基本概率分布Basic Concept of Probability Distributions 5: Hypergemometric Distribution
PMF
Suppose that a sample of size $n$ is to be chosen randomly (without replacement) from an urn containing $N$ balls, of which $m$ are white and $N-m$ are black. If we let $X$ denote the number of white balls selected, then $$f(x; N, m, n) = \Pr(X = x) = {{m\choose x}{N-m\choose n-x}\over {N\choose n}}$$ for $x= 0, 1, 2, \cdots, n$.
Proof:
This is essentially the Vandermonde's identity: $${m+n\choose r} = \sum_{k=0}^{r}{m\choose k}{n\choose r-k}$$ where $m$, $n$, $k$, $r\in \mathbb{N}_0$. Because $$ \begin{align*} \sum_{r=0}^{m+n}{m+n\choose r}x^r &= (1+x)^{m+n} \quad\quad\quad\quad\quad\quad\quad\quad \mbox{(binomial theorem)}\\ &= (1+x)^m(1+x)^n\\ &= \left(\sum_{i=0}^{m}{m\choose i}x^{i}\right)\left(\sum_{j=0}^{n}{n\choose j}x^{j}\right)\\ &= \sum_{r=0}^{m+n}\left(\sum_{k=0}^{r}{m\choose k}{n\choose r-k}\right)x^r \quad\quad\mbox{(product of two binomials)} \end{align*} $$ Using the product of two binomials: $$ \begin{eqnarray*} \left(\sum_{i=0}^{m}a_i x^i\right)\left(\sum_{j=0}^{n}b_j x^j\right) &=& \left(a_0+a_1x+\cdots + a_mx^m\right)\left(b_0+b_1x+\cdots + b_nx^n\right)\\ &=& a_0b_0 + a_0b_1x +a_1b_0x +\cdots +a_0b_2x^2 + a_1b_1x^2 + a_2b_0x^2 +\\ & &\cdots + a_mb_nx^{m+n}\\ &=& \sum_{r=0}^{m+n}\left(\sum_{k=0}^{r}a_{k}b_{r-k}\right)x^{r} \end{eqnarray*} $$ Hence $$ \begin{eqnarray*} & &\sum_{r=0}^{m+n}{m+n\choose r}x^r = \sum_{r=0}^{m+n}\left(\sum_{k=0}^{r}{m\choose k}{n\choose r-k}\right)x^r\\ &\implies& {m+n\choose r} = \sum_{k=0}^{r}{m\choose k}{n\choose r-k}\\ & \implies& \sum_{k=0}^{r}{{m\choose k}{n\choose r-k}\over {m+n\choose r}} = 1 \end{eqnarray*} $$
Mean
The expected value is $$\mu = E[X] = {nm\over N}$$
Proof:
$$ \begin{eqnarray*} E[X^k] &=& \sum_{x=0}^{n}x^kf(x; N, m, n)\\ &=& \sum_{x=0}^{n}x^k{{m\choose x}{N-m\choose n-x}\over {N\choose n}}\\ &=& {nm\over N}\sum_{x=0}^{n} x^{k-1} {{m-1 \choose x-1}{N-m\choose n-x}\over {N-1 \choose n-1}}\\ & & (\mbox{identities:}\ x{m\choose x} = m{m-1\choose x-1},\ n{N\choose n} = N{N-1\choose n-1})\\ &=& {nm\over N}\sum_{x=0}^{n} (y+1)^{k-1} {{m-1 \choose y}{(N-1) - (m - 1)\choose (n-1)-y}\over {N-1 \choose n-1}}\quad\quad(\mbox{setting}\ y=x-1)\\ &=& {nm\over N}E\left[(Y+1)^{k-1}\right] \quad\quad\quad \quad\quad \quad\quad\quad\quad (\mbox{since}\ Y\sim g(y; m-1, n-1, N-1)) \end{eqnarray*} $$ Hence, setting $k=1$ we have $$E[X] = {nm\over N}$$ Note that this follows the mean of the binomial distribution $\mu = np$, where $p = {m\over N}$.
Variance
The variance is $$\sigma^2 = \mbox{Var}(X) = np(1-p)\left(1 - {n-1 \over N-1}\right)$$ where $p = {m\over N}$.
Proof:
$$ \begin{align*} E[X^2] &= {nm\over N}E[Y+1] \quad\quad\quad \quad\quad\quad \quad (\mbox{setting}\ k=2)\\ &= {nm\over N}\left(E[Y] + 1\right)\\ & = {nm\over N}\left[{(n-1) (m-1) \over N-1}+1\right] \end{align*} $$ Hence the variance is $$ \begin{align*} \mbox{Var}(X) &= E\left[X^2\right] - E[X]^2\\ &= {mn\over N}\left[{(n-1) (m-1) \over N-1}+1 - {nm\over N}\right]\\ &= np \left[ (n-1) \cdot {pN-1\over N-1}+1-np\right] \quad\quad \quad \quad \quad\quad(\mbox{setting}\ p={m\over N})\\ &= np\left[(n-1)\cdot {p(N-1) + p -1 \over N-1} + 1 -np\right]\\ &= np\left[(n-1)p + (n-1)\cdot{p-1 \over N-1} + 1-np\right]\\ &= np\left[1-p - (1-p)\cdot {n-1\over N-1}\right] \\ &= np(1-p)\left(1 - {n-1 \over N-1}\right) \end{align*} $$ Note that it is approximately equal to 1 when $N$ is sufficient large (i.e. ${n-1\over N-1}\rightarrow 0$ when $N\rightarrow +\infty$). And then it is the same as the variance of the binomial distribution $\sigma^2 = np(1-p)$, where $p = {m\over N}$.
Examples
1. At a lotto game, seven balls are drawn randomly from an urn containing 37 balls numbered from 0 to 36. Calculate the probability $P$ of having exactly $k$ balls with an even number for $k=0, 1, \cdots, 7$.
Solution:
$$P(X = k) = {{19\choose k}{18\choose 7-k}\over {37 \choose 7}}$$
p = NA; k = 0:7
for (i in k){
+ p[i+1] = round(choose(19, i) * choose(18, 7-i)
+ / choose(37, 7), 3)
+ }
p
# [1] 0.003 0.034 0.142 0.288 0.307 0.173 0.047 0.005
2. Determine the same probabilities as in the previous problem, this time using the normal approximation.
Solution:
The mean is $$\mu = {nm\over N} = {7\times19\over 37} = 3.594595$$ and the standard deviation is $$\sigma = \sqrt{{nm\over N}\left(1-{m\over N}\right)\left(1 - {n-1\over N-1}\right)} = \sqrt{{7\times19\over 37}\left(1 - {19\over 37}\right) \left(1 - {7-1\over 37-1}\right)} = 1.207174$$ The probability of normal approximation is
p = NA; k = 0:7
mu = 7 * 19 / 37
s = sqrt(7 * 19 / 37 * (1 - 19/37) * (1 - 6/36))
for (i in k){
+ p[i+1] = round(dnorm(i, mu, s), 3)
+ }
p
# [1] 0.004 0.033 0.138 0.293 0.312 0.168 0.045 0.006
Reference
- Ross, S. (2010). A First Course in Probability (8th Edition). Chapter 4. Pearson. ISBN: 978-0-13-603313-4.
- Brink, D. (2010). Essentials of Statistics: Exercises. Chapter 11. ISBN: 978-87-7681-409-0.