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Thread: finding variance

  1. #1
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    finding variance

    A set of 200 people consisting of 100 men and 100 women is randomly divided into 100 pairs of 2 each.

    Number the men arbitrarily from 1 to 100, and for $\displaystyle i$= 1,2,...100, let
    $\displaystyle \ X_i=\begin{cases}
    1 & \textrm{if men i is paired with a woman } \\
    0 & \textrm{otherwise }\end{cases}$

    Then X, the number of man-woman pairs, can be expressed as
    $\displaystyle X=\sum\limits_{i = 1}^{100} X_i$

    Because man i is equally likely to be paired with any of the other 199 people, of which 100 are women, we have
    $\displaystyle E[X_i] = P({X_i = 1}) = \frac {100}{199}$

    Similarly, for i not equal to j,
    $\displaystyle E[X_iX_j] = \frac{100}{199}\frac{99}{197}$

    $\displaystyle E[X] = \sum\limits_{i = 1}^{100} E[Xi] = (100)\frac{100}{199}$

    what about variance?
    $\displaystyle Var(X) = \sum\limits_{i = 1}^{100} Var(X_i) + 2 \sum\limits_{i<j}\sum Cov(X_i,X_j)$
    $\displaystyle = (100)\frac{100}{199}\frac{99}{199} + (2) {100 \choose 2} [\frac{100}{199}\frac{99}{197} - (\frac{100}{199})^2]$

    why $\displaystyle Var(X_i)= \frac{100}{199}\frac{99}{199} $ and $\displaystyle \sum\limits_{i<j}\sum Cov(X_i,X_j)= {100 \choose 2} [\frac{100}{199}\frac{99}{197} - (\frac{100}{199})^2] $??
    Last edited by noob mathematician; Apr 16th 2009 at 09:14 AM.
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  2. #2
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    Hello,

    Quote Originally Posted by noob mathematician View Post
    A set of 200 people consisting of 100 men and 100 women is randomly divided into 100 pairs of 2 each.

    Number the men arbitrarily from 1 to 100, and for $\displaystyle i$= 1,2,...100, let
    $\displaystyle \ Xi=\begin{cases}
    1 & \textrm{if men i is paired with a woman } \\
    0 & \textrm{otherwise }\end{cases}$

    Then X, the number of man-woman pairs, can be expressed as
    $\displaystyle X=\sum\limits_{i = 1}^{100} Xi$

    Because man i is equally likely to be paired with any of the other 199 people, of which 100 are women, we have
    $\displaystyle E[Xi] = P({Xi = 1}) = \frac {100}{199}$

    Similarly, for i not equal to j,
    $\displaystyle E[XiXj] = \frac{100}{199}\frac{99}{197}$

    $\displaystyle E[X] = \sum\limits_{i = 1}^{100} E[Xi] = (100)\frac{100}{199}$

    what about variance?
    $\displaystyle Var(X) = \sum\limits_{i = 1}^{100} Var(Xi) + 2 \sum\limits_{i<j}\sum Cov(Xi,Xj)$
    $\displaystyle = (100)\frac{100}{199}\frac{99}{199} + (2) {100 \choose 2} [\frac{100}{199}\frac{99}{197} - (\frac{100}{199})^2]$

    why $\displaystyle Var(Xi)= \frac{100}{199}\frac{99}{199} $ and $\displaystyle \sum\limits_{i<j}\sum Cov(Xi,Xj)= {100 \choose 2} [\frac{100}{199}\frac{99}{197} - (\frac{100}{199})^2] $??
    $\displaystyle Var(X_i)=E(X_i^2)-E(X_i)^2$
    Note that since $\displaystyle X_i^2=0 \text{ or } 1$, because $\displaystyle X_i=0 \text{ or } 1$, $\displaystyle E(X_i^2)=1^2 \cdot P(X_i=1)+0^2 \cdot P(X_i=0)=E(X_i)$


    $\displaystyle Cov(X_i,X_j)=E(X_iX_j)-E(X_i)E(X_j)= \frac{100}{199}\frac{99}{197}-\left(\frac{100}{199}\right)^2$


    And now you have to prove that $\displaystyle \sum_{j=1}^{100}\sum_{i<j} 1={100 \choose 2}$

    $\displaystyle \sum_{i<j} 1=j-1$, because there are j-1 times 1.
    Then note that since $\displaystyle i<j$ and i,j have values in $\displaystyle \{1,\dots,100\}$, then $\displaystyle i \in \{1,\dots,99\}$ and $\displaystyle j \in \{2,\dots,100\}$

    So we have $\displaystyle \sum_{j=2}^{100} (j-1)=\sum_{j=1}^{99} j=\frac{99 \cdot 100}{2}$, which is exactly $\displaystyle {100 \choose 2}$
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  3. #3
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    IC.. I get it thanks!
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