Why is "Omitting the $+G'(1)$ term, writing $\text{Var}(X)=G''(1)-[G'(1)]^2$" a common mistake in PGFs and applications to mean, variance & sums of independent variables?

Why it happens: Treating G''(1) as E(X^2) directly, mirroring the Var=E(X^2)-[E(X)]^2 shortcut. How to avoid it: Remember G''(1)=E[X(X-1)], so E(X^2)=G''(1)+G'(1). The full formula is G''(1)+G'(1)-[G'(1)]^2. Source: 9231 Paper 4 Examiner Reports — most common error

Why is "Assuming $G''(1)=E(X^2)$" a common mistake in PGFs and applications to mean, variance & sums of independent variables?

Why it happens: Expecting the second derivative to give the second moment by analogy with the mean. How to avoid it: G''(1) gives the second FACTORIAL moment E[X(X-1)]. Add E(X)=G'(1) to recover E(X^2). Source: 9231 Paper 4 mark schemes

Why is "Using $G_{X+Y}=G_XG_Y$ without stating (or checking) independence" a common mistake in PGFs and applications to mean, variance & sums of independent variables?

Why it happens: The product rule is so convenient that the condition is forgotten. How to avoid it: State 'X and Y are independent' whenever you multiply PGFs — it is a marked condition of the theorem. Source: 9231 Paper 4 Examiner Reports

Why is "Claiming a sum of binomials is binomial when the values of $p$ differ" a common mistake in PGFs and applications to mean, variance & sums of independent variables?

Why it happens: Over-generalising B(m,p)+B(n,p)=B(m+n,p) to unequal p. How to avoid it: The collapse needs the SAME p so the bases (q+pt) match. With different p, the product is not a single binomial PGF. Source: 9231 Paper 4 Examiner Reports

Why is "Quoting a standard mean/variance without differentiating the PGF" a common mistake in PGFs and applications to mean, variance & sums of independent variables?

Why it happens: The results (e.g. Poisson mean =) are well known, so students skip the working. How to avoid it: The skill examined is the PGF method: show G'(t) and G''(t) explicitly. An unsupported standard result earns little. Source: 9231 Paper 4 mark schemes

Why is "Chain-rule errors when differentiating $e^{\lambda(t-1)}$ or $(q+pt)^n$" a common mistake in PGFs and applications to mean, variance & sums of independent variables?

Why it happens: Forgetting the inner derivative ( for the exponential, p for the binomial base). How to avoid it: Differentiate carefully: d/dte^(t-1)= e^(t-1); d/dt(q+pt)^n=np(q+pt)^n-1. Source: 9231 Paper 4 Examiner Reports

Cambridge International A Levels Further Mathematics (9231)

PGFs and applications to mean, variance & sums of independent variables

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Detailed Study Notes

Detailed notes on Probability Generating Functions for Cambridge International A Levels Further Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

PGFs — Mean, Variance and Sums of Independent Variables (Cambridge International AS & A Level Further Mathematics 9231 Paper 4, 2026-2027 syllabus)

Differentiate a probability generating function and read off the mean from $G'(1)$ and the variance from $G''(1) + G'(1) - [G'(1)]^2$ . Then use the headline result — the PGF of a sum of independent variables is the product of their PGFs — to identify the distribution of a sum.

At a glance

Mean: $E(X) = G'(1)$ — differentiate the PGF once and set $t=1$ .
Variance: $\text{Var}(X) = G''(1) + G'(1) - [G'(1)]^2$ .
Derive standard results: the mean and variance of B, Geo and Po all drop out of their PGFs.
Sum rule (independent): $G_{X+Y}(t) = G_X(t)\,G_Y(t)$ — the PGF of a sum is the PRODUCT of the PGFs.
Identify the sum: if the product PGF matches a standard form, the sum has that distribution (e.g. sum of independent Poissons is Poisson).

What you’ll learn

Mapped to the Cambridge International A Level 9231 syllabus (2026-2027).

