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Probability Generating Functions

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Edexcel A Level Further Maths · Further Statistics 1

Probability Generating Functions

🎯 The Big Idea: A PGF is just a clever way of packing an entire probability distribution into a single algebraic expression — where the power of the dummy variable t tells you the value of X, and the coefficient tells you its probability. Once your distribution is "encoded" this way, you can find means, variances, and even combine random variables using ordinary algebra instead of long sums.

Summary — What This Chapter Covers

Constructing a PGF — turning a probability distribution table into a polynomial in t.
G(1) = 1 — a built-in check that always works, because all probabilities sum to 1.
Finding probabilities from a PGF — expanding (using the Binomial Theorem or Maclaurin series) and reading off coefficients.
E(X) = G′(1) — differentiate once, substitute t = 1.
Var(X) = G″(1) + G′(1) − [G′(1)]² — differentiate twice, substitute t = 1.
PGFs of standard distributions — Binomial, Poisson, Geometric, Negative Binomial — and how to derive each from first principles.
PGF of X + Y — multiply the two PGFs together (only valid when X and Y are independent).
PGF of aX + b — replace t with tᵃ, then multiply the whole thing by t^b.

1. Constructing PGFs

1What actually is a PGF?

Imagine you've got a discrete random variable X with this distribution:

x	0	1	4	5
P(X = x)	0.4	0.3	0.2	0.1

A PGF just glues each value of x onto the power of a "dummy" variable t, and each probability becomes the coefficient sitting in front of it:

Definition G_X(t) = 0.4t⁰ + 0.3t¹ + 0.2t⁴ + 0.1t⁵ = ¹⁄₁₀(4 + 3t + 2t⁴ + t⁵) In words: line up every possible value of X as an exponent on t, and every matching probability as the coefficient in front of it.

Why bother? Because once your distribution is written as a single algebraic expression, you can differentiate it, multiply two of them together, or substitute values into it — and all of those algebraic operations turn out to correspond to meaningful probability operations (finding the mean, combining independent variables, and so on). It's the same trick as using logarithms to turn multiplication into addition — you're changing the "space" you work in to make the maths easier.

⚠️ Don't confuse X and t X is your actual random variable — the thing with real-world meaning (a dice score, a number of successes, etc). t is just a "carrier" variable with no meaning of its own — it only exists to build the polynomial structure. Never try to interpret t itself.

2The golden check: G(1) = 1

If you substitute t = 1 into any PGF, every power of t becomes 1^anything = 1, so you're just left with the sum of all the probabilities — which must equal 1.

Rule G_X(1) = 1 — always, for every valid PGF, no exceptions.

This is genuinely useful, not just a curiosity. If a question gives you a PGF containing an unknown constant (like p or k), you can find it instantly by setting t = 1 and solving.

Worked Example

Q: G_X(t) = k(2t + 3t² + 5t³). Find k.

Since G(1) = 1: k(2 + 3 + 5) = 1

10k = 1 → k = 0.1

3The formal definition: G(t) = E(tˣ)

You've probably met E(X) = Σx·P(X = x) before — the expectation formula. A PGF is exactly the same idea, but instead of weighting each value of x by itself, you weight t^x by its probability:

Formal definition (Formulae Booklet) G_X(t) = E(t^X) = Σ tˣ P(X = x) This is where all the later results (E(X), Var(X), sums of PGFs) come from — everything is derived by manipulating this one equation.

4Going backwards: finding probabilities from a PGF

Sometimes you're handed a PGF and asked to find a specific probability. The strategy is always: fully expand the PGF into a polynomial, then read off the coefficient of the power of t you care about.

The tricky part is that PGFs are rarely handed to you already expanded — you usually need one of two expansion tools:

PGF looks like...	Use this tool
(1 + at)ⁿ for rational n (including negative/fractional)	General Binomial Theorem
A function like ln(1 − at) that isn't a polynomial	Maclaurin Series

Worked Example — Binomial Theorem route

Q: G_X(t) = t² / (5 − 4t)². Find P(X = 4).

