# Anybody know these kinds of statistics and their serious names? Do you know any books

• Nov 25th 2011, 09:20 PM
BookEnquiry
I'm a late learner of science & maths.
These problems origin from Asian examinations.
Do you know any books discussing these kind of problems?

Sample queations

Q1

The mean $\displaystyle \mu$ and variance $\displaystyle \sigma$ of a finite population P =$\displaystyle {x_1, x_2, ..., x_n}$ where

$\displaystyle n>2$, are defined as follows:
$\displaystyle \mu=\frac{1}{n}\sum_{i=1}^{n}x_i\\\sigma^2=\frac{1 }{n}\sum_{i=1}^{n}(x_i-\mu)^2$

Samples of size 2, $\displaystyle \{x_i, x_j\}$, where $\displaystyle i\neqj$, are drawn at random from P. the mean of $\displaystyle x_i$ and $\displaystyle x_j$ is denoted by $\displaystyle x_{ij}$. These means then form a set
$\displaystyle S=\{x_{ij}: i\neqj ; i= 1,2, ..., n; j=1,2, ..., n\}$

a) Show that $\displaystyle \sum_{i=1}^{n-1}\sum_{j=i+1}^{n}(x_i+x_j)=(n-1)\sum_{i=1}^{n}x_i$

b) Show that the mean of all the elemeants of S is equal to $\displaystyle \mu.$

c) Show that the variance of all the elements of S is equal to $\displaystyle \frac{n-2}{2(n-1)}\sigma^2$.

hence deduce that if $\displaystyle n\right arrow\infinity$, this variance tends to $\displaystyle \frac{1}{2}\sigma^2$.

solution
a)
$\displaystyle \sum_{i=1}^{n-1}\sum_{j=i+1}^{n}(x_i+x_j)\\=(n-1)x_1+x_2+x_3+... +x_{n-1}+x_n\\+(n-2)x_2+x_3+... +x_{n-1} +x_n\\+(n-3)x_3+... x_{n-1}+x_n\\+...\\+x_{n-1}+x_n\\\\=(n-1)x_1+(n-1)x_2+(n-1)x_3+... +(n-1)x_{n-1}+(n-1)x_n\\=(n-1)\sum_{i=1}^{n}x_i$

b)
$\displaystyle E(x_{ij})=\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}(\frac{x_i+x_j}{2})/C^{n}_{2}\\=\frac{n-1}{2}\sum_{i=1}^{n}x_i/\frac{n(n-1)}{2}\\=\frac{\sum_{i=1}^{n}x_i}{n}=\mu$

c)
variance =
$\displaystyle \frac{\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}[x_{ij}-E(x_{ij})]^2}{C^n_2}\\=\frac{\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}(\frac{x_i+x_j}{2}-\mu)^2}{\frac{n(n-1)}{2}}\\=\frac{\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}[(x_i-\mu)+(x_j-\mu)]^2}{2n(n-1)}\\=\frac{(n-1)\sum_{=1}^{n}(x_i-\mu)^2+2\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}(x_i-\mu)(x_j-\mu)}{2n(n-1)}$

as $\displaystyle 0=[\sum_{i=1}^{n}(x_i+\mu)]^2=\sum_{i=1}^{n}(x_i-\mu)^2+2\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}(x_i-\mu)(x_j-\mu)$

we have $\displaystyle 2\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}[(x_i-\mu)(x_j-\mu)=-\sum_{i=1}^{n}(x_i-\mu)^2$

$\displaystyle \therefore$ variance = $\displaystyle \frac{(n-2)\sum_{i=1}^{n}(x_i-\mu)^2}{2n(n-1)}=\frac{(n-2)\sigma^2}{2(n-1)}$

as $\displaystyle n\rightarrow\infinity$, variance$\displaystyle \rightarrow\frac{1}{2}\sigma^2$
• Nov 25th 2011, 10:19 PM
BookEnquiry
here is another one without solution

Q2

let $\displaystyle x_1, X_2, ..., x_n$ be observations of variable X and $\displaystyle y_1, y_2, ..., y_n$ be observations of a variable . Let $\displaystyle \bar{X}=\frac{\sum_{i=1}^{n}x_i}{n}$, $\displaystyle \bar{Y}=\frac{\sum_{i=1}^{n}y_i}{n}$, Var(X) = $\displaystyle \frac{\sum_{i=1}^{n}(x_i-\bar{X})^2}{n}$, Var(Y) = $\displaystyle \frac{\sum_{i=1}^{n}(y_i-\bar{Y})^2}{n}$

a) prove that $\displaystyle \overline{aX+bY}=a\bar{X}+b\bar{Y}$, where a, b are constants.

b) prove that Var(aX+b) = $\displaystyle a^2$Var(X), where a, b are constants.

c) prove that Var$\displaystyle (\frac{X+Y}{\sqrt{2}})$+Var$\displaystyle (\frac{X-Y}{\sqrt{2}})$ = Var(X) + Var(Y)

d) given the following observations on (X, Y), verify the equality in (c)

