Session 3 — Discrete Random Variable

Decision Making Statistics — S04

This session introduces discrete random variables and the main distributions used in inferential statistics.

1 Introduction

After performing a random experiment, we are often interested in a function of the outcome rather than in the outcome itself. For example, when we roll two fair dice with two different colors, we may focus only on the sum of the two numbers obtained. Such quantities are called random variables.

Definition

A random variable is a real-valued function that associates a number with each outcome of a random experiment.

In business applications, a random variable can represent:

• the number of defective products in a sample,
• the number of customers who answer a survey,
• the number of calls received in one minute.

2 Discrete random variable

2.1 Probability distribution

Definition

A random variable \(X\) is discrete if it takes a countable number of values \(x_1, x_2, \dots, x_k\).

Its probability distribution associates each value \(x_i\) with a probability

\[ p_i = P(X = x_i) \]

with the properties:

\[ p_i \geq 0 \quad \text{for all } i, \qquad \sum_i p_i = 1 \]

2.2 Expected value, variance, and standard deviation

Key formulas

The main numerical characteristics of a discrete random variable are:

\[ E(X) = \mu = \sum_i x_i p_i \]

\[ \operatorname{Var}(X) = \sigma^2 = \sum_i (x_i - \mu)^2 p_i = E(X^2) - [E(X)]^2 \]

\[ \sigma = \sqrt{\operatorname{Var}(X)} \]

2.3 Bernoulli distribution

Definition

A Bernoulli distribution models one trial with only two outcomes: success or failure.

If \(X \sim B(p)\):

• \(P(X=1) = p\)
• \(P(X=0) = 1-p\)
• \(E(X) = p\)
• \(\operatorname{Var}(X) = p(1-p)\)

2.4 Binomial distribution

Definition

A binomial distribution models the number of successes in \(n\) independent Bernoulli trials with the same success probability \(p\).

If \(X \sim B(n,p)\), then for \(k = 0,1,\dots,n\):

Binomial formula

\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]

\[ E(X) = np, \qquad \operatorname{Var}(X) = np(1-p) \]

2.5 Poisson distribution

Definition

A Poisson distribution models the number of rare events occurring in a fixed interval of time, space, or volume.

If \(X \sim \mathcal{P}(\lambda)\), then for \(k = 0,1,2,\dots\):

Poisson formula

\[ P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} \]

\[ E(X) = \lambda, \qquad \operatorname{Var}(X) = \lambda \]

2.6 Binomial–Poisson approximation

Approximation rule

When \(n \geq 30\) and \(p \leq 0.1\) (or \(np \leq 5\)), the binomial distribution can be approximated by a Poisson distribution:

\[ B(n,p) \approx \mathcal{P}(np) \]

Important remark

Choose the distribution by reading the situation carefully:

• Bernoulli: one trial,
• Binomial: a fixed number \(n\) of repeated independent trials,
• Poisson: counting events in a time or space interval.

Exam tip

Before calculating, always identify:

1. what the random variable \(X\) counts,
2. the possible values of \(X\),
3. the distribution that matches the wording of the problem.

3 Detailed example

A salesperson contacts 4 prospects. Each prospect independently accepts a meeting with probability \(0.3\).

Let \(X\) be the number of accepted meetings. Then:

\[ X \sim B(4,0.3) \]

3.1 Probability of exactly 2 accepted meetings

Using the binomial formula:

\[ P(X=2) = \binom{4}{2}(0.3)^2(0.7)^2 \]

\[ P(X=2) = 6 \times 0.09 \times 0.49 = 0.2646 \]

So the probability of getting exactly 2 accepted meetings is 0.2646.

3.2 Mean and variance

\[ E(X) = np = 4 \times 0.3 = 1.2 \]

\[ \operatorname{Var}(X) = np(1-p) = 4 \times 0.3 \times 0.7 = 0.84 \]

The expected number of accepted meetings is 1.2.

4 Application exercise

Exercise

A call center receives on average 3 calls per minute. Assuming a Poisson distribution:

1. What is the probability of receiving exactly 2 calls in a minute?
2. What is the probability of receiving at most 2 calls in a minute?

Solution

Let \(X\) be the number of calls received in one minute. Then:

\[ X \sim \mathcal{P}(3) \]

1. Probability of exactly 2 calls

\[ P(X=2) = \frac{e^{-3} \cdot 3^2}{2!} = \frac{e^{-3} \cdot 9}{2} \approx 0.2240 \]

2. Probability of at most 2 calls

\[ P(X \leq 2) = P(X=0) + P(X=1) + P(X=2) \]

\[ P(X \leq 2) = e^{-3}\left(1 + 3 + \frac{3^2}{2}\right) = e^{-3}(1 + 3 + 4.5) \]

\[ P(X \leq 2) = e^{-3} \cdot 8.5 \approx 0.4232 \]