Step 1 — rewrite in the form (1 + ...)ⁿ:
G_X(t) = t²(5 − 4t)⁻² = t² × 5⁻²(1 − ⁴⁄₅t)⁻² = ¹⁄₂₅ t²(1 − ⁴⁄₅t)⁻²

Step 2 — expand using the General Binomial Theorem:
G_X(t) = ¹⁄₂₅ t² [ 1 + (−2)(−⁴⁄₅t) + ^(−2)(−3)⁄_2!(−⁴⁄₅t)² + ... ]

Step 3 — simplify each term:
= ¹⁄₂₅ t² + ⁸⁄₁₂₅ t³ + ⁴⁸⁄₆₂₅ t⁴ + ...

Step 4 — read off the coefficient of t⁴:
P(X = 4) = ⁴⁸⁄₆₂₅

Worked Example — Maclaurin Series route

Q: G_X(t) = −¹⁄_{ln 2} ln(1 − 0.5t). Expand it.

Recall the Maclaurin series: ln(1 + x) = x − ^x²⁄₂ + ^x³⁄₃ − ...

Substitute x = −0.5t:
G_X(t) = −¹⁄_{ln 2} [ (−0.5t) − ^(−0.5t)²⁄₂ + ^(−0.5t)³⁄₃ − ... ]

Then simplify term by term to get a genuine polynomial you can read coefficients from.

💡 Exam Tip Always double-check your expansion by confirming coefficients are non-negative and (if you can) that they roughly sum to something sensible near t = 1. Negative "probabilities" are a big red flag that you've made an algebra slip.

Practice Question 1

A discrete random variable X has probability distribution:

x	2	3	7	9
P(X = x)	0.25	0.15	k	0.35

Find k, then write down G_X(t).

Practice Question 2

G_X(t) = ¹⁄₄(1 + t)². Find P(X = 1).

2. E(X) and Var(X) from PGFs

1Finding E(X)

Here's the beautiful part of PGFs: once you have G(t), finding the mean is just one differentiation and one substitution — no sums, no tables.

Formula (Formulae Booklet) E(X) = G′_X(1) In words: differentiate the PGF with respect to t, then plug in t = 1.

Why does this work? Differentiating Σtˣ P(X=x) term by term brings each exponent x down as a multiplier: Σ x t^x−1 P(X=x). Substituting t = 1 turns every t^x−1 into 1, leaving exactly Σ x P(X=x) — which is the definition of E(X).

Worked Example

Q: G_X(t) = ¹⁄₈₁ t³(1 + 2t)⁴. Find E(X).

Step 1 — differentiate using the product rule:
G′_X(t) = ¹⁄₈₁(3t²)(1+2t)⁴ + ¹⁄₈₁ t³ × 4(1+2t)³ × 2

Step 2 — substitute t = 1 (no need to simplify the algebra first — just plug in):
G′_X(1) = ¹⁄₈₁ × 3 × (3)⁴ + ¹⁄₈₁ × 1 × 4 × (3)³ × 2 = 3 + ⁸⁄₃

E(X) = ¹⁷⁄₃

🔧 Technique reminder You'll almost always need the product, quotient, or chain rule here — the PGFs in this topic are rarely simple polynomials you can differentiate term-by-term. Brush up on all three before your exam.

2Finding Var(X)

Variance needs a second derivative too, because we need E(X²) as an ingredient — and getting E(X²) directly from a PGF requires differentiating twice.

Formula (Formulae Booklet) Var(X) = G″_X(1) + G′_X(1) − [G′_X(1)]² The first two terms together equal E(X²); the last term is [E(X)]². This is just the familiar Var(X) = E(X²) − [E(X)]² in disguise.

Where does the "+ G′(1)" come from? Differentiating a second time gives you Σ x(x−1) t^x−2 P(X=x). At t=1 this becomes E(X(X−1)) = E(X²) − E(X). Rearranging: E(X²) = G″(1) + E(X) = G″(1) + G′(1). That's exactly why the formula has three terms instead of the usual two.

Worked Example

Q: G_X(t) = 0.2 + 0.4t + 0.3t⁴ + 0.1t⁵. Find Var(X).

