If X is a hypergeometric rv, show directly from the definition that E(X) = nM/N ( consider only the case n < M). [Hint: Factor nM/N out of the sum for E(X), and show that the terms inside the sum are of the form h( y; n – 1, M – 1, N -1), where y – x -1.]

Answer : Step 1 : Given If X is a hypergeometric rv. Now we have to show that directly from the definition that E(X) = N Here X:hypergeometric row variable. P(X = x)=h(x;n,m,N) Proposition : If X is the number of S’s in a completely random sample of size n drawn from a population consisting of M S’s and (N-M) F’s , then the probability distribution of X, called the hypergeometric distribution, is given by M NM (x)( nx ) P(X = x) = h(x; n, M, N) = N (n) X= the number of successes from a sample size of n, from a population of size N, with M successes in it X~hypergeom( n, M, N ). Where n= number of successes among the population M= sample size N= total population size The expected value is n E(X) = P(X = x) x, x integer x=0 We know that. (x)( nx ) P(X = x) = ( n n M NM (x )(nx ) E(X) = N x x=0 ( n n E(X) = M ! (NM) ! n!(Nn) !x x=1(x!) (Mx)! (nx)! (N+Mn)(N)! n NM nM (M1)! (nx ) (n1)(N1(n1)! E(X) = N (x1)! (M1(x1)! (nx)! (N+Mn)! (N1)! x=1 n (M1)( nx ) E(X) = nM x1 N x=1 (N1) Let y = x-1 Then, n1 M1 NM nM ( y )( n1y E(X) = N (N1) y=0 nM n1 E(X) = N h(y;n 1,M 1,N 1) y=0 nM E(X) = N nM HenceE(X) = N