =\psi_A = 1/16$$
\begin{eqnarray*}
& & p(\mbox{store } B \mbox{ chosen on second draw}| A\mbox{ chosen on
first draw})\\
&=& \ds\frac{\ds\frac 2 {16}}{1-\ds\frac 1{16}}\\
 &=& \ds\frac{\psi_B}{1-\psi_A}
\end{eqnarray*}
In general, 
\begin{eqnarray*}
& &p(\mbox{unit } i \mbox{ chosen first, unit } k
\mbox{
chosen second})\\
 &=& p(\mbox{unit } i \mbox{ chosen first}) p(\mbox{ unit } k \mbox{ chosen second} |
\mbox{ unit } i\mbox{ chosen first})\\
& =& \psi_i \ds\frac{\psi_k}{1-\psi_i}
\end{eqnarray*}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\small
\begin{eqnarray*}
 & & p(\mbox{unit } k \mbox{ chosen first, unit } i
\mbox{
chosen second})\\
&=& \psi_k \ds\frac{\psi_i}{1-\psi_k}
\end{eqnarray*}
\begin{eqnarray*}
&  &p(\mbox{units } i \mbox{ and } k \mbox{ in sample})\\
& =&\pi_{ik} = \psi_i \ds\frac{\psi_k}{1-\psi_i} + \psi_k
\ds\frac{\psi_i}{1-\psi_k}
\end{eqnarray*}
\begin{eqnarray*}
\pi_{AB} &=&\psi_A \ds\frac{\psi_B}{1-\psi_A} + \psi_B \ds\frac{\psi_A}{1-\psi_B}\\
&=& \ds\frac 1{16} \cdot\ds\frac{\frac{2}{16}}{1-\frac 1{16}} +  \ds\frac 2{16} \cdot \ds\frac{\frac{1}{16}}{1-\frac 2{16}}\\
&=& 0.0173
\end{eqnarray*}
 The probability that psu $i$ is in the
sample $\pi_i = \sum_{S: i\in S} P(S)$.
}
\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[fragile]
\begin{verbatim}
supermarket<-data.frame(store=c('A','B','C','D'),
area=c(100,200,300,1000), ti=c(11,20,24,245))
supermarket$psi<-supermarket$area/sum(supermarket$area)
psii<-supermarket$area/sum(supermarket$area)
piik<- psii %*% t(psii/(1-psii)) + 
(psii/(1-psii)) %*% t(psii)
diag(piik)<-rep(0,4) # diagonal entries: zero
piik  # joint inclusion probabilities
pii<-apply(piik,2,sum)
pii # inclusion probabilities
\end{verbatim}
\end{frame}


\begin{frame}[fragile]