Step 1 — first derivative: G′_X(t) = 0.4 + 1.2t³ + 0.5t⁴

Step 2 — second derivative: G″_X(t) = 3.6t² + 2t³

Step 3 — substitute t = 1 into both:
G′_X(1) = 0.4 + 1.2 + 0.5 = 2.1
G″_X(1) = 3.6 + 2 = 5.6

Step 4 — plug into the variance formula:
Var(X) = 5.6 + 2.1 − 2.1² = 5.6 + 2.1 − 4.41

Var(X) = 3.29

❌ Common mistake Students frequently forget the "+ G′_X(1)" term and just compute Var(X) = G″(1) − [G′(1)]² — this is wrong! You must add G′(1) back in before subtracting the square. Write the full three-term formula out every single time rather than trying to remember a shortcut.

Practice Question 3

G_X(t) = ¹⁄₆(1 + 2t + 3t²). Find E(X) and Var(X).

3. PGFs of Standard Distributions

These four are given to you in the Formulae Booklet, but if a question says "derive" or "from first principles", you must prove them yourself — quoting the result gets you zero marks.

Distribution	P(X = x)	PGF
Binomial B(n, p)	ⁿC_x pˣ(1−p)ⁿ⁻ˣ	(1 − p + pt)ⁿ
Poisson Po(λ)	e⁻λ λˣ/x!	e^λ(t−1)
Geometric Geo(p) on 1,2,...	p(1−p)ˣ⁻¹	pt / [1 − (1−p)t]
Negative Binomial on r, r+1,...	ˣ⁻¹C_r−1 pʳ(1−p)ˣ⁻ʳ	[pt / (1−(1−p)t)]ʳ

1Binomial — derivation logic

The trick every derivation shares: write the PGF as a polynomial using the probability formula, then spot a known series hiding inside it. For Binomial, that known series is the Binomial Expansion itself.

1. Build the table of P(X=x) for x = 0 to n using ⁿC_x pˣq^n−x

2. Write G(t) = Σ ⁿC_x pˣq^n−x tˣ

3. Group p and t together as (pt)ˣ, leaving: G(t) = Σ ⁿC_x (pt)ˣ q^n−x

4. Recognise this is exactly the expansion of (q + pt)ⁿ

G_X(t) = (q + pt)ⁿ, where q = 1 − p

2Poisson — derivation logic

Here the known series you're hunting for is the Maclaurin expansion of eˣ.

1. G(t) = Σ e⁻λ ^λˣ⁄_x! tˣ = e⁻λ Σ ^(λt)ˣ⁄_x!

2. Spot that Σ ^(λt)ˣ⁄_x! is exactly the Maclaurin series for e^λt

3. So G(t) = e⁻λ · e^λt = e^{λt − λ}

G_X(t) = e^λ(t−1)

3Geometric — derivation logic

This one hides an infinite geometric series (S_∞ = a/(1−r)).

1. G(t) = pt + pqt² + pq²t³ + ... — first term pt, common ratio qt

2. Apply S_∞ = a/(1−r) with a = pt, r = qt

G_X(t) = pt / (1 − qt), where q = 1 − p

4Negative Binomial — derivation logic

This proof is normally done in sigma notation, and you'll be given the summation identity Σ (x−1 choose r−1) aˣ⁻ʳ = (1−a)⁻ʳ to use — you don't need to know it from memory.

1. Write t^x as t^x−rt^r to separate out the part that depends on the summation index

2. Factorise out pʳtʳ (it doesn't depend on x)

3. Apply the given identity with a = qt

G_X(t) = (pt / (1−qt))ʳ

Worked Example — Using a standard PGF

Q: Write down the PGF for X ~ Geo(p) and use it to prove that Var(X) = (1−p)/p².

