1. ## 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
$\ X_i=\begin{cases}
1 & \textrm{if men i is paired with a woman } \\
0 & \textrm{otherwise }\end{cases}$

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

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,
$E[X_iX_j] = \frac{100}{199}\frac{99}{197}$

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

$Var(X) = \sum\limits_{i = 1}^{100} Var(X_i) + 2 \sum\limits_{i
$= (100)\frac{100}{199}\frac{99}{199} + (2) {100 \choose 2} [\frac{100}{199}\frac{99}{197} - (\frac{100}{199})^2]$

why $Var(X_i)= \frac{100}{199}\frac{99}{199}$ and $\sum\limits_{i??

2. Hello,

Originally Posted by noob mathematician
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
$\ Xi=\begin{cases}
1 & \textrm{if men i is paired with a woman } \\
0 & \textrm{otherwise }\end{cases}$

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

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,
$E[XiXj] = \frac{100}{199}\frac{99}{197}$

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

$Var(X) = \sum\limits_{i = 1}^{100} Var(Xi) + 2 \sum\limits_{i
$= (100)\frac{100}{199}\frac{99}{199} + (2) {100 \choose 2} [\frac{100}{199}\frac{99}{197} - (\frac{100}{199})^2]$

why $Var(Xi)= \frac{100}{199}\frac{99}{199}$ and $\sum\limits_{i??
$Var(X_i)=E(X_i^2)-E(X_i)^2$
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

$\sum_{i, because there are j-1 times 1.
Then note that since $i 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 ${100 \choose 2}$

3. IC.. I get it thanks!