Binomial and geometric distributions Knowledge-Based Study

Study notes

1. Bernoulli Trials and the Binomial Setting

   A Bernoulli trial is a random experiment with exactly two mutually exclusive outcomes, conventionally labeled success (with probability $p$) and failure (with probability $q = 1 - p$).

   A discrete random variable $X$ follows a Binomial distribution, written as $X \sim \text{Bin}(n, p)$, when it counts the total number of successes across a fixed number of independent and identical Bernoulli trials.

   Key Conditions for a Binomial Experiment

   • Fixed number of trials: The experiment consists of a predetermined number of trials, $n$.
   • Two possible outcomes: Each trial results in either success or failure.
   • Constant probability: The probability of success, $p$, remains the same for every trial.
   • Independence: The outcome of any trial does not influence any other trial.

   Probability Mass Function of the Binomial Distribution

   If $X \sim \text{Bin}(n, p)$, the probability of observing exactly $k$ successes in $n$ trials is given by the probability mass function (PMF):

   $$P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}, \quad \text{for } k = 0, 1, 2, \dots, n$$

   where $\binom{n}{k} = \frac{n!}{k!(n - k)!}$ denotes the binomial coefficient representing the number of ways to arrange $k$ successes among $n$ trials.

   Expectation and Variance

   • Expected Value (Mean): $E[X] = \mu = np$
   • Variance: $\text{Var}(X) = \sigma^2 = np(1 - p)$
   • Standard Deviation: $\sigma = \sqrt{np(1 - p)}$

   Example

   A fair six-sided die is rolled 5 times. Let $X$ be the number of times a 4 appears. Here, $n = 5$ and $p = \frac{1}{6}$.

   The probability of getting a 4 exactly twice is:

   $$P(X = 2) = \binom{5}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^3 = 10 \cdot \frac{1}{36} \cdot \frac{125}{216} = \frac{625}{3888} \approx 0.1608$$

   Waiting for the First Success

   Unlike the Binomial distribution where the number of trials is fixed, the Geometric distribution models the number of independent Bernoulli trials needed until the first success occurs.

   Let $Y$ be the trial number on which the first success occurs. We write $Y \sim \text{Geo}(p)$, where $p$ is the constant probability of success on each trial.

   Probability Mass Function

   For the first success to occur on trial $k$, the preceding $k - 1$ trials must all be failures, followed by a success on trial $k$:

   $$P(Y = k) = (1 - p)^{k - 1} p, \quad \text{for } k = 1, 2, 3, \dots$$

   Cumulative Distribution Function

   The probability that the first success requires more than $k$ trials is simply the probability of getting $k$ consecutive failures:

   $$P(Y > k) = (1 - p)^k$$

   Consequently, the cumulative probability of achieving the first success on or before trial $k$ is:

   $$P(Y \le k) = 1 - P(Y > k) = 1 - (1 - p)^k$$

   Summary of the Geometric Distribution

   • Expected Value (Mean): $E[Y] = \frac{1}{p}$
   • Variance: $\text{Var}(Y) = \frac{1 - p}{p^2}$
   • Memoryless Property: For positive integers $s$ and $t$, $$P(Y > s + t \mid Y > s) = P(Y > t)$$ Past failures do not alter the likelihood of future success.

   Comparing Binomial and Geometric Models

   • What is fixed? In a Binomial distribution, the number of trials $n$ is fixed. In a Geometric distribution, the number of successes (namely 1) is fixed.
   • What is counted? Binomial counts the number of successes $X \in \{0, 1, \dots, n\}$. Geometric counts the trial number of the first success $Y \in \{1, 2, 3, \dots\}$.
   • Support: Binomial random variables are bounded (finite support), whereas Geometric random variables can take arbitrarily large positive integer values (countably infinite support).