Aqui añado el lirbo que e usado para ayudarme: Introduction to Bayesian Statistics Bolstad, William M. Editorial: Hoboken, New Jersey, U.S.A.: Wiley-Interscience, 2004

1. Ejercicio-4.2
Isidoro Gléz.-Adalid Pemartín
Métodos Matemáticos I
Primero de Grado en Física
Curso 2013-14
Sabado, 26 de octubre.
Enunciado:
Calcula la media y la varianza de la funcion de distribucion hipergeométrica.
p(X|C) =
B
x
R
r
=
N
n
B
x
N −B
n−x
N
n
M edia :
n
x=1
x =
=
n
x=1
xp(X|C) =
n
x=1
x
(B)(N −B)
n−x
x
=
(N )
n
B(B−1)!
(N −B)!
(x−1)!(B−1−(x−1))! (n−x)!(N −B−(n−x))!
N (N −1)!
n(n−1)!(N −1−(n−1))!
=
n
x=1
Bn
N
B!
(N −B)!
x x!(B−x)! (n−x)!(N −B−(n−x))! =
N!
n!(N −n)!
n
x=1
(B−1)!
(N −1−(B−1))!
(x−1)!(B−1−(x−1))! (n−1−(x−1))!(N −1−(B−1)−(n−1−(x−1)))!
(N −1)!
(n−1)!(N −1−(n−1))!
Llamando a N − 1 = N , n − 1 = n , B − 1 = B y x − 1 = y; la formula queda:
n
x=1
Bn
N
Siendo
x =
(N −B )!
B !
y!(B −y)! (n −y)!(N −B −(n −y))!
N !
n !(N −n )!
n
x=1
(N −B )!
B !
y!(B −y)! (n −y)!(N −B −(n −y))!
N !
n !(N −n )!
= 1 nos queda:
nB
N
V arianza :
2
σx = x2 − x
2
= x2 − x
2
Primero consideramos: x(x − 1) = x2 − x = x2 − x
x(x − 1) =
=
n
x=1
n
x=1
n
x=1
x(x − 1)p(X|C) =
B(B−1)(B−2)!
(N −B)!
(x−2)!(B−2−(x−2))! (n−x)!(N −B−(n−x))!
N (N −1)(N −2)!
n(n−1)(n−2)!(N −2−(n−2))!
1
=
x(x − 1)
(B)(N −B)
x
n−x
=
(N )
n
n
x=1
B!
(N −B)!
x(x − 1) x!(B−x)! (n−x)!(N −B−(n−x))! =
N!
n!(N −n)!

2. =
n
x=1
B(B−1)n(n−1)
N (N −1)
(N −2−(B−2))!
(B−2)!
(x−2)!(B−2−(x−2))! (n−2−(x−2))!(N −2−(B−2)−(n−2−(x−2)))!
(N −2)!
(n−2)!(N −2−(n−2))!
Llamando a N − 2 = N , n − 2 = n , B − 2 = B y x − 2 = y; la formula queda:
n
x=1
B(B−1)n(n−1)
N (N −1)
Siendo
(N −B )!
B !
y!(B −y)! (n −y)!(N −B −(n −y))!
N !
n !(N −n )!
n
x=1
x(x − 1) =
(N −B )!
B !
y!(B −y)! (n −y)!(N −B −(n −y))!
N !
n !(N −n )!
= 1 nos queda:
B(B−1)n(n−1)
N (N −1)
2
σx = x(x − 1) + x − x
2
= x2 − x + x − x
2
σx = x(x − 1) + x − x
2
=
=
B(B−1)n(n−1)
N (N −1)
=
nB N (B−1)(n−1)+N (N −1)−nB(N −1)
]
N [
N (N −1)
=
nB N 2 −N n
N [ N (N −1)
2
σx =
nB
N
∗
+
−
N −n
N −1
nB
N
−
( nB )2
N
BN −Bn
N (N −1) ]
∗ (1 −
=
=
2
= x2 − x
2
+
N (N −1)
N (N −1)
nB N Bn−N B−N n+N +N 2 −N −N Bn+Bn
]
N [
N (N −1)
=
−
= x2 − x + x − x−
nB N (B−1)(n−1)
N [
N (N −1)
=
nB N 2 −N n−BN +Bn
]
N [
N (N −1)
nB (B−1)(n−1)
N [
N −1
nB N −n
N [ N −1
2
B N −n
N (N −1) ]
B
N)
2
+1−
=
nB
N ]
=
nB N −n
N [ N −1 (1
−
B
N )]
=
nB
N
∗
N −n
N −1
∗ (1 −
−
nB(N −1)
N (N −1) ]
B
N)
=
=