$\displaystyle \begin{array}{ccc}O&X&Y\\1&1&1\\2&1&-1\\3&-1&1\\4&-1&-1\end{array}$

e)Let Z= aX +bY and W = cX +dY so that in matrix notation
$\displaystyle [\begin{array}{c}Z\\W\end{array}]=[\begin{array}{cc}a&b\\c&d\end{array}][\begin{array}{c}X\\Y\end{array}]$

Show that Var(Z) +Var(W) = Var(X) +Var(Y)
if $\displaystyle [\begin{array}{cc}a&c\\b&d\end{array}][\begin{array}{cc}a&b\\c&d\end{array}]=[\begin{array}{cc}1&0\\0&1\end{array}]$
• Nov 26th 2011, 05:21 AM
BookEnquiry
Q3

The mean $\displaystyle \mu$ and the variance $\displaystyle \sigma^2$ of a finite production A = $\displaystyle \{x_1, x_2, ..., x_n\}$ are defined as follows:
$\displaystyle \mu=\frac{1}{n}\sum_{i=1}^{n}x_i$ and $\displaystyle \sigma^2=\frac{1}{n}\sum_{i=1}^{n}(x_i-\mu)^2$

Samples of size 2, $\displaystyle \{x_i, x_j\}$, are randomly drawn from A with replacement so that i=j is allowed. he mean $\displaystyle \bar{x_{ij}}$ of the sample $\displaystyle \{x_i, x_j\}$ is defined as $\displaystyle \bar{x_{ij}}=\frac{1}{2}(x_i+x_j)}$ These sample means then form a set S = $\displaystyle \{\bar{x_{ij}}:i=1,2,...,n;j=1,2,,...,n\}$

a) Show that $\displaystyle \sum_{i=1}^{n}\sum_{j=1}^{n}(x_i+x_j)=2n\sum_{i=2} ^{n}x_i$

b) Show hat

i) the mean of all the elements of S is $\displaystyle \mu$

ii) the variance of all the elements of S is $\displaystyle \frac{1}{2}\sigma^2$

c) Suppose A = $\displaystyle \[x_i=i:i=1,2,...,10\}$
Find the mean and variance of all the elements of S.

d) Suppose A =$\displaystyle \{x_i=3i+4:i=1,2,...,10\}$
Using (c), find the mean and variance of all the elements of S.

solution

a)
$\displaystyle \sum_{i=1}^{n}\sum_{j=1}^{n}(x_i+x_j)=\sum_{i=1}^{ n}nx_i+n\sum_{j=1}^{n}x_j\\+\2n\sum_{i=1}^{n}x_i$

b)
i) mean of all the elements of S
$\displaystyle =E(\bar{x_{ij}})=\frac{1}{n^2}\sum_{i=1}^{n}\sum-{j=1}^{n}\bar{x_{ij}}$ (S has $\displaystyle n^2$ elements)

$\displaystyle =\frac{1}{n^2}\sum_{i=1}^{n}\sum_{j=1}^{n}\frac{1} {2}(x_i+x_j)\\=\frac{1}{2n^2}(2n\sum_{i=1}^{n}x_i) \\=\frac{1}{2n^2}(2n)(n\mu)=\mu$

ii) Variance of all the elements of S
= Var(\bar{x_{ij}}=$\displaystyle \frac{1}{n^2}\sum_{i=1}^{n}\sum_{j=1}^{n}(\bar{x_{ ij}}-\mu)^2$ (S has $\displaystyle n^2$ elements)

$\displaystyle =\frac{1}{n^2}\sum_{i=1}^{n}\sum_{j=1}^{n}(\frac{x _i+x_j}{2}-\mu)^2\\=\frac{1}{4n^2}\sum_{i=1}^{n}\sum_{j=1}^{n }[(x_i-\mu)+(x_j-\mu)]^2\\=\frac{1}{4n^2}\sum_{i=1}^{n}\sum_{j=1}^{n}[(x_i-\mu)^2+2(x_i-\mu)(x_j-\mu)+(x_j-\mu)^2]\\=\frac{1}{4n^2}[n\sum_{i=1}^{n}(x_i-\mu)^2+2\sum_{i=1}^{n}(x_i-\mu)\sum_{j=1}^{n}(x_j-\mu)+n\sum_{j=1}^{n}(x_j-\mu)^2]\\=\frac{1}{4n^2}[n(n\sigma^2)+2(0)(0)+n(n\sigma^2)]\\=\frac{1}{2}\sigma^2$

b) ii) alternative method

c) mean =5.5
variance = 8.25
$\displaystyle \thereforeE(\bar{x_{ij}})=5.5$, Var($\displaystyle \bar{x_{ij}}$) = $\displaystyle \frac{1}{2}(8.25)=4.125$

d) Using (c), mean = $\displaystyle 3\times5.5+4=20.5$
Variance = $\displaystyle 3^2\times8.25=74.25$

$\displaystyle \therefore$$\displaystyle E(\bar{x_{ij}})=20.5$. Var$\displaystyle (\bar{x_{ij}})$=$\displaystyle \frac{1}{2}(74.25)=37.125$