Probability Density Function

We first check to see that f(x) is a valid pmf. This requires that it is non-negative everywhere and that its total sum is equal to 1. The first condition is obvious. For the second condition we will start with Vandermonde's identity

${\displaystyle \sum _{x=0}^{n}{a \choose x}{b \choose n-x}={a+b \choose n}}$

${\displaystyle \sum _{x=0}^{n}{{a \choose x}{b \choose n-x} \over {a+b \choose n}}=1}$

We now see that if a=m and b=N-m that the condition is satisfied.

Mean

We derive the mean as follows:

${\displaystyle \operatorname {E} [X]=\sum _{x=0}^{n}x\cdot f(x;n,m,N)=\sum _{x=0}^{n}x\cdot {{{m \choose x}{{N-m} \choose {n-x}}} \over {N \choose n}}}$

${\displaystyle \operatorname {E} [X]=0\cdot {{{m \choose 0}{{N-m} \choose {n-0}}} \over {N \choose n}}+\sum _{x=1}^{n}x\cdot {{{m \choose x}{{N-m} \choose {n-x}}} \over {N \choose n}}}$

We use the identity ${\displaystyle {\binom {a}{b}}={\frac {a}{b}}{\binom {a-1}{b-1}}}$ in the denominator.

${\displaystyle \operatorname {E} [X]=0+\sum _{x=1}^{n}x\cdot {{{m \choose x}{{N-m} \choose {n-x}}} \over {{N \over n}{{N-1} \choose {n-1}}}}}$

${\displaystyle \operatorname {E} [X]={n \over N}\sum _{x=1}^{n}x\cdot {{{m \choose x}{{N-m} \choose {n-x}}} \over {{N-1} \choose {n-1}}}}$

Next we use the identity ${\displaystyle b{\binom {a}{b}}=a{\binom {a-1}{b-1}}}$ in the first binomial of the numerator.

${\displaystyle \operatorname {E} [X]={n \over N}\sum _{x=1}^{n}{m{{m-1 \choose x-1}{{N-m} \choose {n-x}}} \over {{N-1} \choose {n-1}}}}$

Next, for the variables inside the sum we define corresponding prime variables that are one less. So N′=N−1, m′=m−1, x′=x−1, n′=n-1.

${\displaystyle \operatorname {E} [X]={mn \over N}\sum _{x'=0}^{n'}{{{m' \choose x'}{{N'-m'} \choose {n'-x'}}} \over {{N'} \choose {n'}}}}$

${\displaystyle \operatorname {E} [X]={mn \over N}\sum _{x'=0}^{n'}f(x';n',m',N')}$

Now we see that the sum is the total sum over a Hypergeometric pmf with modified parameters. This is equal to 1. Therefore

${\displaystyle \operatorname {E} [X]={nm \over N}}$

Variance

We first determine E(X²).

${\displaystyle \operatorname {E} [X^{2}]=\sum _{x=0}^{n}f(x;n,m,N)\cdot x^{2}=\sum _{x=0}^{n}{{{m \choose x}{{N-m} \choose {n-x}}} \over {N \choose n}}\cdot x^{2}}$

${\displaystyle \operatorname {E} [X^{2}]={{{m \choose 0}{{N-m} \choose {n-0}}} \over {N \choose n}}\cdot 0^{2}+\sum _{x=1}^{n}{{{m \choose x}{{N-m} \choose {n-x}}} \over {N \choose n}}\cdot x^{2}}$

${\displaystyle \operatorname {E} [X^{2}]=0+\sum _{x=1}^{n}{{m{m-1 \choose x-1}{{N-m} \choose {n-x}}} \over {{N \over n}{{N-1} \choose {n-1}}}}\cdot x}$

${\displaystyle \operatorname {E} [X^{2}]={mn \over N}\sum _{x=1}^{n}{{{m-1 \choose x-1}{{N-m} \choose {n-x}}} \over {{N-1} \choose {n-1}}}\cdot x}$

We use the same variable substitution as when deriving the mean.

${\displaystyle \operatorname {E} [X^{2}]={mn \over N}\sum _{x'=0}^{n'}{{{m' \choose x'}{{N'-m'} \choose {n'-x'}}} \over {{N'} \choose {n'}}}(x'+1)}$

${\displaystyle \operatorname {E} [X^{2}]={mn \over N}\left[\sum _{x'=0}^{n'}{{{m' \choose x'}{{N'-m'} \choose {n'-x'}}} \over {{N'} \choose {n'}}}x'+\sum _{x'=0}^{n'}{{{m' \choose x'}{{N'-m'} \choose {n'-x'}}} \over {{N'} \choose {n'}}}\right]}$

The first sum is the expected value of a hypergeometric random variable with parameteres (n',m',N'). The second sum is the total sum that random variable's pmf.

${\displaystyle \operatorname {E} [X^{2}]={mn \over N}\left[{n'm' \over N'}+1\right]}$

${\displaystyle \operatorname {E} [X^{2}]={mn \over N}\left[{(n-1)(m-1) \over (N-1)}+1\right]={mn \over N}\left[{{(n-1)(m-1)+(N-1)} \over (N-1)}\right]}$

We then solve for the variance

${\displaystyle \operatorname {Var} (X)=\operatorname {E} [X^{2}]-(\operatorname {E} [X])^{2}}$

${\displaystyle \operatorname {Var} (X)={mn \over N}\left[{{(n-1)(m-1)+(N-1)} \over (N-1)}\right]-\left({mn \over N}\right)^{2}}$

${\displaystyle \operatorname {Var} (X)={Nmn \over N^{2}}\left[{{(n-1)(m-1)+(N-1)} \over (N-1)}\right]-{(N-1)(mn)^{2} \over (N-1)N^{2}}}$

${\displaystyle \operatorname {Var} (X)={nm(N-n)(N-m) \over N^{2}(N-1)}}$