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- 1. Mean and Variance of the HyperGeometric Distribution Page 1 Al Lehnen Madison Area Technical College 11/30/2011 In a drawing of n distinguishable objects without replacement from a set of N (n < N) distinguishable objects, a of which have characteristic A, (a < N) the probability that exactly x objects in the draw of n have the characteristic A is given by then number of different ways the x objects can be chosen from the a available times the number of different ways the n-x objects in the draw which don’t have A can be chosen from the N-a available divided by the number of different ways n distinguishable objects can be chosen from a set of N. The resulting probability distribution for the random variable x is called the hypergeometric distribution. In symbols, ( ) a N a x n x f x N n −⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ −⎝ ⎠⎝ ⎠= ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ . The binomial coefficient ( ) ! ! ! k k j j k j ⎛ ⎞ =⎜ ⎟ −⎝ ⎠ is defined to be zero if either j or k-j is negative, so that the probability of the null event of drawing more objects than those available is zero. To prove that ( ) 0 0 1 n n x x a N a x n x f x N n = = −⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ −⎝ ⎠⎝ ⎠= = ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ∑ ∑ , consider the factorization ( ) ( ) ( )N a N a B C B C B C − + = + + . From the binomial theorem, ( ) ( ) ( ) 0 0 0 0 a N a a N a a j j N a l l j l a N a N l j l j j l a N a B C B C B C B C j l a N a B C j l − − − − − = = − − + + = = −⎛ ⎞ ⎛ ⎞ + + = ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ −⎛ ⎞⎛ ⎞ = ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ∑ ∑ ∑ ∑ Using the diagonal rearrangement suggested by the figure below with l n j= − , with the intercept n running from 0 to N and j running from 0 to a. This generates more than the ( )( )1 1a N a+ − + terms in the above sum. However, all of the new terms generated vanish since they have l N a> − . ( ) ( ) 0 0 N a a N a N n n n j a N a B C B C B C j n j − − = = −⎛ ⎞⎛ ⎞ + + = ⎜ ⎟⎜ ⎟ −⎝ ⎠⎝ ⎠ ∑ ∑ Now, for n a> extending the sum over j to n because of the a j ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ factor would only add terms which are zero. Similarly, if n a< , the terms in the sum over j from j = n + 1 to j = a are all zero due to the N a n j −⎛ ⎞ ⎜ ⎟ −⎝ ⎠ factor. Thus, ( ) ( ) 0 0 0 0 N a N n a N a N n n N n n n j n j a N a a N a B C B C B C B C j n j j n j − − − = = = = − −⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ + + = =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ − −⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ∑ ∑ ∑ ∑ .
- 2. Mean and Variance of the HyperGeometric Distribution Page 2 Al Lehnen Madison Area Technical College 11/30/2011 But from a second use of the binomial theorem, ( ) ( ) ( ) 0 0 0 N n N a N a NN n n N n n n j n a N a N B C B C B C B C B C j n j n − − − = = = −⎛ ⎞⎛ ⎞ ⎛ ⎞ + + = = + =⎜ ⎟⎜ ⎟ ⎜ ⎟ −⎝ ⎠⎝ ⎠ ⎝ ⎠ ∑ ∑ ∑ . The only way the two sums can be equal for all values of B and C is for 0 n j a N a N j n j n= −⎛ ⎞⎛ ⎞ ⎛ ⎞ =⎜ ⎟⎜ ⎟ ⎜ ⎟ −⎝ ⎠⎝ ⎠ ⎝ ⎠ ∑ . (1) This in turn implies that the hypergeometric probabilities do indeed construct a valid probability distribution, i.e. ( ) 0 0 1 n n x x a N a x n x f x N n = = −⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟ −⎝ ⎠⎝ ⎠= = ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ∑ ∑ . The mean or expected value of the hypergeometric random variable is given by ( ) 1 0 0 n n x x x N a N a x x f x x n x n x μ − = = −⎛ ⎞ ⎛ ⎞⎛ ⎞ = = =⎜ ⎟ ⎜ ⎟⎜ ⎟ −⎝ ⎠ ⎝ ⎠⎝ ⎠ ∑ ∑ . Now, using Equation (1),
