Geometric Distribution (Discrete)

Last modified: April 21, 2021

This article is written in: 🇺🇸

Geometric Distribution (Discrete)

A discrete random variable X follows a geometric distribution if it represents the number of trials needed to get the first success in a sequence of Bernoulli trials. The geometric distribution is denoted as $X \sim Geometric (p)$ , where p is the probability of success on each trial.

Probability Mass Function (PMF)

The PMF of a geometric distribution is given by:

$P (X = k) = (1 - p)^{k - 1} p$

where $k \in 1, 2, 3, \dots$ , representing the number of trials until the first success.

output(24)

Cumulative Distribution Function (CDF)

The CDF of a geometric distribution is given by:

$F (k) = P (X \leq k) = 1 - (1 - p)^{k}$

output(23)

Expected Value and Variance

The expected value (mean) of a geometric distribution is:

$E [X] = \frac{1}{p}$

The variance of a geometric distribution is:

$Var (X) = \frac{1 - p}{p^{2}}$

Moment Generating Function (MGF)

The moment generating function (MGF) of a geometric distribution is:

$M_{X} (t) = \frac{p e^{t}}{1 - (1 - p) e^{t}}$

Example: Flipping a Coin

Consider flipping a fair coin (probability of heads p = 0.5) until the first head appears.

I. What is the probability that the first head appears on the 3rd flip?

For the first head on the 3rd flip:

$P (X = 3) = (1 - 0.5)^{3 - 1} \times 0.5 = 0.125$

II. What is the expected number of flips to get the first head?

Expected number of flips:

$E [X] = \frac{1}{0.5} = 2$

III. What is the variance in the number of flips?

Variance in the number of flips:

$Var (X) = \frac{1 - 0.5}{{0.5}^{2}} = 2$

Applications

Geometric distributions are useful in scenarios where you are waiting for the first success in a series of independent trials, such as in quality control testing, network packet delivery analysis, and reliability testing in engineering.

Table of Contents