{\small
\begin{verbatim}
> cbind(piik,pii)
        A            B            C          D        pii
[1,] A 0.0000000 0.01726190 0.02692308 0.1458333 0.1900183
[2,] B 0.0172619 0.00000000 0.05563187 0.2976190 0.3705128
[3,] C 0.0269230 0.05563187 0.00000000 0.4567308 0.5392857
[4,] D 0.1458333 0.29761905 0.45673077 0.0000000 0.9001832
\end{verbatim}
}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
\begin{figure}[!htp]
       \centering
       \includegraphics[width=0.90\textwidth,
       height=1.5\textheight]{./plots/p253.pdf}
\caption{Example 3.10}
       \label{archers}
     \end{figure}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 {\bf Unequal probability sampling without replacement (one-stage)}
%\item $\psi_i = p(\mbox{select unit}\quad i \quad\mbox{on first draw})$
\bi
\item $\pi_i= p(\mbox{unit }i \mbox{ in sample})$
\item $\pi_i/n$ is the average probability that a unit will be
selected on one of the draws: the probability we would assign to the
$i$th unit's being selected on draw $k (k=1,\cdots,n)$ if we did not
know the true probabilities
\item the estimator $t_i /\psi_i$ is then estimated by $
t_i/(\pi_i/n)$
\ei
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
 {\bf Horvitz-Thompson (HT) Estimator for One-Stage Sampling}
Horvitz and Thompson 1952
\begin{eqnarray*}
\hat t_{\mbox{HT}} &=& \ds\frac 1 n \sum\limits_{i \in S} \ds\frac {
t_i}{\pi_i /n}\\
&=& \sum\limits_{i \in S} \ds\frac{t_i}{\pi_i}\\
&=& \sum\limits_{i=1}^N Z_i \ds\frac {t_i}{\pi_i}
\end{eqnarray*}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
Unbiased: $E[\hat t_{\mbox{HT}}] = \sum\limits_{i=1}^N \pi_i
\ds\frac {t_i}{\pi_i} = t$\\
 Variance:
\begin{eqnarray*}
V[\hat t_{\mbox{HT}}] &=& \sum\limits_{i=1}^N \ds\frac{1-
\pi_i}{\pi_i} { t_i^2} + \sum\limits_{i=1}^N \sum\limits_{k\neq i}^N
\ds\frac{\pi_{ik} - \pi_i \pi_k}{\pi_{i}\pi_k} { t_i}
t_k\\
&=& \ds\frac 1 2 \sum\limits_{i=1}^N \sum\limits_{k=1, k\neq i}^N
(\pi_i \pi_k - \pi_{ik})\left(\ds\frac{t_i}{\pi_i}
-\ds\frac{t_k}{\pi_k}\right)^2
\end{eqnarray*}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
Horvitz-Thompson (HT) Estimator
\begin{eqnarray*}
\hat V_1[\hat t_{\mbox{HT}}] &=& \sum\limits_{i\in S} (1- \pi_i)
\ds\frac{ t_i^2}{\pi_i^2} + \sum\limits_{i\in S} \sum\limits_{k\in
S, k\neq i}  \ds\frac{\pi_{ik} - \pi_i \pi_k}{\pi_{ik}} \cdot
\ds\frac{ t_i}{\pi_i} \cdot \ds\frac{ t_k}{\pi_k}
\end{eqnarray*}

Sen-Yates-Grundy Estimator\\
$$\hat V_2 [\hat t_{\mbox{HT}}] = \ds\frac  1 2\sum\limits_{i\in S}
\sum\limits_{k\in S, k\neq i} \ds\frac{\pi_i \pi_k -
\pi_{ik}}{\pi_{ik}} \left(\ds\frac{ t_i}{\pi_i} -\ds\frac{
t_k}{\pi_k}\right)^2
$$
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Example 6.9}, The HT estimator for a sample of 2 supermarkets in
Example 6.8 with joint inclusion probabilities given in Table 6.7.
\bi
\item To select
the first psu, we generate a random integer from ${1, \cdots, 16}$,
the random integer we generate is 12, which tells us that store $D$
is selected on the first draw.
\item We then remove the values ${7, \cdots, 16}$ corresponding to store $D$, and generate a second random integer
from ${1,\cdots, 6}$, we generate 6, which tells us to select store
$C$ on the second draw.
\ei
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[fragile]
\begin{verbatim}
> supermarket2<-supermarket[3:4,]
# these are the unit inclusion probs when n=2
> supermarket2$pii <- pii[3:4] 
# joint probability matrix for stores C and D
> jointprob<-piik[3:4,3:4]  
# set diagonal entries equal to pii
> diag(jointprob)<-supermarket2$pii 
> jointprob
          [,1]      [,2]
[1,] 0.5392857 0.4567308
[2,] 0.4567308 0.9001832
\end{verbatim}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[fragile]
\begin{verbatim}

> dht<- svydesign(id=~1,  fpc=~pii, data=supermarket2, 
+                    pps=ppsmat(jointprob),variance="HT")  
> dht
Sparse-matrix design object:
 svydesign(id = ~1, fpc = ~pii, data = supermarket2, 
pps = ppsmat(jointprob), 
    variance = "HT")
> svytotal(~ti,dht)
    total     SE
ti 316.67 82.358
\end{verbatim}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}[fragile]
\begin{verbatim}