Let q = 1 − p. G_X(t) = pt / (1 − qt)

Differentiate using the quotient rule:
G′_X(t) = [(1−qt)p − pt(−q)] / (1−qt)² = p / (1−qt)² = p(1−qt)⁻²

Differentiate again using the chain rule:
G″_X(t) = −2p(1−qt)⁻³(−q) = 2pq / (1−qt)³

Substitute t = 1:
G′_X(1) = p/(1−q)² = p/p² = 1/p
G″_X(1) = 2pq/(1−q)³ = 2pq/p³ = 2q/p²

Substitute into Var(X) = G″(1) + G′(1) − [G′(1)]²:
Var(X) = 2q/p² + 1/p − 1/p² = (2q + p − 1)/p² = (2(1−p) + p − 1)/p² = (2 − 2p + p − 1)/p²

Var(X) = (1 − p)/p² ✓

Practice Question 4

Use G_X(t) = (q + pt)ⁿ to show that E(X) = np for X ~ B(n,p).

4. PGFs of Sums & Transformations

1PGF of X + Y (independent variables)

This is arguably the single most powerful trick in the whole chapter: if X and Y are independent, the PGF of their sum is simply the product of their individual PGFs.

Formula (Formulae Booklet) If Z = X + Y, then G_Z(t) = G_X(t) × G_Y(t) Only valid when X and Y are independent. Multiply the whole expressions together like ordinary algebra.

This extends naturally: if Z = X₁ + X₂ + ... + Xₙ where each Xᵢ is an independent copy of the same distribution X (e.g. "playing the same game 5 times"), then:

G_Z(t) = [G_X(t)]ⁿ

Worked Example — Poisson + Poisson = Poisson

Q: Prove that the sum of two independent Poisson distributions is itself Poisson.

Let X ~ Po(λ), Y ~ Po(μ), independent. G_X(t) = e^λ(t−1), G_Y(t) = e^μ(t−1)

Z = X + Y has PGF: G_Z(t) = e^λ(t−1) × e^μ(t−1)

Add the powers: G_Z(t) = e^{(λ+μ)(t−1)}

This is exactly the PGF of Po(λ + μ). So Z ~ Po(λ + μ). ✓

✅ Why this matters This is a genuinely elegant proof technique — instead of convolving two probability distributions by hand (which is painful), you just multiply two algebraic expressions and pattern-match the result. This is the whole point of PGFs existing.

Practice Question 5

A fair 3-sided spinner labelled 2, 4, 8 is spun. A biased coin labelled 4, 6 is flipped (25% chance of landing on 4). Use PGFs to find P(sum of scores = 8).

2PGF of aX + b (linear transformation)

This one is not in the Formulae Booklet — you need to memorise it. If Y = aX + b is a linear transformation of X:

Formula — must be memorised G_Y(t) = t^b G_X(t^a) In words: everywhere you see t inside G_X, replace it with tᵃ, then multiply the entire resulting expression by t^b.

Why does it work? G_Y(t) = E(t^Y) = E(t^aX+b) = E(t^aXt^b) = t^bE((t^a)^X) = t^bG_X(t^a). It's just algebra on the exponent laws.

Worked Example

Q: G_X(t) = t / (5 − 4t). Find G_Y(t) where Y = 2(X + 5).

Step 1 — write Y in the form aX + b:
Y = 2X + 10 → a = 2, b = 10

Step 2 — apply the rule G_Y(t) = t^bG_X(t^a):
G_Y(t) = t¹⁰ G_X(t²)

Step 3 — replace every t in G_X(t) with t²:
G_Y(t) = t¹⁰ × t² / (5 − 4t²)

Step 4 — simplify:
G_Y(t) = t¹² / (5 − 4t²)

💡 Exam Tip Always rewrite Y in the exact form "aX + b" first before applying the rule — if the question gives you something like Y = 2(X + 5), expand it to Y = 2X + 10 so you can clearly identify a and b. Mixing this up is the #1 source of errors on this topic.

Practice Question 6

G_X(t) = (0.3 + 0.7t)⁵. Let Y = 3X − 2. Find G_Y(t).

