91Ó°ÊÓ

The hypergeometric distribution is a discrete probability distribution in probability theory and statistics that describes the probability of\({\rm{k}}\)successes (random draws for which the object drawn has a specified feature) in\({\rm{n}}\)draws, without replacement, from a finite population of size\({\rm{N}}\)that contains exactly\({\rm{K}}\)objects with that feature, where each draw is either successful or unsuccessful.

Proposition: Assume that population has\({\rm{M}}\)successes\({\rm{(S)}}\),\({\rm{N - M}}\)failures\({\rm{(F)}}\). If random variable\({\rm{X}}\)is –

\({\rm{X = }}\)number of successes in a random sample size\({\rm{n}}\),

Then it has probability mass function –

\({\rm{(x;n,M,N) = P(X = x) = }}\frac{{\left( {\begin{array}{*{20}{c}}{\rm{M}}\\{\rm{x}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{N - M}}}\\{{\rm{n - x}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{\rm{N}}\\{\rm{n}}\end{array}} \right)}}\)

For all integers\({\rm{x}}\)for which –

\({\rm{max\{ 0,n - N + M\} }} \le {\rm{x}} \le {\rm{min\{ n,M\} }}\)

The probability distribution is called hypergeometric distribution.

The Expected Value (mean value) of a discrete random variable\(X\)with set of possible values\({\rm{S}}\)and\({\rm{pmf p(x)}}\)is –

\({\rm{E(X) = \mu x = }}\sum\limits_{{\rm{x}} \in {\rm{S}}} {{\rm{x}} \cdot {\rm{p(x)}}} \)

From the definition, using the hint, the following is holds –

\(\begin{aligned}E(X) &= \sum\limits_{x = 0}^n x \cdot h(x;n,M,N) = \sum\limits_{x = 0}^n x \cdot \frac{{\left({\begin{array}{*{20}{c}}M \\ x \end{array}} \right)\left( {\begin{array}{*{20}{c}}{N - M} \\{n - x} \end{array}} \right)}}{{\left( {\begin{array}{*{20}{l}} N \\ n \end{array}} \right)}} \\ &= \sum\limits_{x = 1}^n {\not x} \cdot \frac{{\frac{{M!}}{{(x - 1)!(M - x)!}}}}{{\left({\begin{array}{*{20}{l}}N \\ n \end{array}} \right)}} \cdot \left( {\begin{array}{*{20}{c}}{N - M} \\ {n - x}\end{array}} \right) \\ &= n\frac{M}{N}\sum\limits_{x = 1}^n {\left( {\begin{array}{*{20}{c}}{M - 1} \\ {x - 1}\end{array}} \right)} \frac{1}{{\left( {\begin{array}{*{20}{c}}{N - 1} \\ {n - 1}\end{array}} \right)}} \cdot \left( {\begin{array}{*{20}{c}}{N - M} \\ {n - x}\end{array}} \right) \\ &= n\frac{M}{N}\sum\limits_{k = 0}^{n - 1} {\left( {\begin{array}{*{20}{c}}{M - 1} \\ k \end{array}} \right)} \frac{1}{{\left( {\begin{array}{*{20}{c}}{N - 1} \\ {n - 1} \end{array}} \right)}} \cdot \left( {\begin{array}{*{20}{c}}{N - M} \\ {n - 1 - k} \end{array}} \right) \\ \end{aligned} \)

Notice that –

\({\rm{h(k;n - 1,M - 1,N - 1) = }}\frac{{\left( {\begin{array}{*{20}{c}}{{\rm{M - 1}}}\\{\rm{x}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\rm{N - 1 - (M - 1)}}}\\{{\rm{n - 1 - k}}}\end{array}} \right)}}{{\left( {\begin{array}{*{20}{c}}{{\rm{N - 1}}}\\{{\rm{n - 1}}}\end{array}} \right)}}\)

And it is needed to create that in order to do that we just need to add and subtract \({\rm{1}}\) from \({\rm{N - M}}\) and we get \({\rm{N - 1 - (M - 1)}}\). Hence –

\(\begin{aligned} E(X) &= n\frac{M}{N}\sum\limits_{k = 0}^{n - 1} {\left( {\begin{array}{*{20}{c}}{M - 1} \\ k \end{array}} \right)} \frac{1}{{\left( {\begin{array}{*{20}{c}} {N - 1} \\ {n - 1} \end{array}} \right)}} \cdot \left( {\begin{array}{*{20}{c}} {N - 1 - (M - 1)} \\ {n - 1 - k} \end{array}} \right) \\ &= n\frac{M}{N} \cdot \sum\limits_{k = 0}^{n - 1} h (k;n - 1,M - 1,N - 1) \\ &= n\frac{M}{N} \\ \end{aligned} \)

Because the sum of the \({\rm{pmf}}\) is equal to \({\rm{1}}\).

Therefore, the theorem is proved.