SD05 — Discrete Distributions
Course: Statistics and Probability for Data Science (DS14270).
Source deck: Samer El Kababji, SD05 — Discrete Distributions.
This lesson develops Bernoulli trials and their binomial sum. Each formula is followed by executable base R, including the deck’s coin and credit-card examples. Simulations are labeled as extensions.
A Bernoulli random variable records failure or success:
\[ X\in\{0,1\},\qquad P(X=1)=p,\qquad P(X=0)=1-p. \]
Its moments are
\[ E(X)=p,\qquad \operatorname{Var}(X)=p(1-p). \]
p <- 0.20
x <- 0:1
probability <- c(1 - p, p)
mean_X <- sum(x * probability)
variance_X <- sum((x - mean_X)^2 * probability)
data.frame(x, probability) x probability
1 0 0.8
2 1 0.2c(mean = mean_X, variance = variance_X) mean variance
0.20 0.16 For example, if a visitor buys a desktop computer with probability \(0.20\), then the purchase indicator has mean \(0.20\) and variance \(0.16\).
Suppose there are \(n\) trials, each trial has two outcomes, the trials are independent, and the success probability \(p\) is constant. The total number of successes
\[ X=X_1+\cdots+X_n \]
has a binomial distribution, written \(X\sim\operatorname{Bin}(n,p)\), with PMF
\[ P(X=k)=\binom nkp^k(1-p)^{n-k},\qquad k=0,1,\ldots,n. \]
n <- 5
p <- 0.5
k <- 0:n
pmf <- choose(n, k) * p^k * (1 - p)^(n - k)
data.frame(k, formula = pmf, R = dbinom(k, size = n, prob = p)) k formula R
1 0 0.03125 0.03125
2 1 0.15625 0.15625
3 2 0.31250 0.31250
4 3 0.31250 0.31250
5 4 0.15625 0.15625
6 5 0.03125 0.03125sum(pmf)[1] 1For five fair coin tosses, the probability of exactly three heads is
\[ P(X=3)=\binom53(0.5)^3(0.5)^2=0.3125. \]
choose(5, 3) * 0.5^3 * 0.5^2[1] 0.3125dbinom(3, size = 5, prob = 0.5)[1] 0.3125The binomial CDF adds the masses up to \(k\):
\[ F(k)=P(X\leq k)=\sum_{j=0}^{k}\binom njp^j(1-p)^{n-j}. \]
# At most 3 heads in 5 fair tosses
pbinom(3, size = 5, prob = 0.5)[1] 0.8125# Smallest k for which P(X <= k) is at least 0.90
qbinom(0.90, size = 5, prob = 0.5)[1] 4Because a binomial variable is a sum of independent Bernoulli variables,
\[ E(X)=np,\qquad \operatorname{Var}(X)=np(1-p),\qquad \operatorname{SD}(X)=\sqrt{np(1-p)}. \]
In the deck’s credit-card example, 10 customers independently pay by credit card with probability \(0.75\). Thus
\[ E(X)=7.5,\qquad \operatorname{Var}(X)=1.875, \qquad \operatorname{SD}(X)\approx1.37. \]
The integer counts within one standard deviation of the mean are 7 and 8.
n <- 10
p <- 0.75
mu <- n * p
variance <- n * p * (1 - p)
sigma <- sqrt(variance)
within_one_sd <- (0:n)[abs((0:n) - mu) <= sigma]
c(mean = mu, variance = variance, sd = sigma) mean variance sd
7.500000 1.875000 1.369306 within_one_sd[1] 7 8sum(dbinom(within_one_sd, size = n, prob = p))[1] 0.5318499set.seed(14270)
simulated_counts <- rbinom(100000, size = 10, prob = 0.75)
c(
simulated_mean = mean(simulated_counts),
theoretical_mean = 10 * 0.75,
simulated_variance = var(simulated_counts),
theoretical_variance = 10 * 0.75 * 0.25
) simulated_mean theoretical_mean simulated_variance
7.508100 7.500000 1.879993
theoretical_variance
1.875000 dbinom(4, 12, 0.30) gives approximately \(0.2311\).pbinom(4, 12, 0.30) gives approximately \(0.7237\), and \(P(X\geq1)=1-P(X=0)\approx0.9862\).