Statistics 1 for Economics: Discrete Probability Distributions Study Notes

Discrete Probability Distributions Overview

• Discrete probability distributions are used to model variables with countable outcomes.
• The primary distributions covered in this course include:
  • Hypergeometric Distribution: Used for sampling without replacement.
  • Bernoulli Distribution: Used for a single trial with two outcomes.
  • Binomial Distribution: Used for sampling with replacement (or multiple independent Bernoulli trials).
  • Poisson Distribution: Used to model the number of events occurring within a specific interval of time or space.

Hypergeometric Distribution

• Conceptual Definition: This distribution is used for sampling without replacement from a finite population. The probability of success changes with each draw because the items are not returned to the population.
• Parameters:
  • $N$: Population size.
  • $n$: Sample size.
  • $k$: Total number of successes in the population.
  • $x$: Number of successes in the sample ($X$: number of successes).
• Formula:
  • $P(X = x) = \frac{\binom{k}{x} \binom{N-k}{n-x}}{\binom{N}{n}}$
• Expected Value and Variance:
  • Mean: $E[X] = n \times p = n \times \frac{k}{N}$
  • Variance: $V[X] = n \times p(1 - p) \times \frac{N - n}{N - 1}$, where $p = \frac{k}{N}$. The term $\frac{N - n}{N - 1}$ is known as the finite population correction factor.
• Example: Stock Index Analysis:
  • Population size $N = 500$ stocks.
  • Successes in population $k = 400$ (stocks that increased in value).
  • Sample size $n = 4$ stocks selected randomly.
  • Probability that exactly 3 increased: $P(X = 3) = \frac{\binom{400}{3} \binom{100}{1}}{\binom{500}{4}} = 0.411$
  • Probability that at least 3 increased: $P(X \ge 3) = P(X = 3) + P(X = 4) = \frac{\binom{400}{3} \binom{100}{1}}{\binom{500}{4}} + \frac{\binom{400}{4} \binom{100}{0}}{\binom{500}{4}} = 0.411 + 0.409 = 0.820$
• Example: Marble Sample Analysis:
  • Population: 5 blue and 5 green marbles ($N=10$, $k=5$ for green).
  • Sample size $n = 3$ without replacement.
  • Probability of 0 green marbles: $X \sim Hyp(3, 10, 5)$.
  • $P(X = 0) = \frac{\binom{5}{0} \binom{5}{3}}{\binom{10}{3}} = 0.083$

Bernoulli Distribution

• Conceptual Definition: Represents a single experiment with exactly two possible outcomes: success ($X=1$) and failure ($X=0$).
• Probability Mass Function:
  • $P(X = 1) = p$
  • $P(X = 0) = 1 - p$
• Expected Value and Variance:
  • $E[X] = \sum P(X=x) \cdot x = (1-p) \cdot 0 + p \cdot 1 = p$
  • $V[X] = E[x - \mu]^2 = (1-p) \cdot (0-p)^2 + p \cdot (1-p)^2 = p(1-p)$

Binomial Distribution

• Conceptual Definition: Describes the number of successes in $n$ independent Bernoulli trials, where the probability of success $p$ remains constant (sampling with replacement).
• Parameters:
  • $n$: Sample size (number of trials).
  • $p$: Probability of success in a single trial.
• Formula:
  • $P(X = x) = \binom{n}{x} p^x (1 - p)^{n - x}$
• Expected Value and Variance:
  • Mean: $E[X] = n \times p$
  • Variance: $V[X] = n \times p(1 - p)$
• Derivation from Bernoulli:
  • If $X = X_1 + X_2 + ... + X_n$ (where each $X_i$ is a Bernoulli trial):
  • $E[X] = E[X_1] + ... + E[X_n] = p + p + ... + p = n \times p$
  • $V[X] = V[X_1] + ... + V[X_n] = p(1-p) + ... + p(1-p) = n \times p(1-p)$
• Example: Multiple-Choice Exam:
  • A student guesses on 20 questions ($n=20$) with 4 options each ($p=0.25$).
  • Goal: At least 60% correct ($0.60 \times 20 = 12$ questions).
  • $P(X \ge 12) = 1 - P(X \le 11)$.
  • Using tables: $1 - 0.999 = 0.001$.

