subsets probability question Consider a set $\Omega$ with $N$ distinct members, and a function $f$ defined on $\Omega$ that takes the values 0,1 such that $ \frac{1}{N} \sum_{x \in \Omega } f(x)=p$. For a subset $S⊆Ω$ of size n, define the sample proportion
$p:= p(S)= \frac{1}{n} \sum_{x\in S} f(x)$.
If each subset of size $n$ is chosen with equal probability, calculate the expectation and standard deviation of the random variable $p$.
 A: It helps to introduce indicator random variables here.
For each $x\in\Omega$, let $Z_x$ be the indicator random variable 
that takes the value 1 if $x\in S$, and value 0  otherwise. 
We can express $$p(S)={1\over n}\sum_{x\in\Omega} Z_x\cdot f(x),$$
where the sum is no longer over the random set $S$. Since all points
are equally likely to be elements of $S$, it is not hard to calculate
$$\mathbb{E}(Z_x)={n\over N},\quad \text{Var}(Z_x)={n\over N}\left({1-{n\over N}}\right),
\quad \text{cov}(Z_x,Z_y)={-n\over N^2} {N-n \over N-1}\text{ for }x\neq y.$$
Using linearity of expectation, and bilinearity of covariance, after some 
calculation  we get
$$\mathbb{E}(p(S))={1\over N}\sum_{x\in\Omega} f(x),$$
and 
$$\text{Var}(p(S))={1\over n} {N-n \over N-1} \left[{1\over N}\sum_{x\in\Omega} f(x)^2-
\left( {1\over N}\sum_{x\in\Omega} f(x)\right)^2\right].$$
A: I think the answer is:
a) $E[\bar{p}] = p$,
b) Var$[\bar{p}] = \frac{\sqrt{p(1-p)}}{\sqrt{n}}$.
I believe the answer can be found on page 10 of
http://math.arizona.edu/~faris/stat.pdf
~JD
