20.1.2 PGFs of Standard Distributions

Introduction

A Probability Generating Function (PGF) represents a discrete random variable by encoding its probabilities into a function. For a given distribution, the PGF allows us to find key properties, such as the mean and variance.

This note covers:

Definitions and derivations of PGFs for standard distributions.
Applications of PGFs to calculate the mean and variance.
Worked examples with step-by-step solutions.

PGFs for Standard Distributions

Binomial Distribution

For $X \sim B(n, p)$

G_X(t) = [pt + (1-p)]^n

Mean:

\mathbb{E}[X] = np.

Variance:

\text{Var}(X) = np(1-p)

Poisson Distribution

For $X \sim \text{Po}(\lambda)$

G_X(t) = e^{\lambda(t-1)}

Mean:

\mathbb{E}[X] = \lambda

Variance:

\text{Var}(X) = \lambda

Geometric Distribution

For $X \sim \text{Geom}(p)$ (number of ailures before the first success):

G_X(t) = \frac{p}{1 - (1-p)t}, \quad |t| < \frac{1}{1-p}

Mean:

\mathbb{E}[X] = \frac{1-p}{p}

Variance:

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

Negative Binomial Distribution

For $X \sim \text{NegBin}(r, p)$ (number of failures before rr successes):

G_X(t) = \left(\frac{p}{1 - (1-p)t}\right)^r, \quad |t| < \frac{1}{1-p}

Mean:

\mathbb{E}[X] = \frac{r(1-p)}{p}

Variance:

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

Worked Examples

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Example 1: Mean and Variance for a Binomial Distribution

Problem

For $X \sim B(5, 0.4)$ , use the PGF to find the mean and variance of $X$ .

Part 1: Write the PGF

For $X \sim B(n, p)$ , the PGF is:

G_X(t) = [pt + (1-p)]^n.

Substitute $n = 5$ and $p = 0.4$

G_X(t) = [0.4t + 0.6]^5

Part 2: Find the Mean

The mean is given by $\mathbb{E}[X] = G_X'(1)$

Step 1: Differentiate $G_X(t)$

G_X'(t) = 5[0.4t + 0.6]^4 \times 0.4

Step 2: Evaluate at $t = 1$

G_X'(1) = 5[0.4(1) + 0.6]^4 \times 0.4

= 5[1]^4 \times 0.4 = 5 \times 0.4 = 2

Thus, $\mathbb{E}[X] = :highlight[2]$

Part 3: Find the Variance

The variance is:

\text{Var}(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2

Step 1: Differentiate $G_X'(t)$ to get $G_X''(t)$

G_X''(t) = 5 \times 4[0.4t + 0.6]^3 \times 0.4^2

= 20[0.4t + 0.6]^3 \times 0.16

Step 2: Evaluate $G_X''(1)$

G_X''(1) = 20[0.4(1) + 0.6]^3 \times 0.16

= 20[1]^3 \times 0.16 = 20 \times 0.16 = 3.2

Step 3: Use the variance formula:

\text{Var}(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2

= 3.2 + 2 - 2^2 = 3.2 + 2 - 4 = 1.2

Final Answer:

Mean: $\mathbb{E}[X] = :highlight[2]$
Variance: $\text{Var}(X) = :highlight[1.2]$

infoNote

Example 2: Mean and Variance for a Poisson Distribution

Problem

For $X \sim \text{Po}(3)$ , use the PGF to find the mean and variance of $X$

Part 1: Write the PGF

For $X \sim \text{Po}(\lambda)$ , the PGF is:

G_X(t) = e^{\lambda(t-1)}.

Substitute $\lambda = 3:$

G_X(t) = e^{3(t-1)}

Part 2: Find the Mean

The mean is $\mathbb{E}[X] = G_X'(1)$

Step 1: Differentiate $G_X(t)$

G_X'(t) = 3e^{3(t-1)}

Step 2: Evaluate at $t=1$

G_X'(1) = 3e^{3(1-1)} = 3e^0 = 3

Thus, $\mathbb{E}[X] = :highlight[3]$

Part 3: Find the Variance

The variance is:

\text{Var}(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2

Step 1: Differentiate $G_X'(t)$ to get $G_X''(t)$

G_X''(t) = 9e^{3(t-1)}

Step 2: Evaluate $G_X''(1)$

G_X''(1) = 9e^{3(1-1)} = 9e^0 = 9

Step 3: Use the variance formula:

\text{Var}(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2

= 9 + 3 - 3^2 = 9 + 3 - 9 = 3

Final Answer:

Mean: $\mathbb{E}[X] = :highlight[3]$
Variance: $\text{Var}(X) = :highlight[3]$

Note Summary

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Common Mistakes

Forgetting differentiation rules: Ensure accurate differentiation of PGFs to find $G_X'(t)$ and $G_X''(t)$
Misinterpreting PGF formulas: Use the correct PGF for the given distribution.
Neglecting conditions for convergence: Ensure that $|t| < \frac{1}{1-p}$ for geometric and negative binomial PGFs.

infoNote

Key Formulas

Definition of PGF:

G_X(t) = \sum_{x=0}^\infty P(X = x) t^x

Common PGFs:

Binomial: $G_X(t) = [pt + (1-p)]^n$
Poisson: $G_X(t) = e^{\lambda(t-1)}$
Geometric: $G_X(t) = \frac{p}{1-(1-p)t}$
Negative Binomial: $G_X(t) = \left(\frac{p}{1-(1-p)t}\right)^r$

PGFs of Standard Distributions Simplified Revision Notes for A-Level Edexcel Further Maths Further Statistics 1

20.1.2 PGFs of Standard Distributions

Introduction

PGFs for Standard Distributions

Binomial Distribution

Poisson Distribution

Geometric Distribution

Negative Binomial Distribution

Worked Examples

Example 1: Mean and Variance for a Binomial Distribution

Example 2: Mean and Variance for a Poisson Distribution

Note Summary

Common Mistakes

Key Formulas

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