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This implies (among other things) that all the components $X(\omega)$ have the same expectation. … Since these components sum to the sample size $M,$ that common expectation must be $$E[X(\omega)] = \frac{M}{\binom{N}{n}}.$$ In this case we can use the sample statistics $t_j$ to estimate the measures …

mathcal Y)\times 0 + \Pr(j \in \bigcup \mathcal Y)\times 1 = 1 - \pi_j.$$ The number of elements in $\bigcup \mathcal Y$ common to $A$ is just the sum of the $X_j$ for $j=1,2,\ldots, m,$ whence the expectation … For sampling with replacement, $p = m/n.$ For sampling without replacement, $p = 1-\left(1 - 1/n)\right)^m \approx 1-e^{-m/n}.$ The last approximation is excellent when $sm \ll n.$ …

So, the probability that a particular unit is not sampled, $P(Y_{i} = 0)$, is calculated from the binomial mass function as $$ P(Y_{i} = 0) = (1 - 1/N)^{k} $$ So, the indicator of subject $i$ not being … Regardless of whether or not they are, linearity of expectation still holds so the expected proportion of the population that is not sampled is $$ \mu_{k} = E \left( \frac{1}{N} \sum_{i=1}^{N} X_{i}\ …