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Math Help - Anybody know these kinds of statistics and their serious names? Do you know any books

  1. #1
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    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 \mu and variance \sigma of a finite population P = {x_1, x_2, ..., x_n} where

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

    a) Show that \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 \mu.

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

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

    solution
    a)
    \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)
    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 =
    \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 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 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

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

    as n\rightarrow\infinity, variance \rightarrow\frac{1}{2}\sigma^2
    Last edited by BookEnquiry; November 26th 2011 at 05:38 PM.
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  2. #2
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    here is another one without solution

    Q2

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

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

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

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

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

    \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
    [\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 [\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}]
    Last edited by BookEnquiry; November 26th 2011 at 05:22 AM.
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  3. #3
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    Q3

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

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

    a) Show that \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 \mu

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

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

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

    solution

    a)
    \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
    =E(\bar{x_{ij}})=\frac{1}{n^2}\sum_{i=1}^{n}\sum-{j=1}^{n}\bar{x_{ij}} (S has n^2 elements)

    =\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}}= \frac{1}{n^2}\sum_{i=1}^{n}\sum_{j=1}^{n}(\bar{x_{  ij}}-\mu)^2 (S has n^2 elements)

    =\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
    \thereforeE(\bar{x_{ij}})=5.5, Var( \bar{x_{ij}}) = \frac{1}{2}(8.25)=4.125

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

    \therefore E(\bar{x_{ij}})=20.5. Var (\bar{x_{ij}})= \frac{1}{2}(74.25)=37.125
    Last edited by BookEnquiry; November 26th 2011 at 06:52 AM.
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