4.5d — Use a probability generating function to find the mean $G'(1)$ and variance $G''(1)+G'(1)-[G'(1)]^2$ of a distribution (syllabus 4.5).
4.5e — Use the result that the PGF of the sum of independent random variables is the product of their PGFs (syllabus 4.5).
4.5f — Identify the distribution of a sum of independent variables from its probability generating function (syllabus 4.5).

The mean: E(X) = G'(1)

Differentiate the PGF once and substitute t = 1.

Where the result comes from. Differentiate $G_X(t)=\sum_x P(X=x)\,t^x$ term by term with respect to $t$ : $G_X'(t)=\sum_x x\,P(X=x)\,t^{x-1}.$ Now set $t=1$ : $G_X'(1)=\sum_x x\,P(X=x)=E(X).$ That sum is exactly the definition of the expectation. So differentiate once, put $t=1$ , and you have the mean.

The procedure:

Write down (or derive) $G_X(t)$ .
Differentiate to get $G_X'(t)$ .
Substitute $t=1$ to obtain $E(X)=G_X'(1)$ .

Worked logic for the Poisson $G_X(t)=e^{\lambda(t-1)}$ : $G_X'(t)=\lambda\,e^{\lambda(t-1)} \quad\Rightarrow\quad G_X'(1)=\lambda\,e^{0}=\lambda.$ So $E(X)=\lambda$ — the familiar Poisson mean, recovered from the PGF.

One differentiation, then t = 1, gives the mean.

Cambridge tip. Show the differentiation explicitly — write $G'(t)=\ldots$ before substituting $t=1$ . An unsupported " $E(X)=\lambda$ " risks losing the method mark.

$E(X)=G_X'(1)$ : differentiate the PGF once, then set $t=1$ .
Always write $G'(t)$ out before substituting (method mark).
Poisson: $G'(t)=\lambda e^{\lambda(t-1)}\Rightarrow E(X)=\lambda$ .

See the full worked example for pgfs and applications to mean, variance & sums of independent variables →

The variance: G''(1) + G'(1) − [G'(1)]²

Differentiate twice; G''(1) gives E[X(X−1)], then assemble the variance.

The second derivative gives a factorial moment. Differentiate $G_X(t)$ twice: $G_X''(t)=\sum_x x(x-1)\,P(X=x)\,t^{x-2} \quad\Rightarrow\quad G_X''(1)=\sum_x x(x-1)P(X=x)=E[X(X-1)].$ So $G_X''(1)=E[X(X-1)]=E(X^2)-E(X)$ , which rearranges to $E(X^2)=G_X''(1)+G_X'(1)$ .

Assemble the variance using $\text{Var}(X)=E(X^2)-[E(X)]^2$ : $\boxed{\;\text{Var}(X)=G_X''(1)+G_X'(1)-\left[G_X'(1)\right]^2\;}$

The procedure:

Differentiate the PGF twice to get $G_X''(t)$ .
Compute $G_X'(1)$ and $G_X''(1)$ .
Substitute into $\text{Var}(X)=G''(1)+G'(1)-[G'(1)]^2$ .

Worked logic for the Poisson $G_X(t)=e^{\lambda(t-1)}$ :

$G_X'(t)=\lambda e^{\lambda(t-1)}\Rightarrow G_X'(1)=\lambda.$
$G_X''(t)=\lambda^2 e^{\lambda(t-1)}\Rightarrow G_X''(1)=\lambda^2.$
$\text{Var}(X)=\lambda^2+\lambda-\lambda^2=\lambda.$

So $E(X)=\text{Var}(X)=\lambda$ , the well-known Poisson property — derived cleanly from the PGF.

The two derivatives at t = 1 combine into the variance. The +G'(1) term is the one candidates most often forget.

Cambridge tip. Do not forget the $+G'(1)$ term. $G''(1)$ gives $E[X(X-1)]$ , not $E(X^2)$ — you must add $G'(1)$ back to recover $E(X^2)$ .

$G''(1)=E[X(X-1)]=E(X^2)-E(X)$ .
$\text{Var}(X)=G''(1)+G'(1)-[G'(1)]^2$ .
The $+G'(1)$ term is essential — $G''(1)$ alone is NOT $E(X^2)$ .

