1 2014-Spring
1.2 Problem 1
1.2.1 Part 1(a)
Let \(S\) represent the \(n\) sized set chosen:
\(P(S=s_n) = \frac{1}{\binom{N}{n}}\)
1.2.2 Part 1(b)
\(E[I_i] = P(I_i=1) = \frac{\binom{N-1}{n-1}}{\binom{N}{n}} = \frac{n}{N}\)
\(E[I_iI_j] = \frac{\binom{N-2}{n-2}}{\binom{N}{n}} = \frac{n(n-1)}{N(N-1)}\)
\(Cov(I_i, I_j) = E[I_iI_j]-E[I_i]E[I_j] = \frac{n(n-1)}{N(N-1)} - (\frac{n}{N})^2 = \frac{n(n-N)}{N(N-1)}\)
1.2.3 Part 1(c)
\(EW = E\sum_{i=1}^N a_iI_i = \sum_{i=1}^NE[a_iI_1] = \sum a_i\frac{n}{N} = \frac{\sum a_in}{N}\)
$EW^2 = E(_i a_iI_i)^2 = E[i a_i^2I_i + 2{i>j}a_ia_jI_iI_j] =i a_i^2 + 2 {i>j}a_ia_j $
\(Var(W) = \sum_i a_i^2 \frac{n}{N} + 2 \sum_{i>j}a_ia_j \frac{n(n-1)}{N(N-1)} - (\frac{\sum a_in}{N})^2\)
$Var(W) = $ ### Part 1(d)
For \(W\) to be hypergeomettrc, \(a_i = \{0, 1\}\) like marbles in an urn (red = 0, black = 1) and all distinct.
Then \(W\) represents the number of black marbles among the picked \(n\) marbles from the original \(B\) black and \(N-B\) red marbles
\(P(W=x) = \frac{\binom{B}{x}\binom{N-B}{n-x}}{\binom{N}{n}}\)
\(EW = \frac{Bn}{N}\)
\(Var(W) = \sum_i a_i^2 \frac{n}{N} + 2 \sum_{i>j}a_ia_j \frac{n(n-1)}{N(N-1)} - (\frac{\sum a_in}{N})^2 = B\frac{n}{N} + 2 \binom{B}{2} \frac{n(n-1)}{N(N-1)} - (\frac{Bn}{N})^2 = \frac{BnN(N-1) + n(n-1)B(B-1)-B^2n^2(N-1)}{N(N-1)}\)
… Incorrect