4.6.3 problem 86

(a)

$P(X=x,Y=y,Z=z)=\frac{\left(\frac{n_A}{x}\right)\left(\frac{n_B}{y}\right)\left(\frac{n_C}{z}\right)}{\left(\frac{n}{m}\right)}$

(b)

Let $I_j$ be the indicator variable for person $j$ in the sample being a member of party $A$. Then, $X={\sum }_{i=1}^m{I}_i\Rightarrow \text{E}\left(X\right)=m\frac{n_A}{n}$ by symmetry.

(c)

Let’s find $E\left(X^2\right)$. If we square the expression for the sum of $X$’s constituent indicator r.v.s, we get

$E\left(X^2\right)={\sum }_{i=1}^{m}E\left(I_i^2\right)+2\ast {\sum }_{i<j,1\le j\le m}E\left(I_i{I}_j\right)$

Since $I_i^2=I_i$, we have ${\sum }_{i=1}^{m}E\left(I_i^2\right)=\frac{m\ast n_A}{n}$

Additionally, for any pair $i,j$, the r.v. $I_i{I}_j$ equals 1 only when some pair of samples are both members of party A, which occurs with probability $\frac{n_A(n_A-1)}{n(n-1)}$. There are $\left(\frac{m}{2}\right)$ pairs $i,j$. Therefore, the expression $2\ast {\sum }_{i<j,1\le j\le m}E\left(I_i{I}_j\right)$ evaluates to $\frac{n_{A}m(m-1)(n_A-1)}{n(n-1)}$.

Finally, we have $E\left(X^2\right)$, so now we can write

$$V\mathrm{ar}\left(X\right)=E\left(X^2\right)-E{X}^2=\frac{m\ast n_A}{n}+\frac{n_{A}m(m-1)(n_A-1)}{n(n-1)}-\frac{m^2{n}_A^2}{n^2}$$

When $m=1$, $\text{Var}\left(X\right)=\frac{n_A}{n}-{\left(\frac{n_A}{n}\right)}^2=\frac{n_A}{n}(1-\frac{n_A}{n})=\frac{n_A}{n}\times \frac{n_B+n_C}{n}$.

When $m=n$, Var(X) = 0. This makes sense, as if the sample is the entire population, we always get the same number of members of party A in our sample (all of them), so there is no variation.