See the full worked example for pgfs and applications to mean, variance & sums of independent variables →

Sums of independent variables: multiply the PGFs

For independent X and Y, the PGF of X + Y is the product of their PGFs.

The headline theorem. If $X$ and $Y$ are independent, the PGF of their sum is the product of their PGFs: $G_{X+Y}(t)=G_X(t)\,G_Y(t).$

Why it works. $G_{X+Y}(t)=E\!\left(t^{X+Y}\right)=E\!\left(t^X\,t^Y\right)$ . Because $X$ and $Y$ are independent, the expectation of the product factorises: $E\!\left(t^X t^Y\right)=E\!\left(t^X\right)E\!\left(t^Y\right)=G_X(t)\,G_Y(t).$

Independence is essential — the factorisation $E(t^X t^Y)=E(t^X)E(t^Y)$ holds only when $X$ and $Y$ are independent.

This extends to any number of variables. For independent $X_1,X_2,\ldots,X_n$ : $G_{X_1+\cdots+X_n}(t)=\prod_{i=1}^{n}G_{X_i}(t).$ If they are also identically distributed with common PGF $G(t)$ , then $G_{\text{sum}}(t)=[G(t)]^n$ .

Power of the method. Adding distributions directly (a convolution) is messy; multiplying their PGFs is easy. Then you identify the resulting distribution by recognising the product as a standard PGF.

Multiplying the two PGFs gives the PGF of the sum — provided X and Y are independent.

Cambridge tip. Always state " $X$ and $Y$ are independent" when you use this result — it is a stated condition of the theorem and examiners look for it.

Independent $X, Y$ : $G_{X+Y}(t)=G_X(t)\,G_Y(t)$ .
Extends to $n$ variables: product of all PGFs; if identical, $[G(t)]^n$ .
Independence is required — state it when you use the rule.

See the full worked example for pgfs and applications to mean, variance & sums of independent variables →

Identifying the distribution of a sum

Recognise the product PGF as a standard form to name the sum's distribution.

Once you have multiplied the PGFs, compare the product with the standard PGFs to read off the distribution of the sum.

Sum of independent Poissons. Let $X\sim\text{Po}(\lambda)$ and $Y\sim\text{Po}(\mu)$ be independent. Then $G_{X+Y}(t)=e^{\lambda(t-1)}\,e^{\mu(t-1)}=e^{(\lambda+\mu)(t-1)}.$ This is the PGF of a Poisson with parameter $\lambda+\mu$ . Therefore $X+Y\sim\text{Po}(\lambda+\mu)$ — Poissons add, and the parameters add.

Sum of independent binomials with the same $p$ . Let $X\sim\text{B}(m,p)$ and $Y\sim\text{B}(n,p)$ be independent. Then $G_{X+Y}(t)=(q+pt)^m(q+pt)^n=(q+pt)^{m+n}.$ This is $\text{B}(m+n,p)$ . So $X+Y\sim\text{B}(m+n,p)$ — the trials simply combine. (This requires the same $p$ ; if $p$ differs the product is not a single binomial PGF.)

The general technique:

Write each PGF in its standard form.
Multiply them and simplify (combine exponents / indices).
Match the result against a standard PGF.
Name the distribution and state its parameter(s).

Recognising the product PGF as a standard form names the distribution of the sum.

Cambridge tip. A binomial sum only collapses to a single binomial when the success probability $p$ is identical. Different values of $p$ leave a product that is not binomial — say so if the question tries to catch you.

Match the product PGF against a standard form to name the sum's distribution.
$\text{Po}(\lambda)+\text{Po}(\mu)=\text{Po}(\lambda+\mu)$ (independent).
$\text{B}(m,p)+\text{B}(n,p)=\text{B}(m+n,p)$ — only if $p$ is the same.