What to Memorise

Definition

G_X(t) = E(tˣ) = Σ tˣ P(X=x)

Check

G_X(1) = 1, always

Mean

E(X) = G′_X(1)

Variance

Var(X) = G″_X(1) + G′_X(1) − [G′_X(1)]²

Binomial B(n,p)

G(t) = (q + pt)ⁿ, q = 1−p

Poisson Po(λ)

G(t) = e^λ(t−1)

Geometric Geo(p)

G(t) = pt / (1 − qt)

Negative Binomial

G(t) = (pt / (1−qt))ʳ

Sum (independent)

G_X+Y(t) = G_X(t) × G_Y(t)

Linear transform (NOT in booklet!)

G_aX+b(t) = t^b G_X(t^a)

Concepts Checklist

I can build a PGF directly from a probability distribution table. I understand why G(1) = 1 always, and can use it to find unknown constants. I can expand a non-polynomial PGF using the General Binomial Theorem. I can expand a PGF using a Maclaurin series when it involves ln, e, etc. I can read off P(X = x) as the coefficient of tˣ after expanding. I can find E(X) = G′(1), using the product/quotient/chain rule as needed. I can find Var(X) using the full three-term formula (not just two terms!). I know the PGF formula for Binomial, Poisson, Geometric, and Negative Binomial. I can derive each standard PGF from first principles, spotting the hidden series (binomial expansion / eˣ Maclaurin series / geometric S∞). I can find the PGF of X + Y for independent X and Y by multiplying. I can prove that the sum of two independent Poissons is Poisson. I have memorised G_aX+b(t) = t^bG_X(t^a) — since it's NOT given in the booklet. I always rewrite Y in the exact form aX + b before applying transformation rules.

Exam Tips & Common Mistakes

Forgetting the "+ G′(1)" in the variance formula

The single most common error on this whole topic. Write out Var(X) = G″(1) + G′(1) − [G′(1)]² in full every time — don't try to shortcut it from memory.

Not memorising G_aX+b(t) = t^bG_X(t^a)

This formula is genuinely not in the Formulae Booklet, unlike almost everything else in this chapter. If you don't know it cold, you cannot answer transformation questions.

Use G(1) = 1 as a free sanity check

Whenever you derive or are given a PGF, quickly substitute t = 1 to confirm you get exactly 1. This catches algebra slips before they cost you marks on the rest of the question.

"Derive" or "from first principles" means prove it, not quote it

If a question uses either of these words, you must show the full derivation (building the table, writing the sum, spotting the hidden series) — simply writing down the known PGF result earns no marks.

G(X+Y) = G(X)·G(Y) only works for independent variables

Never multiply two PGFs together to find the PGF of a sum unless the question confirms (or you can assume) independence — this is a fundamental requirement of the formula.

Negative or coefficients that don't make sense = an algebra error

Since coefficients in a PGF are always probabilities, they can never be negative and can never exceed 1. If your expansion produces one, go back and check your working.

Rewrite Y = a(X + c) style expressions before transforming

Always expand brackets first to get Y into the clean form aX + b so you can clearly read off a and b before applying the transformation rule.

Also in the full note

4. PGFs of Sums & Transformations
Exam Tips & Common Mistakes

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Probability Generating Functions

Probability Generating Functions

Summary — What This Chapter Covers

1. Constructing PGFs

1What actually is a PGF?

2The golden check: G(1) = 1

3The formal definition: G(t) = E(tˣ)

4Going backwards: finding probabilities from a PGF

2. E(X) and Var(X) from PGFs

1Finding E(X)

2Finding Var(X)

3. PGFs of Standard Distributions

1Binomial — derivation logic

2Poisson — derivation logic

3Geometric — derivation logic

4Negative Binomial — derivation logic

4. PGFs of Sums & Transformations

1PGF of X + Y (independent variables)

2PGF of aX + b (linear transformation)

What to Memorise

Concepts Checklist

Exam Tips & Common Mistakes

Forgetting the "+ G′(1)" in the variance formula

Not memorising GaX+b(t) = tbGX(ta)

Use G(1) = 1 as a free sanity check

"Derive" or "from first principles" means prove it, not quote it

G(X+Y) = G(X)·G(Y) only works for independent variables

Negative or coefficients that don't make sense = an algebra error

Rewrite Y = a(X + c) style expressions before transforming

Not memorising G_aX+b(t) = t^bG_X(t^a)