Can anyone shed some light on the below statistical theory questions?

Question

Can anyone shed some light on the below:

1. Consider a set with $N$ distinct members, and a function $f$ defined on $\mathbb Q$ that takes the values $0$, $1$ such that $\frac1N\sum_{x\in\mathbb Q} f(x) = p$. For a subset $S$ of $\mathbb Q$ of size $n$, define the sample proportion $$p = p(S) = \frac1N\sum_{x\in S} f(x)$$ If each subset of size $n$ is chosen with equal probability, calculate the expectation and standard deviation of the random variable $p$.

2. 1. Let $X\sim \mathcal N(0, 1)$ be a normally distributed random variable with mean 0 and variance 1. Suppose that $x \in \mathbb R, x > 0$. Find upper and lower bounds for the conditional expectation $E(X \mid X >x)$
   2. Now suppose that $X$ has a power law distribution, $P(X >x) = ax^{-b}$, for $x>x_0>0$, and some $a> 0, b> 1$. Calculate the conditional expectation $E(X\mid X>x), x >x_0$

Many thanks in advance.

Answer 1

The way I see it, $f \left( x \right)$ follows a Bernoulli distribution with success probability $p$. Assuming this is correct, then the distribution of $n \hat{p}$ given $S$ would be a binomial random variable with success probability $p$ and $n$ trials. \begin{equation} \left. n \hat{p} \middle| S \right. \sim \mathcal{B} \left( p , n \right) \text{.} \end{equation} The conditional mean and variance of $n \hat{p}$ are \begin{align*} \mathrm{E} \left[ n \hat{p} \middle| S \right] = & n p \text{,} \\ \mathrm{Var} \left[ n \hat{p} \middle| S \right] = & n p \left( 1 - p \right) \text{,} \end{align*} and from the properties of the expectation it follows that \begin{align*} \mathrm{E} \left[ \hat{p} \middle| S \right] = & p \text{,} \\ \mathrm{Var} \left[ \hat{p} \middle| S \right] = & \frac{p \left( 1 - p \right)}{n} \text{.} \end{align*} Now it's time to marginalize out $S$, \begin{align*} \mathrm{E} \left[ \hat{p} \right] = & p \text{,} \\ \mathrm{Var} \left[ \hat{p} \right] = & p \left( 1 - p \right) \sum_{n = 1}^N \frac{P \left( n \right)}{n} \text{,} \end{align*} where $P \left( n \right)$ is $n$'s probability mass function. Since $S$ is chosen with equal probability, and the number of distinct $n$-element subsets in an $N$-element set such as $\Omega$ is \begin{equation*} \left( \begin{array}{c} n \\ N \end{array} \right) = \frac{N!}{n! \left( N - n \right)!} \text{,} \end{equation*} I'd be tempted to say that $P \left( n \right)$ is the reciprocal of this coefficient. But this doesn't lead anywhere.