- 3. Mean and Variance of the HyperGeometric Distribution Page 3 Al Lehnen Madison Area Technical College 11/30/2011 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 1 1 1 1 0 0 1 11 !! 1 1! ! 1 ! 1 1 ! 1 1 1 11 ! 1 1 1! 1 ! 1 1 n n n x x x n n x x N aa aa N a N axa x n xx n x n xx n x x n x N a N aa a a a n x n xxx n x N a n = = = − − = = ⎛ ⎞− − −−− −⎛ ⎞⎛ ⎞ ⎛ ⎞ = = ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− − −− −− ⎡ ⎤− − − −⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎛ ⎞ ⎛ ⎞− − − − − −− −⎛ ⎞ = =⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟− − − −⎡ ⎤− − ⎝ ⎠⎝ ⎠ ⎝ ⎠⎣ ⎦ −⎛ ⎞ = ⎜ ⎟ −⎝ ⎠ ∑ ∑ ∑ ∑ ∑ This gives that ( ) ( ) ( ) ( ) ( )1 0 1 ! ! !1 1 1 ! ! ! n x x N n N nN N na x x f x a a n n n N n N N μ − = − −−⎛ ⎞ ⎛ ⎞ = = = = ⋅ =⎜ ⎟ ⎜ ⎟ − − −⎝ ⎠ ⎝ ⎠ ∑ . Using the notation of the binomial distribution that a p N = , we see that the expected value of x is the same for both drawing without replacement (the hypergeometric distribution) and with replacement (the binomial distribution). x na x np N μ = = = (2) The variance of the hypergeometric distribution can be computed from the generic formula that 2 22 2 x x x x xσ ⎡ ⎤= − = −⎣ ⎦ . Again from Equation (1), ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 2 2 2 2 0 0 2 21 ! 1 2 ! 1 2 2! ! 2 ! 2 2 ! 2 2 2 22 ! 2 1 1 2 2! 2 ! n n n x x x n n x x N ax x a a a aa N a N a x x n xx n x n xx n x x n x N a N aa a a a a a n x n xxx n x = = = − − = = ⎛ ⎞− − −− − −− −⎛ ⎞⎛ ⎞ ⎛ ⎞ − = = ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟− − −− −− ⎡ ⎤− − − −⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎛ ⎞ ⎛ ⎞− − − − − −− −⎛ ⎞ = − = −⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟− − − −⎡ ⎤− − ⎝ ⎠⎝ ⎠ ⎝ ⎠⎣ ⎦ ∑ ∑ ∑ ∑ ( ) 2 1 2 N a a n −⎛ ⎞ = − ⎜ ⎟ −⎝ ⎠ ∑ So, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 0 2 1 1 1 2 2 ! ! ! 1 1 1 2 ! ! ! 1 n x N a N a N N x x x x a a n x n x n n N N n n a a n n a a n N n N N N − − = − −⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞ − = − = −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ − −⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠ − − − − = − ⋅ = − − − ∑ and ( ) ( ) ( ) ( ) ( )( ) ( ) 2 1 1 1 1 1 1 1 1 a a n n a nan an x x x x N N N N N ⎡ ⎤− − − − = − + = + = +⎢ ⎥ − −⎢ ⎥⎣ ⎦ .
- 4. Mean and Variance of the HyperGeometric Distribution Page 4 Al Lehnen Madison Area Technical College 11/30/2011 Thus, ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) 22 2 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x a n N a n N N an Nan an an x x N N N N N N N N N N an Nan Na Nn N N N Nan an an N Na Nn an N N N N N N N N a n N a N n N aan an an N a N N N N N N N N σ ⎡ ⎤⎡ ⎤− − − − − − = − = + − = + −⎢ ⎥⎢ ⎥ − − − −⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎡ ⎤ ⎡ ⎤− − + + − − + − − + = =⎢ ⎥ ⎢ ⎥ − −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎡ ⎤ ⎡ ⎤− − − − − −⎛ = = =⎢ ⎥ ⎢ ⎥ ⎜− − ⎝⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ( ) 1 1 1 1 1 N n N an a N n N n np p N N N N −⎞⎛ ⎞ ⎟⎜ ⎟−⎠⎝ ⎠ − −⎛ ⎞⎛ ⎞ ⎛ ⎞ = − = −⎜ ⎟⎜ ⎟ ⎜ ⎟ − −⎝ ⎠⎝ ⎠ ⎝ ⎠ The last factor 1 N n N −⎛ ⎞ ⎜ ⎟−⎝ ⎠ is called the “finite population correction” and is the reason that the variance of the binomial distribution ( )1np p− differs from the hypergeometric distribution. For N large compared to the sample size n, the two distributions are essentially identical.