See the full worked example for pgfs and applications to mean, variance & sums of independent variables →

Quick recap

$E(X)=G'(1)$ — differentiate the PGF once and set $t=1$ .
$\text{Var}(X)=G''(1)+G'(1)-[G'(1)]^2$ — never drop the $+G'(1)$ .
$G''(1)=E[X(X-1)]$ , so $E(X^2)=G''(1)+G'(1)$ .
Independent $X,Y$ : $G_{X+Y}(t)=G_X(t)\,G_Y(t)$ (product rule).
$\text{Po}(\lambda)+\text{Po}(\mu)=\text{Po}(\lambda+\mu)$ ; $\text{B}(m,p)+\text{B}(n,p)=\text{B}(m+n,p)$ (same $p$ ).
Identify a sum's distribution by matching its product PGF to a standard form.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Mean: $E(X)=G'(1)$ .
Variance: $\text{Var}(X)=G''(1)+G'(1)-[G'(1)]^2$ .
$G''(1)=E[X(X-1)]$ (a factorial moment, NOT $E(X^2)$ ).
Sum rule (independent): $G_{X+Y}(t)=G_X(t)\,G_Y(t)$ .
Identical independent variables: $G_{\text{sum}}(t)=[G(t)]^n$ .

How it’s examined

This is the high-value half of Paper 4's PGF question (syllabus 4.5), often 8-12 marks. A typical question gives or asks for a PGF, then requires $E(X)=G'(1)$ and $\text{Var}(X)=G''(1)+G'(1)-[G'(1)]^2$ with full differentiation shown, before adding two independent variables via $G_{X+Y}=G_X G_Y$ and identifying the resulting distribution. The most penalised errors are (i) omitting the $+G'(1)$ term in the variance, (ii) treating $G''(1)$ as $E(X^2)$ directly, and (iii) failing to state the independence condition for the product rule. Show every derivative — unsupported standard results score little, since the skill being tested is the PGF method itself.

Sources: Cambridge International AS & A Level Further Mathematics 9231 syllabus (2026-2027); 9231 Examiner Reports 2022-2024 (Paper 4 — Further Probability & Statistics); 9231/41, 9231/42, 9231/43 Further Probability & Statistics — past papers, mark schemes and MF19 list of formulae. Last reviewed 2026-06-23.

Take this whole topic with you

Step-by-step worked examples — PGFs and applications to mean, variance & sums of independent variables

Step-by-step solutions to past-paper-style questions on pgfs and applications to mean, variance & sums of independent variables, written exactly the way a tutor would explain them at the board.

Question type:

