Essentials of Mathematical Probability and Statistics

For the following, we will presume thatSince you are sampling without replacement and trying only measure the number of items from your desired group in the sample, then the space of X will include R = {0, 1, ..., r} assuming \(n_1 \ge r\) and \(n-n_1 \ge r\text{.}\) In the case when r is too large for either of these, the formulas below will follow noting that binomial coefficients are zero if the top is smaller than the bottom or if the bottom is negative.

So f(x) = P(X = x) = P(x from the sample are from the target group and the remainder are not). Breaking these up gives

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.

For skewness, take the limit of the skewness result above as \(n_1, n_2\text{,}\) and \(r\) are uniformly increased together (i.e. all are doubled, tripled, etc.) To model this behavior one can simply scale each of those term by some variable \(k\) and then let \(k\) increase. Asymptotically, the result is

Similarly for kurtosis. However, since we can't find a nice formula for the kurtosis then taking the limit can be difficult. So, we will have to appeal to general results proved in a course that would follow up this one that would prove this fact must be true. Sadly, not here.