problem 54

(a)

Since

P (W_{j} = y_{k})

for

1 \leq k \leq N

\frac{(\frac{N - 1}{n - 1})}{(\frac{N}{n})} \frac{1}{n} = \frac{n}{N} \frac{1}{n} = \frac{1}{N}

E (W_{j}) = \frac{1}{N} N \sum k = 1 y_{k} .

Thus,

\begin{matrix} E ({W ¯}) & = \frac{1}{n} \sum_{j = 1}^{n} (\frac{1}{N} \sum_{k = 1}^{N} y_{k}) = \frac{1}{N} \sum_{k = 1}^{N} y_{k} = {y ¯} \end{matrix}

(b)

Since

{W ¯} = \frac{1}{n} N \sum j = 1 (I_{j} y_{j})

where $I_{j}$ is the indicator variable for the $j$ -th person being in the sample. Then,

\begin{matrix} E ({W ¯}) & = \frac{1}{n} \sum_{j = 1}^{N} \frac{n}{N} y_{j} = \frac{1}{N} \sum_{k = 1}^{N} y_{k} = {y ¯} \end{matrix}