1Mean of a Poisson from its PGF
Getting startedDirect calculation• mean, poisson, easy
Question
The random variable $X\sim\text{Po}(\lambda)$ has PGF $G_X(t)=e^{\lambda(t-1)}$ . Use $E(X)=G_X'(1)$ to find the mean.
Step-by-step solution
1. Step 1
  Differentiate the PGF with respect to $t$ (chain rule: derivative of $\lambda(t-1)$ is $\lambda$ ).
  $G_X'(t) = \lambda\,e^{\lambda(t-1)}$
2. Step 2
  Set $t=1$ , so the exponent $\lambda(1-1)=0$ and $e^0=1$ .
  $G_X'(1) = \lambda\,e^{0} = \lambda$
3. Step 3
  Identify the mean.
  $E(X) = G_X'(1) = \lambda$
Answer
$E(X)=\lambda$ .
Examiner tip
Write $G_X'(t)$ out explicitly before substituting $t=1$ — an unsupported ' $E(X)=\lambda$ ' risks the method mark even though the answer is standard.
2Mean of a geometric distribution
Getting startedDirect calculation• mean, geometric, easy
Question
The random variable $X$ has PGF $G_X(t)=\dfrac{pt}{1-qt}$ with $q=1-p$ . Find $E(X)$ using $E(X)=G_X'(1)$ .
Step-by-step solution
1. Step 1
  Differentiate using the quotient rule on $\dfrac{pt}{1-qt}$ .
  $G_X'(t) = \frac{p(1-qt) - pt(-q)}{(1-qt)^2} = \frac{p(1-qt) + pqt}{(1-qt)^2}$
2. Step 2
  Simplify the numerator: $p(1-qt)+pqt = p$ .
  $G_X'(t) = \frac{p}{(1-qt)^2}$
3. Step 3
  Set $t=1$ . The denominator becomes $(1-q)^2=p^2$ .
  $E(X) = G_X'(1) = \frac{p}{p^2} = \frac{1}{p}$
Answer
$E(X)=\dfrac{1}{p}$ .
Examiner tip
This is the well-known geometric mean $\frac1p$ . Note $1-q=p$ at $t=1$ — keep $q$ symbolic until the final substitution to keep the algebra clean.
3Variance of a binomial from its PGF
Building confidenceMulti-step problem• variance, binomial, medium
Question
The random variable $X\sim\text{B}(n,p)$ has PGF $G_X(t)=(q+pt)^n$ with $q=1-p$ . Show that $E(X)=np$ and $\text{Var}(X)=npq$ .
Step-by-step solution
1. Step 1
  First derivative (chain rule). At $t=1$ , $q+p=1$ .
  $G_X'(t) = np\,(q+pt)^{n-1} \;\Rightarrow\; G_X'(1) = np\cdot 1^{n-1} = np$
2. Step 2
  So the mean is $E(X)=G_X'(1)=np$ .
  $E(X) = np$
3. Step 3
  Second derivative. Differentiate $G_X'(t)$ again.
  $G_X''(t) = n(n-1)p^2(q+pt)^{n-2} \;\Rightarrow\; G_X''(1) = n(n-1)p^2$
4. Step 4
  Assemble the variance $\text{Var}(X)=G''(1)+G'(1)-[G'(1)]^2$ .
  $\text{Var}(X) = n(n-1)p^2 + np - (np)^2$
5. Step 5
  Simplify: $n^2p^2 - np^2 + np - n^2p^2 = np - np^2 = np(1-p) = npq$ .
  $\text{Var}(X) = np(1-p) = npq$
Answer
$E(X)=np$ and $\text{Var}(X)=npq$ .
Examiner tip
Keep the $+G'(1)$ term — dropping it gives $n(n-1)p^2-(np)^2=-np^2$ , which is the classic wrong answer. The $np$ from $+G'(1)$ is exactly what rescues the result.
4Sum of two independent Poissons
Building confidenceMulti-step problem• sum, poisson, medium
Question
Independent random variables $X\sim\text{Po}(3)$ and $Y\sim\text{Po}(5)$ . Use PGFs to find the distribution of $X+Y$ , and state $P(X+Y=0)$ to 3 s.f.
Step-by-step solution
1. Step 1
  Write each PGF in standard form: $G_X(t)=e^{3(t-1)}$ , $G_Y(t)=e^{5(t-1)}$ .
2. Step 2
  Multiply (valid because $X,Y$ are independent): $G_{X+Y}(t)=G_X(t)\,G_Y(t)$ .
  $G_{X+Y}(t) = e^{3(t-1)}\,e^{5(t-1)} = e^{8(t-1)}$
3. Step 3
  Recognise $e^{8(t-1)}$ as the PGF of $\text{Po}(8)$ . So $X+Y\sim\text{Po}(8)$ .
4. Step 4
  Find $P(X+Y=0)$ using the Poisson formula $P(0)=e^{-8}$ .
  $P(X+Y=0) = e^{-8} = 0.000335$
Answer
$X+Y\sim\text{Po}(8)$ , and $P(X+Y=0)=e^{-8}=0.000335$ (3 s.f.).
Examiner tip
State that $X$ and $Y$ are independent when you multiply the PGFs — it is a condition of the theorem. The parameters add: $3+5=8$ .
5Identifying the distribution of a binomial sum
StretchMulti-step problem• sum, binomial, hard, identify
Question
Independent variables $X\sim\text{B}(4,\,0.3)$ and $Y\sim\text{B}(6,\,0.3)$ . Find the PGF of $X+Y$ , identify its distribution, and hence state $E(X+Y)$ and $\text{Var}(X+Y)$ .
Step-by-step solution
1. Step 1
  Write each PGF with $p=0.3$ , $q=0.7$ : $G_X(t)=(0.7+0.3t)^4$ , $G_Y(t)=(0.7+0.3t)^6$ .
2. Step 2
  Multiply (independent), adding the indices since the base is identical.
  $G_{X+Y}(t) = (0.7+0.3t)^4(0.7+0.3t)^6 = (0.7+0.3t)^{10}$
3. Step 3
  Recognise $(q+pt)^{10}$ as the binomial PGF with $n=10$ , $p=0.3$ . So $X+Y\sim\text{B}(10,\,0.3)$ .
4. Step 4
  Use the standard binomial results $E=np$ , $\text{Var}=npq$ .
  $E(X+Y) = 10(0.3) = 3,\quad \text{Var}(X+Y) = 10(0.3)(0.7) = 2.1$
Answer
$X+Y\sim\text{B}(10,\,0.3)$ , with $E(X+Y)=3$ and $\text{Var}(X+Y)=2.1$ .
Examiner tip
The collapse to a single binomial works ONLY because $p=0.3$ is the same for both. If the success probabilities differed, $(q+pt)^4(q'+p't)^6$ would not be a single binomial PGF.
6Mean and variance from a polynomial PGF
StretchMulti-step problem• mean, variance, hard
Question
A random variable $X$ has PGF $G_X(t)=\dfrac{1}{6}\left(1 + 2t + 3t^2\right)$ . Find $E(X)$ and $\text{Var}(X)$ , giving each to 3 s.f.
Step-by-step solution
1. Step 1
  Differentiate once.
  $G_X'(t) = \tfrac{1}{6}\left(2 + 6t\right) \;\Rightarrow\; G_X'(1) = \tfrac{1}{6}(2+6) = \tfrac{8}{6} = \tfrac{4}{3}$
2. Step 2
  So the mean is $E(X)=G_X'(1)=\dfrac{4}{3}$ .
  $E(X) = \tfrac{4}{3} = 1.33$
3. Step 3
  Differentiate again.
  $G_X''(t) = \tfrac{1}{6}(6) = 1 \;\Rightarrow\; G_X''(1) = 1$
4. Step 4
  Assemble the variance $\text{Var}(X)=G''(1)+G'(1)-[G'(1)]^2$ .
  $\text{Var}(X) = 1 + \tfrac{4}{3} - \left(\tfrac{4}{3}\right)^2 = \tfrac{7}{3} - \tfrac{16}{9} = \tfrac{21-16}{9} = \tfrac{5}{9}$
Answer
$E(X)=\dfrac{4}{3}=1.33$ and $\text{Var}(X)=\dfrac{5}{9}=0.556$ (3 s.f.).
Examiner tip
Here $G''(t)$ is a constant. Work the variance as exact fractions before converting — $\frac{7}{3}-\frac{16}{9}=\frac{5}{9}$ — to avoid rounding error.