> dsyg<- svydesign(id=~1,  fpc=~pii, data=supermarket2, 
+                    pps=ppsmat(jointprob),variance="YG")  
> dsyg
Sparse-matrix design object:
 svydesign(id = ~1, fpc = ~pii, data = supermarket2, 
pps = ppsmat(jointprob), 
    variance = "YG")
> svytotal(~ti,dsyg)
    total     SE
ti 316.67 57.094
\end{verbatim}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\begin{itemize}
\item Given the sample \(\{C, D\}\), the HT estimate of the total sales is:
\[
\hat{t}_{HT} = \sum_{i \in S} \frac{t_i}{\pi_i} = \frac{24}{0.5393} + \frac{245}{0.9002} = 316.6639
\]

\item Variance and standard error of the HT estimate:
\[
\hat{V}(\hat{t}_{HT}) = 6782.8, \quad SE(\hat{t}_{HT}) = 82.358
\]

\item SYG (Sen-Yates-Grundy) approximation for variance and standard error:
\[
\hat{V}_{SYG}(\hat{t}_{HT}) = 3259.8, \quad SE_{SYG}(\hat{t}_{HT}) = 57.094
\]
\end{itemize}

\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}

{\bf Example: Basu’s Elephant}

Debabrata Basu (1971) famously critiqued the Horvitz-Thompson (HT) estimator in his essay *“An Essay on the Logical Foundations of Survey Sampling, Part I”*. He illustrated his point with the following story:

\begin{itemize}
\item A circus owner plans to ship 50 elephants and needs an estimate of their total weight.  
\item She decides to weigh only one elephant, Sambo (a middle-sized elephant), and uses $50 \cdot y_{\text{Sambo}}$ (where $y_{\text{Sambo}}$ is Sambo’s weight) as the total weight estimate.  
\item The circus statistician is horrified because this method gives zero probability of sampling the other elephants.  
\item To address this, the statistician proposes a different plan:
  \begin{itemize}
    \item Assign a 99\% probability of selecting Sambo.
    \item Assign the remaining 1\% probability equally among the other 49 elephants.
  \end{itemize}
\end{itemize}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf The Sampling Results}

As expected, Sambo is selected.  

\begin{itemize}
\item The circus owner asks, "Since Sambo was selected, isn’t the total weight estimate just $50 \cdot y_{\text{Sambo}}$?"  
\item The statistician replies, No, the HT estimate is:  
\[
\hat{t}_{HT} = \frac{y_{\text{Sambo}}}{0.99} = 1.01 \cdot y_{\text{Sambo}}
\]

\item The owner then asks, "What if the largest elephant, Jumbo, had been selected instead?"  
\item The statistician explains, The HT estimate would then be:  
\[
\hat{t}_{HT} = \frac{y_{\text{Jumbo}}}{\frac{0.01}{49}} = 4900 \cdot y_{\text{Jumbo}}
\]

\item Upon hearing this, the circus owner immediately fires the statistician!  
\end{itemize}
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf What’s Wrong with Basu’s Elephant Example?}

The example highlights several critical issues with this approach:  

\begin{itemize}
\item **Sample size is too small**: Only one elephant is selected, making the estimate unreliable.  
\item **Selection probabilities are too extreme**: The probabilities heavily favor Sambo, making the design inherently biased and prone to large variability.  
\item **Huge standard error**: Extreme selection probabilities lead to large variability in the HT estimator.  
\item **Unstable estimates**: While the HT estimator is unbiased in theory (over repeated samples, the average estimate is close to the truth), individual estimates can deviate drastically, especially with such extreme designs.  
\end{itemize}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}
{\bf The Horvitz-Thompson (HT) Estimator for Two-stage Sampling}
$$
\hat t_{\mbox{HT}} = \sum\limits_{i \in S} \ds\frac{\hat t_i}{\pi_i}
= \sum\limits_{i=1}^N Z_i \ds\frac {\hat t_i}{\pi_i}
$$
If $E(\hat t_i) =t_i$, $E[\hat t_{\mbox{HT}}] = t$\\
Variance Estimator (HT form)

