Proof

  1. \begin{align*} \sum_{x=0}^n \binom{n}{x} y^x & = (1+y)^n, \text{ by the Binomial Theorem}\\ & = (1+y)^{n_1} \cdot (1+y)^{n_2} \\ & = \sum_{x=0}^{n_1} \binom{n_1}{x} y^x \cdot \sum_{x=0}^{n_2} \binom{n_2}{x} y^x \\ & = \sum_{x=0}^n \sum_{t=0}^r \binom{n_1}{r} \binom{n_2}{r-t} y^x \end{align*}
    Equating like coefficients for the various powers of y gives
    \begin{equation*} \binom{n}{r} = \sum_{t=0}^r \binom{n_1}{r} \binom{n_2}{r-t}. \end{equation*}
    Dividing gives
    \begin{equation*} 1 = \sum_{x=0}^r f(x). \end{equation*}
  2. For the mean
    \begin{align*} \sum_{x=0}^n x \frac{\binom{n_1}{x} \binom{n-n_1}{r-x}}{\binom{n}{r}} & = \frac{1}{\binom{n}{r}} \sum_{x=1}^n \frac{n_1(n_1-1)!}{(x-1)!(n_1-x)!} \binom{n-n_1}{r-x}\\ & = \frac{n_1}{\binom{n}{r}} \sum_{x=1}^n \frac{(n_1-1)!}{(x-1)!((n_1-1)-(x-1))!} \binom{n-n_1}{r-x} \\ & = \frac{n_1}{\frac{n(n-1)!}{r!(n-r)!}} \sum_{x=1}^n \binom{n_1-1}{x-1} \binom{n-n_1}{r-x} \end{align*}
    Consider the following change of variables for the summation:
    \begin{gather*} y = x-1\\ n_3 = n_1-1\\ s = r-1\\ m = n-1 \end{gather*}
    Then, this becomes
    \begin{align*} \mu = \sum_{x=0}^n x \frac{\binom{n_1}{x} \binom{n-n_1}{r-x}}{\binom{n}{r}} & = r \frac{n_1}{n} \sum_{y=0}^m \frac{\binom{n_3}{y} \binom{m-n_3}{s-y}}{\binom{m}{s}}\\ & = r \frac{n_1}{n} \cdot 1 \end{align*}
    noting that the summation is in the same form as was show yields 1 above.
  3. For variance, we will use an alternate form of the definition that is useful when looking for cancellation options with the numerous factorials in the hypergeometric probability function 6.4. Indeed, you can easily notice that
    \begin{equation*} \sigma^2 = E[X^2] - \mu^2 = E[X^2-X]+E[X] -\mu^2 = E[X(X-1)] + \mu - \mu^2. \end{equation*}
    Since we have \(\mu = r \frac{n_1}{n}\) from above then let’s focus on the first term only and use the substitutions
    \begin{gather*} y = x-2\\ n_3 = n_1-2\\ s = r-2\\ m = n-2 \end{gather*}
    to get
    \begin{gather*} E[X(X-1)] = \sum_{x=0}^n x(x-1) \frac{\binom{n_1}{x} \binom{n-n_1}{r-x}}{\binom{n}{r}}\\ = \sum_{x=2}^n x(x-1) \frac{\frac{n_1!}{x(x-1)(x-2)!(n_1-x)! } \binom{n-n_1}{r-x}}{\binom{n}{r}}\\ = \sum_{x=2}^n \frac{\frac{n_1!}{(x-2)!(n_1-x)! } \frac{n_2!}{(r-x)!(n_2-r+x)!}}{\binom{n}{r}}\\ = n_1 \cdot (n_1-1) \cdot \sum_{x=2}^n \frac{\frac{(n_3)!}{(x-2)!(n_3 -(x-2))! } \frac{n_2!}{((r-2)-(x-2))!(n_2-(r-2)+(x-2))!}}{\binom{n}{r}}\\ = n_1 \cdot (n_1-1) \sum_{y=0}^{m} \frac{\frac{(n_3)!}{y!(n_3 -y)! } \frac{n_2!}{(s-y)!(n_2-s+y)!}}{\binom{n}{r}}\\ = \frac{n_1 \cdot (n_1-1) \cdot r \cdot (r-1)}{n (n-1)} \sum_{y=0}^{m} \frac{\binom{(n_3)}{y} \binom{n_2}{s-y}}{\binom{m}{s}}\\ = \frac{n_1 \cdot (n_1-1) \cdot r \cdot (r-1)}{n (n-1)} \end{gather*}
    where we have used the summation formula above that showed that f(x) was a probability function.
    Putting this together with the earlier formula gives
    \begin{equation*} \sigma^2 = \frac{n_1 \cdot (n_1-1) \cdot r \cdot (r-1)}{n (n-1)} + r \frac{n_1}{n} - \left ( r \frac{n_1}{n} \right )^2. \end{equation*}
  4. The formula and resulting proof of kurtosis is even more messy and we won’t bother with proving it for this distribution! If you start searching on the web for the formula then you will find many places just ignore the kurtosis for the hypergeometric. So we will as well here and just note that as the parameters increase the resulting graph sure looks bell-shaped. That is, approaching a kurtosis of 3.
in-context