Model Answers — PGFs and applications to mean, variance & sums of independent variables

High-scoring sample answers for pgfs and applications to mean, variance & sums of independent variables on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
3 marks
The random variable $X$ has a Poisson distribution with mean $\lambda$ and PGF $G_X(t)=e^{\lambda(t-1)}$ . Use the PGF to show that $E(X)=\lambda$ .
Model answer
Differentiate the PGF (chain rule — the derivative of $\lambda(t-1)$ is $\lambda$ ): $G_X'(t) = \lambda\,e^{\lambda(t-1)}$

Substitute $t=1$ , where the exponent $\lambda(1-1)=0$ , so $e^0=1$ : $G_X'(1) = \lambda\,e^{0} = \lambda$

Therefore $E(X)=G_X'(1)=\lambda$ .
Why this scores
M1 for the correct derivative, M1 for substituting $t=1$ , A1 for $E(X)=\lambda$ with the result quoted as $G'(1)$ . The differentiation must be shown — quoting the standard mean earns nothing.
Question 2
3 marks
A random variable $X$ has PGF $G_X(t)=\dfrac{1}{10}\left(2 + 3t + 4t^2 + t^3\right)$ . Find $E(X)$ .
Model answer
Differentiate term by term: $G_X'(t) = \frac{1}{10}\left(3 + 8t + 3t^2\right)$