$$
\hat V_1[\hat t_{\mbox{HT}}] = \sum\limits_{i\in S} (1- \pi_i)
\ds\frac{\hat t_i^2}{\pi_i^2} + \sum\limits_{i\in S}
\sum\limits_{k\in S, k\neq i} \ds\frac{\pi_{ik} - \pi_i
\pi_k}{\pi_{ik}} \ds\frac{\hat t_i}{\pi_i} \ds\frac{\hat t_k}{\pi_k}
+ \sum\limits_{i\in S} \ds\frac {\hat V(\hat t_i)}{\pi_i}
$$
Variance Estimator (SYG form)
$$\hat V_2 [\hat t_{\mbox{HT}}] = \ds\frac 1 2 \sum\limits_{i\in S}
\sum\limits_{k\in S, k\neq i} \ds\frac{\pi_i \pi_k -
\pi_{ik}}{\pi_{ik}} \left(\ds\frac{\hat t_i}{\pi_i} -\ds\frac{\hat
t_k}{\pi_k}\right)^2 + \sum\limits_{i\in S} \ds\frac {\hat V(\hat
t_i)}{\pi_i}$$
\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
{\bf Recommendation: Variance Estimation Using With-Replacement Sampling}

{\small For most situations, we recommend using the with-replacement sampling variance estimator to avoid potential instability and computational complexity.

\begin{eqnarray*}
\hat V_{\mbox{WR}}(\hat t_{\mbox{HT}}) & =& \frac{1}{n} \cdot
\frac{1}{n-1} \sum\limits_{i \in S} \left(\frac{n \hat t_i}{\pi_i} - \hat
t_{\mbox{HT}}\right)^2\\
&=& \frac{n}{n-1} \sum\limits_{i \in S} \left(\frac{\hat
t_i}{\pi_i} - \frac{\hat t_{\mbox{HT}}}{n}\right)^2
\end{eqnarray*}

\vspace{0.3cm}
{\bf Notes:}
\begin{itemize}
\item In practice, samples are often drawn without replacement, but the variance is calculated assuming with-replacement sampling.
\item Using the with-replacement estimator generally results in a larger variance than the true variance (overestimation), which is conservative and ensures robustness.
\end{itemize}
}
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}

{\bf Unequal Probability Sampling in Practice}

\bi
\item Many government surveys rely on unequal probability sampling techniques.
\item Stratification is often used first to reduce the variation in probabilities ($\pi_i$).
\ei

{\bf Example: Random Digit Dialing (RDD)}

\bi
\item Construct a frame using area codes and prefixes (e.g., 505-243-).
\item Draw a random sample of suffixes (0000-9999), dial numbers, and check if they are residential. If not, discard.
\ei

For psu $i$, let $M_i$ be the number of residential numbers:
\begin{eqnarray*}
p(\mbox{dial number})& = & p(\mbox{psu } i \mbox{ selected}) \times p(\mbox{number selected in psu } i) \\
& = & \frac{M_i}{\sum_{j=1}^N M_j} \cdot \frac{m_i}{M_i}
\end{eqnarray*}

If all $m_i$'s are equal, the sampling becomes self-weighting.
\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}
\frametitle{Disclaimer}

Slides are intended for a course based on the book: \textit{Sampling: Design and Analysis, Third Edition} by Sharon L. Lohr

All examples, datasets, and pages directly scanned and included in this chapter are from the textbook.

References to papers or books cited in these slides can be found in the Bibliography section of the textbook.
\end{frame}
\end{document}
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