Then EX = nM/N

DX = nM/N *( 1-M/N)*(N-N)/(N- 1)

In fact, it can be compared with binomial distribution Binomial distribution is the limit of hypergeometric distribution

(1) if the random variable x obeys the binomial distribution with parameters n and p, then EX=np, DX=np( 1-p).

② If the random variable X obeys the hypergeometric distribution with parameters of n, m and n, then ex = nm/n..

Variance of hypergeometric distribution

(1) if the random variable x obeys the binomial distribution with parameters n and p, then EX=np, DX=np( 1-p).

② If the random variable X obeys the hypergeometric distribution with parameters of n, m and n, then ex = nm/n..

Variance of hypergeometric distribution

D(X)=np( 1-p)*

(N-n)/(N- 1)

Extended data:

Prove:

Lemma 1: ∑ {c (x, a) * c (d-x, b), x = 0..min {a, d}} = c (d, a+b), investigate (1+x) a * (/) (In addition, we can also get ... n from hypergeometric distribution 1 = ∑ p (x = k), k = 0, 1, 2).

Lemma 2: k * c (k, n) = n * c (k- 1, n- 1), easy to obtain.

Formal proof:

EX=∑{k*C(k，M)*C(n-k，N-M)/C(n，N)，k=0..min{M，n}}

= 1/C(n，N)*∑{M*C(k- 1，M- 1)*C(n-k，N-M)，k= 1..min{M，n}}

//(extract the common factor and deform with Lemma 2 at the same time, pay attention to the change of the value of k)

= m/c (n, n) * ∑ {c (k- 1, m- 1) * c (n-k, n-m), k = 1..min {m, n}} (extract and sort out lemmas.

= m * c (n- 1, n- 1)/c (n, n) (using lemma 1)

=Mn/N (simplified)

Baidu Encyclopedia-Hypergeometric Distribution