Substitute $t=1$ : $G_X'(1) = \frac{1}{10}(3 + 8 + 3) = \frac{14}{10} = 1.4$

Therefore $E(X)=G_X'(1)=1.4$ .
Why this scores
M1 for differentiating correctly, M1 for $t=1$ , A1 for $E(X)=1.4$ . The derivative of $\frac{4}{10}t^2$ is $\frac{8}{10}t$ — a slip in the power-rule coefficient is the usual error.
Question 3
5 marks
The random variable $X\sim\text{Po}(\lambda)$ has PGF $G_X(t)=e^{\lambda(t-1)}$ . Use the PGF to show that $\text{Var}(X)=\lambda$ .
Model answer
First derivative: $G_X'(t) = \lambda\,e^{\lambda(t-1)} \quad\Rightarrow\quad G_X'(1) = \lambda$

Second derivative (differentiate $G_X'(t)$ again): $G_X''(t) = \lambda^2\,e^{\lambda(t-1)} \quad\Rightarrow\quad G_X''(1) = \lambda^2$

Assemble the variance using $\text{Var}(X)=G_X''(1)+G_X'(1)-[G_X'(1)]^2$ : $\text{Var}(X) = \lambda^2 + \lambda - \lambda^2 = \lambda$

So $E(X)=\text{Var}(X)=\lambda$ , confirming the standard Poisson property.
Why this scores
M1 A1 for $G''(1)=\lambda^2$ , M1 for quoting the variance formula with the $+G'(1)$ term, A1 for $\lambda$ . Omitting $+G'(1)$ gives $\lambda^2-\lambda^2=0$ — the single most penalised slip in this topic.
Question 4
5 marks
The independent random variables $X$ and $Y$ have Poisson distributions with means $2$ and $4$ respectively. Use probability generating functions to find the distribution of $X+Y$ , and hence find $P(X+Y=1)$ to 3 significant figures.
Model answer
Write the PGFs in standard form: $G_X(t) = e^{2(t-1)}, \qquad G_Y(t) = e^{4(t-1)}$

Multiply — valid because $X$ and $Y$ are independent: $G_{X+Y}(t) = G_X(t)\,G_Y(t) = e^{2(t-1)}\,e^{4(t-1)} = e^{6(t-1)}$

Identify $e^{6(t-1)}$ as the PGF of a Poisson with parameter $6$ : $X+Y \sim \text{Po}(6)$

Hence $P(X+Y=1)=\dfrac{e^{-6}\,6^{1}}{1!}$ : $P(X+Y=1) = 6e^{-6} = 0.0149 \;(3\text{ s.f.})$
Why this scores
M1 for both standard PGFs, M1 for multiplying with independence stated, A1 for $\text{Po}(6)$ , M1 A1 for $6e^{-6}=0.0149$ . State independence explicitly — it is the condition for $G_{X+Y}=G_XG_Y$ .
Question 5
7 marks
The random variable $X$ has PGF $G_X(t)=(0.6+0.4t)^8$ . (a) Name the distribution of $X$ . (b) Use the PGF to find $E(X)$ and $\text{Var}(X)$ .
Model answer
(a) Comparing $(0.6+0.4t)^8$ with the binomial PGF $(q+pt)^n$ gives $n=8$ , $p=0.4$ , $q=0.6$ , so $X\sim\text{B}(8,\,0.4)$ .

