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

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 $p$.

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 $E(X\mid X>x), x >x_0$

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Two different exercises (nearly three). Please ask them as two separate questions, add to each your own thoughts, what you tried, which related results you know, and the tag (homework). –  Did Jan 12 '13 at 15:54
...and use of LaTeX would greatly increase legibility –  Hagen von Eitzen Jan 12 '13 at 16:05
The only thing I can think of for the first one is expectation of \hat{p} = p and the limit of the variance of \hat{p} as n -> N is 0. The variance of p is 0. Don't think this is correct though. –  SilvaSquarred Jan 13 '13 at 0:38

Taking on the second question.. $$E(X|X>x) = \int_{x}^{\infty}tP(t,X>x)dt = \int_{x}^{\infty}t\frac{P(t)}{P(X>x)}dt\tag{1}$$ where $$P(x) = \mathcal N(0, 1) = \frac{1}{\sqrt{2\pi}}\mathrm{e}^{-\frac{x^2}{2}}$$ In Eq.1 I use dummy variables to lessen the confusion. $$E(X|X>x) = \frac{1}{\sqrt{2\pi}}\frac{1}{P(X>x)}\int_{x}^{\infty}t\mathrm{e}^{-\frac{t^2}{2}}dt$$ the integral is trivial results in $$E(X|X>x) =\frac{1}{\sqrt{2\pi}}\frac{1}{P(X>x)}\mathrm{e}^{-x^2/2}$$ remembering that $$P(X>x) = \frac{1}{\sqrt{2\pi}}\int_x^{\infty}\mathrm{e}^{-t^2/2}dt,$$ which is also equal to $\frac{1}{2}\mathrm{erfc}(x/\sqrt{2})$

From Abramowitz and Stegun you can obtain an upper and lower bound to the erfc function as $$\sqrt{\frac{2}{\pi}}\frac{\mathrm{e}^{-x^2/2}}{x+\sqrt{x^2+4}}< \frac{1}{2}\mathrm{erfc}(x/\sqrt{2}) < \sqrt{\frac{2}{\pi}}\frac{\mathrm{e}^{-x^2/2}}{x+\sqrt{x^2+\frac{8}{\pi}}}$$ and further more $$\sqrt{\frac{\pi}{2}}\frac{ x+\sqrt{x^2+\frac{8}{\pi}}}{\mathrm{e}^{-x^2/2}}< \frac{1}{\frac{1}{2}\mathrm{erfc}}< \sqrt{\frac{\pi}{2}}\frac{ x+\sqrt{x^2+4}}{\mathrm{e}^{-x^2/2}}$$ putting it all together we find $$\frac{x+\sqrt{x^2+\frac{8}{\pi}}}{2} < E(X|X>x) < \frac{x+\sqrt{x^2+4}}{2}$$

$\textbf{Part 2}$ We have to determine the probability of $P(x)$ first to be able to compute the conditional expectation. $$P(x) = \frac{\alpha-1}{x_0}\left(\frac{x}{x_0}\right)^{-\alpha}$$ here $x_0$ is the min value as outlined in the OP. $$\int_x^{\infty}\frac{\alpha-1}{x_0}\left(\frac{t}{x_0}\right)^{-\alpha}dt =ax^{-b}$$ we find that $$\alpha = b+1,\\ \left(\frac{1}{x_0}\right)^{1-\alpha} = a.$$ this yields that the original distribution should be $$P(x) = abx^{-(b+1)}.$$

Then to compute $$\begin{eqnarray} E(X|X>x) &=& \frac{1}{P(X>x)}\int_{x}^{\infty}t\left(abt^{-(b+1)}\right)dt \\ &=& \frac{ab}{ax^{-b}}\int_x^{\infty}t^{-b}dt \\ &=& \frac{b}{x^{-b}}\frac{x^{1-b}}{b-1}\\ &=& \frac{b}{b-1}x \end{eqnarray}$$

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not sure I agree with you. Isn't eq (1) the folloing: $$E(X|X>x) = \int_{x}^{\infty} t P(t|t>x)dt = \int_{x}^{\infty}t\frac{P(t)}{P(t>x)}dt\tag{1}$$ ? Thanks! –  IcannotFixThis Oct 1 '14 at 13:42
@icannotfixthis it is confusing notation on my part. The denominator is already an integrated quantity so does not factor into the expectation I.e can be treated as a constant. Does that make it clearer? –  Chinny84 Oct 1 '14 at 18:33

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. $$\left. n \hat{p} \middle| S \right. \sim \mathcal{B} \left( p , n \right) \text{.}$$ 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.

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