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Probability Generating Functions (PGFs) Simplified Revision Notes

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20.1.1 Probability Generating Functions (PGFs)

Introduction

A Probability Generating Function (PGF) provides a compact way to represent the probabilities of a discrete random variable. It is especially useful in studying the properties of random variables, such as the mean and variance, and in analysing distributions like the binomial, Poisson, geometric, and negative binomial distributions.

This note covers:

  1. Definitions and derivations of PGFs.
  2. Key properties of PGFs.
  3. Applications to common distributions.

Definition of a Probability Generating Function

The Probability Generating Function (PGF) of a discrete random variable XX with probabilities P(X=x)=pxP(X = x) = p_x is defined as:

GX(t)=E[tX]=x=0pxtxG_X(t) = \mathbb{E}[t^X] = \sum_{x=0}^\infty p_x t^x

where:

  • GX(t)G_X(t) is the PGF of XX
  • px=P(X=x)p_x = P(X = x) is the probability mass function (PMF) of XX
  • tt is a dummy variable (not necessarily a probability).

Properties of PGFs

Normalisation:

GX(1)=x=0px=1G_X(1) = \sum_{x=0}^\infty p_x = 1

Finding Probabilities:

The probability P(X=x)P(X = x) can be obtained by differentiating GX(t)G_X(t) and evaluating at t=0t = 0

P(X=x)=1x!dxdtxGX(t)t=0P(X = x) = \frac{1}{x!} \frac{d^x}{dt^x} G_X(t) \bigg|_{t=0}

PGFs for Common Distributions

Binomial Distribution

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

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

Poisson Distribution

For XPo(λ)X \sim \text{Po}(\lambda)

GX(t)=eλ(t1)G_X(t) = e^{\lambda(t-1)}

Geometric Distribution

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

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

Negative Binomial Distribution

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

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

Worked Example

infoNote

Example: Finding Probabilities with a Poisson PGF


Problem

If XPo(3)X \sim \text{Po}(3), use the PGF to find P(X=2)P(X = 2)


Part 1: Write the PGF

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

GX(t)=eλ(t1)G_X(t) = e^{\lambda(t-1)}

Substitute λ=3\lambda = 3

GX(t)=e3(t1)G_X(t) = e^{3(t-1)}

Part 2: Find P(X=2)P(X = 2)

To find P(X=2)P(X = 2), use the formula:

P(X=x)=1x!dxdtxGX(t)t=0P(X = x) = \frac{1}{x!} \frac{d^x}{dt^x} G_X(t) \bigg|_{t=0}

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

GX(t)=3e3(t1),GX(t)=9e3(t1)G_X'(t) = 3e^{3(t-1)}, \quad G_X''(t) = 9e^{3(t-1)}

Step 2: Evaluate GX(t)G_X''(t) at t=0t = 0

GX(0)=9e3(01)=9e3G_X''(0) = 9e^{3(0-1)} = 9e^{-3}

Step 3: Find P(X=2)P(X = 2)

P(X=2)=GX(0)2!=9e32P(X = 2) = \frac{G_X''(0)}{2!} = \frac{9e^{-3}}{2}

Using e30.0498e^{-3} \approx 0.0498

P(X=2)=9×0.04982=0.224P(X = 2) = \frac{9 \times 0.0498}{2} = 0.224

Final Answer:

P(X=2):highlight[0.224]P(X = 2) \approx :highlight[0.224]

Note Summary

infoNote

Common Mistakes

  1. Forgetting to differentiate correctly: Ensure derivatives are calculated carefully when finding probabilities.
  2. Misinterpreting tt: tt is a dummy variable, not the probability or random variable.
  3. Incorrect substitution of PGFs: Ensure you use the correct PGF for the given distribution.
infoNote

Key Formulas

  1. Definition of PGF:
GX(t)=x=0P(X=x)txG_X(t) = \sum_{x=0}^\infty P(X = x) t^x
  1. Common PGFs:
  • Binomial: GX(t)=[pt+(1p)]nG_X(t) = [pt + (1-p)]^n
  • Poisson: GX(t)=eλ(t1)G_X(t) = e^{\lambda(t-1)}
  • Geometric: GX(t)=p1(1p)tG_X(t) = \frac{p}{1-(1-p)t}
  • Negative Binomial: GX(t)=(p1(1p)t)rG_X(t) = \left(\frac{p}{1-(1-p)t}\right)^r
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