PGFs of Standard Distributions (Edexcel A-Level Further Mathematics): Revision Notes

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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:

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

PGFs for Standard Distributions

Binomial Distribution

For X∼B(n,p)X \sim B(n, p)

GX(t)=[pt+(1−p)]nG_X(t) = [pt + (1-p)]^n

Mean:

E[X]=np.\mathbb{E}[X] = np.

Variance:

Var(X)=np(1−p)\text{Var}(X) = np(1-p)

Poisson Distribution

For X∼Po(λ)X \sim \text{Po}(\lambda)

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

Mean:

E[X]=λ\mathbb{E}[X] = \lambda

Variance:

Var(X)=λ\text{Var}(X) = \lambda

Geometric Distribution

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

GX(t)=p1−(1−p)t,∣t∣<11−pG_X(t) = \frac{p}{1 - (1-p)t}, \quad |t| < \frac{1}{1-p}

Mean:

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

Variance:

Var(X)=1−pp2\text{Var}(X) = \frac{1-p}{p^2}

Negative Binomial Distribution

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

GX(t)=(p1−(1−p)t)r,∣t∣<11−pG_X(t) = \left(\frac{p}{1 - (1-p)t}\right)^r, \quad |t| < \frac{1}{1-p}

Mean:

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

Variance:

Var(X)=r(1−p)p2\text{Var}(X) = \frac{r(1-p)}{p^2}

Worked Examples

infoNote

Example 1: Mean and Variance for a Binomial Distribution

Problem

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

Part 1: Write the PGF

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

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

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

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

Part 2: Find the Mean

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

Step 1: Differentiate GX(t)G_X(t)

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

Step 2: Evaluate at t=1t = 1

GX′(1)=5[0.4(1)+0.6]4×0.4G_X'(1) = 5[0.4(1) + 0.6]^4 \times 0.4 =5[1]4×0.4=5×0.4=2= 5[1]^4 \times 0.4 = 5 \times 0.4 = 2

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

Part 3: Find the Variance

The variance is:

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

Step 1: Differentiate GX′(t)G_X'(t) to get GX′′(t)G_X''(t)

GX′′(t)=5×4[0.4t+0.6]3×0.42G_X''(t) = 5 \times 4[0.4t + 0.6]^3 \times 0.4^2 =20[0.4t+0.6]3×0.16= 20[0.4t + 0.6]^3 \times 0.16

Step 2: Evaluate GX′′(1)G_X''(1)

GX′′(1)=20[0.4(1)+0.6]3×0.16G_X''(1) = 20[0.4(1) + 0.6]^3 \times 0.16 =20[1]3×0.16=20×0.16=3.2= 20[1]^3 \times 0.16 = 20 \times 0.16 = 3.2

Step 3: Use the variance formula:

Var(X)=GX′′(1)+GX′(1)−[GX′(1)]2\text{Var}(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2 =3.2+2−22=3.2+2−4=1.2= 3.2 + 2 - 2^2 = 3.2 + 2 - 4 = 1.2

Final Answer:

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

Example 2: Mean and Variance for a Poisson Distribution

Problem

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

Part 1: Write the PGF

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

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

Substitute λ=3:\lambda = 3:

GX(t)=e3(t−1)G_X(t) = e^{3(t-1)}

Part 2: Find the Mean

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

Step 1: Differentiate GX(t)G_X(t)

GX′(t)=3e3(t−1)G_X'(t) = 3e^{3(t-1)}

Step 2: Evaluate at t=1t=1

GX′(1)=3e3(1−1)=3e0=3G_X'(1) = 3e^{3(1-1)} = 3e^0 = 3

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

Part 3: Find the Variance

The variance is:

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

Step 1: Differentiate GX′(t)G_X'(t) to get GX′′(t)G_X''(t)

GX′′(t)=9e3(t−1)G_X''(t) = 9e^{3(t-1)}

Step 2: Evaluate GX′′(1)G_X''(1)

GX′′(1)=9e3(1−1)=9e0=9G_X''(1) = 9e^{3(1-1)} = 9e^0 = 9

Step 3: Use the variance formula:

Var(X)=GX′′(1)+GX′(1)−[GX′(1)]2\text{Var}(X) = G_X''(1) + G_X'(1) - [G_X'(1)]^2 =9+3−32=9+3−9=3= 9 + 3 - 3^2 = 9 + 3 - 9 = 3

Final Answer:

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

Note Summary

infoNote

Common Mistakes

  1. Forgetting differentiation rules: Ensure accurate differentiation of PGFs to find GX′(t)G_X'(t) and GX′′(t)G_X''(t)
  2. Misinterpreting PGF formulas: Use the correct PGF for the given distribution.
  3. Neglecting conditions for convergence: Ensure that ∣t∣<11−p|t| < \frac{1}{1-p} for geometric and negative binomial PGFs.
infoNote

Key Formulas

  1. Definition of PGF:
GX(t)=∑x=0∞P(X=x)txG_X(t) = \sum_{x=0}^\infty P(X = x) t^x
  1. Common PGFs:
  • Binomial: GX(t)=[pt+(1−p)]nG_X(t) = [pt + (1-p)]^n
  • Poisson: GX(t)=eλ(t−1)G_X(t) = e^{\lambda(t-1)}
  • Geometric: GX(t)=p1−(1−p)tG_X(t) = \frac{p}{1-(1-p)t}
  • Negative Binomial: GX(t)=(p1−(1−p)t)rG_X(t) = \left(\frac{p}{1-(1-p)t}\right)^r

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