Poisson Distribution

• Conceptual Definition: Models the number of occurrences of an event in a fixed interval of time or space when these events occur with a known constant mean rate and independently of the time since the last event.
• Model Assumptions:
  • Independence: The number of successes in one interval is independent of successes in another.
  • Stationarity: The probability distribution is the same for all intervals of the same size.
  • Rarity in small intervals: For extremely small intervals, the probability of two or more successes drops to zero (events cannot happen at the exact same instant).
• Parameters:
  • $\mu$: Mean number of successes per interval.
• Formula:
  • $P(X = x) = \frac{e^{-\mu} \mu^x}{x!}$
• Expected Value and Variance:
  • $E[X] = \mu$
  • $V[X] = \mu$
• Application Examples:
  • Number of claims per year.
  • Number of clients in 20 minutes.
  • Number of accidents per day.
  • Number of flaws per square meter of fabric.
  • Number of WhatsApp messages per hour.
• Example: Defects in Boxes:
  • Average defects per box $\mu = 6$.
  • Probability of at most 3 defects: $P(X \le 3) = e^{-6} \left[ \frac{6^0}{0!} + \frac{6^1}{1!} + \frac{6^2}{2!} + \frac{6^3}{3!} \right] = 0.151$.
  • Probability of exactly 3 defects: $P(X = 3) = P(X \le 3) - P(X \le 2) = 0.151 - 0.062 = 0.089$.

Distribution Approximations

1. Hypergeometric to Binomial

• Condition: Use Binomial to approximate Hypergeometric when the sample size is small relative to the population ($\frac{n}{N} < 0.05$ or 5%). In this case, sampling without replacement behaves similarly to sampling with replacement.
• Calculation: Set $p = \frac{k}{N}$.
• Case Comparison (Clients in Debt):
  • Scenario A: $N=20$, $k=6$, $n=5$. $\frac{n}{N} = 0.25$ (Too large).
    • Hypergeometric: $P(X=1) = 0.387$
    • Binomial: $P(X=1) = 0.360$
    • Difference is large.
  • Scenario B: $N=200$, $k=60$, $n=5$. $\frac{n}{N} = 0.025$ (Under 5%).
    • Hypergeometric: $P(X=1) = 0.363$
    • Binomial: $P(X=1) = 0.360$
    • Difference is negligible.

2. Binomial to Poisson

• Condition: Use Poisson to approximate Binomial when success is sporadic and the number of trials is high. Specifically, when $p < 0.05$ (or 5%).
• Calculation: Set $\mu = n \times p$.
• Example: Sunglasses Quality Control:
  • $n = 50$, $p = 0.02$. Shipment accepted if $X \le 2$.
  • Exact Binomial: $P(X \le 2) = 0.9216$.
  • Poisson Approximation: $\mu = 50 \times 0.02 = 1$. $P(Y \le 2) = 0.9197$.
• Example: Hospital Biopsy Complications:
  • $n = 4000$, $p = 0.001$. Find probability of at least four complications ($X \ge 4$).
  • Condition: $p = 0.001 < 0.05$.
  • Mean: $\mu = 4000 \times 0.001 = 4$.
  • Poisson: $P(X_{poi} \ge 4) = 1 - P(X_{poi} \le 3) = 1 - 0.433 = 0.567$.
  • Exact Binomial comparison: $1 - (0.1954 + 0.1465 + 0.0732 + 0.0183) = 0.5666$.

Student Practice Exercises and Solutions

• Problem 1 (Multiple Choice Guessing):
  • $n=20$, $p=0.25$.
  • $P(X \ge 12) = 1 - P(X \le 11) = 1 - 0.999 = 0.001$.
• Problem 2 (WhatsApp Texts):
  • Mean $\mu = 1$ text/hour.
  • a. 2 messages in 1 hour: $P(X = 2) = \frac{e^{-1} 1^2}{2!} = 0.184$.
  • b. 6 messages in 4 hours: New $\mu = 4$. $P(Y = 6) = P(Y \le 6) - P(Y \le 5) = 0.889 - 0.785 = 0.104$.
  • c. At least 18 messages in 12 hours: New $\mu = 12$. $P(Z \ge 18) = 1 - P(Z \le 17) = 1 - 0.937 = 0.063$.
• Problem 3 (Marbles Sample):
  • Population: 5 blue, 5 green marbles. Sample size 3.
  • Without replacement (Hypergeometric): $P(X=0) = \frac{\binom{5}{0} \binom{5}{3}}{\binom{10}{3}} = 0.083$.
  • With replacement (Binomial): $p = 0.5, n = 3$. $P(Y=0) = \binom{3}{0} (0.5)^0 (0.5)^3 = 0.125$.