SD05 — Discrete Distributions
Course: Statistics and Probability for Data Science (DS14270).
Source deck: Samer El Kababji, SD05 — Discrete Distributions.
The hypergeometric distribution counts successes in sampling without replacement. The mathematics below is paired with base R and the deck’s tagged-animal and light-bulb examples.
Let a population contain \(N\) objects, of which \(M\) are successes. Draw \(n\) objects without replacement and let \(X\) be the number of successes drawn. Then
\[ P(X=x)=\frac{\binom Mx\binom{N-M}{n-x}}{\binom Nn}, \]
where
\[ \max(0,n-N+M)\leq x\leq\min(n,M). \]
N <- 25
M <- 5
n <- 10
x <- 0:min(n, M)
formula_pmf <- choose(M, x) * choose(N - M, n - x) / choose(N, n)
R_pmf <- dhyper(x, m = M, n = N - M, k = n)
data.frame(x, formula_pmf, R_pmf) x formula_pmf R_pmf
1 0 0.056521739 0.056521739
2 1 0.256916996 0.256916996
3 2 0.385375494 0.385375494
4 3 0.237154150 0.237154150
5 4 0.059288538 0.059288538
6 5 0.004743083 0.004743083sum(R_pmf)[1] 1In R, m is the number of successes, n is the number of failures, and k is the sample size. Those argument names differ from the mathematical symbols.
Among \(N=25\) animals, \(M=5\) are tagged. If \(n=10\) are recaptured, then
\[ P(X=2)=\frac{\binom52\binom{20}8}{\binom{25}{10}}\approx0.385. \]
p_exactly_2 <- dhyper(2, m = 5, n = 20, k = 10)
p_at_most_2 <- phyper(2, m = 5, n = 20, k = 10)
components <- dhyper(0:2, m = 5, n = 20, k = 10)
c(P_X_equals_2 = p_exactly_2, P_X_at_most_2 = p_at_most_2) P_X_equals_2 P_X_at_most_2
0.3853755 0.6988142 data.frame(x = 0:2, probability = components) x probability
1 0 0.05652174
2 1 0.25691700
3 2 0.38537549The terms for \(X=0,1,2\) are approximately \(0.057\), \(0.257\), and \(0.385\); their sum is approximately \(0.699\).
For \(X\sim\operatorname{Hypergeometric}(N,M,n)\),
\[ E(X)=n\frac MN, \]
and
\[ \operatorname{Var}(X)= \frac{N-n}{N-1}\,n\frac MN\left(1-\frac MN\right). \]
The factor \((N-n)/(N-1)\) is the finite-population correction. It reduces variance because draws without replacement are dependent.
finite_population_correction <- (N - n) / (N - 1)
mu <- n * M / N
variance <- finite_population_correction * n * (M / N) * (1 - M / N)
c(mean = mu, variance = variance, sd = sqrt(variance)) mean variance sd
2 1 1 When the population is large compared with the sample, sampling without replacement changes the success probability only slightly. A binomial model with \(p=M/N\) can then approximate the hypergeometric model.
The deck considers \(N=10000\) bulbs, \(M=200\) defective bulbs, and a sample of \(n=5\) bulbs. For exactly two defective bulbs,
N <- 10000
M <- 200
n <- 5
p <- M / N
c(
hypergeometric = dhyper(2, m = M, n = N - M, k = n),
binomial = dbinom(2, size = n, prob = p)
)hypergeometric binomial
0.003748545 0.003764768 The two values are about \(0.00377\) because the sampling fraction is only \(5/10000\).
set.seed(14270)
simulated_X <- rhyper(100000, m = 5, n = 20, k = 10)
c(
simulated_P_X_at_most_2 = mean(simulated_X <= 2),
exact_P_X_at_most_2 = phyper(2, 5, 20, 10),
simulated_mean = mean(simulated_X),
theoretical_mean = 10 * 5 / 25
)simulated_P_X_at_most_2 exact_P_X_at_most_2 simulated_mean
0.7012400 0.6988142 1.9940500
theoretical_mean
2.0000000 dhyper(2, 8, 32, 6) gives approximately \(0.2623\).phyper(1, 8, 32, 6) gives approximately \(0.6558\).