Results 1 to 3 of 3

Math Help - finding variance

  1. #1
    Member
    Joined
    Oct 2008
    Posts
    206

    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 i= 1,2,...100, let
    \ X_i=\begin{cases}<br />
1 & \textrm{if men i is paired with a woman } \\<br />
0 & \textrm{otherwise }\end{cases}

    Then X, the number of man-woman pairs, can be expressed as
    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
    E[X_i] = P({X_i = 1}) = \frac {100}{199}

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

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

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

    why Var(X_i)= \frac{100}{199}\frac{99}{199} and \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; April 16th 2009 at 09:14 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    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 i= 1,2,...100, let
    \ Xi=\begin{cases}<br />
1 & \textrm{if men i is paired with a woman } \\<br />
0 & \textrm{otherwise }\end{cases}

    Then X, the number of man-woman pairs, can be expressed as
    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
    E[Xi] = P({Xi = 1}) = \frac {100}{199}

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

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

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

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


    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 \sum_{j=1}^{100}\sum_{i<j} 1={100 \choose 2}

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

    So we have \sum_{j=2}^{100} (j-1)=\sum_{j=1}^{99} j=\frac{99 \cdot 100}{2}, which is exactly {100 \choose 2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2008
    Posts
    206
    IC.. I get it thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Finding the Variance of (1/S)
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: November 6th 2011, 08:56 AM
  2. Finding Mean/Variance from Sum of fi
    Posted in the Statistics Forum
    Replies: 1
    Last Post: March 10th 2011, 09:08 PM
  3. Finding mean and variance of RV
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: April 22nd 2010, 12:22 AM
  4. finding variance?
    Posted in the Statistics Forum
    Replies: 2
    Last Post: February 21st 2010, 01:32 PM
  5. Finding mean and variance from the mean?
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: November 18th 2008, 02:56 AM

Search Tags


/mathhelpforum @mathhelpforum