(b) First derivative (chain rule): $G_X'(t) = 8(0.4)(0.6+0.4t)^7 = 3.2(0.6+0.4t)^7$ At $t=1$ , $0.6+0.4=1$ : $E(X) = G_X'(1) = 3.2(1)^7 = 3.2$

Second derivative: $G_X''(t) = 8\cdot7\cdot(0.4)^2(0.6+0.4t)^6 = 8.96(0.6+0.4t)^6$ $G_X''(1) = 8.96$

Variance: $\text{Var}(X) = G_X''(1)+G_X'(1)-[G_X'(1)]^2 = 8.96 + 3.2 - (3.2)^2 = 12.16 - 10.24 = 1.92$
Why this scores
(a) B1 for $\text{B}(8,0.4)$ . (b) M1 A1 for $E(X)=3.2$ , M1 A1 for $G''(1)=8.96$ , M1 for the variance formula (with $+G'(1)$ ), A1 for $1.92$ . Check: $npq=8(0.4)(0.6)=1.92$ — agrees.
Question 6
6 marks
Independent random variables $X\sim\text{B}(3,p)$ and $Y\sim\text{B}(7,p)$ , where $0<p<1$ and $q=1-p$ . (a) Find the PGF of $X+Y$ and hence name its distribution. (b) Write down $E(X+Y)$ and $\text{Var}(X+Y)$ in terms of $p$ and $q$ .
Model answer
(a) The PGFs are $G_X(t)=(q+pt)^3$ and $G_Y(t)=(q+pt)^7$ . Since $X$ and $Y$ are independent: $G_{X+Y}(t) = G_X(t)\,G_Y(t) = (q+pt)^3(q+pt)^7 = (q+pt)^{10}$

This is the PGF of a binomial with $n=10$ and the same $p$ , so $X+Y \sim \text{B}(10,\,p).$

(b) Using the standard binomial results $E=np$ and $\text{Var}=npq$ with $n=10$ : $E(X+Y) = 10p, \qquad \text{Var}(X+Y) = 10pq.$
Why this scores
(a) M1 for multiplying the PGFs (independence stated), M1 for combining the indices, A1 for $\text{B}(10,p)$ . (b) B1 B1 for the mean and variance. The collapse to one binomial requires the SAME $p$ — emphasise this; different $p$ would not give a single binomial PGF.

Key Formulae — PGFs and applications to mean, variance & sums of independent variables

The formulae you need to memorise for pgfs and applications to mean, variance & sums of independent variables on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Mean from the PGF
$E(X) = G_X'(1)$
$G_X'$
first derivative of the PGF
When to use
Differentiate the PGF once and set $t=1$ to get the mean. Show the derivative explicitly.
Variance from the PGF
$\text{Var}(X) = G_X''(1) + G_X'(1) - \left[G_X'(1)\right]^2$
$G_X''$
second derivative of the PGF
$G_X'$
first derivative of the PGF
When to use
After computing $G'(1)$ and $G''(1)$ . Never omit the $+G'(1)$ term.
Second factorial moment
$G_X''(1) = E[X(X-1)] = E(X^2) - E(X)$
$E[X(X-1)]$
the second factorial moment
When to use
Bridges $G''(1)$ to $E(X^2)$ : rearranges to $E(X^2)=G''(1)+G'(1)$ .
PGF of a sum (independent)
$G_{X+Y}(t) = G_X(t)\,G_Y(t)$
$X, Y$
independent random variables
When to use
The PGF of a sum of INDEPENDENT variables is the product of their PGFs. State the independence.
Sum of n identical independent variables
$G_{X_1+\cdots+X_n}(t) = \left[G(t)\right]^n$
$G(t)$
common PGF of each $X_i$
$n$
number of variables summed
When to use
When $n$ independent variables share the same distribution — raise the common PGF to the power $n$ .
Sums of standard distributions
$\text{Po}(\lambda)+\text{Po}(\mu)=\text{Po}(\lambda+\mu), \qquad \text{B}(m,p)+\text{B}(n,p)=\text{B}(m+n,p)$
$\lambda, \mu$
Poisson means (add)
$m, n$
binomial trial counts (add, same $p$ )
When to use
Read off the distribution of a sum after multiplying PGFs. Binomial collapse needs identical $p$ .

Key Definitions and Keywords — PGFs and applications to mean, variance & sums of independent variables

Definitions to memorise and the exact keywords mark schemes credit for pgfs and applications to mean, variance & sums of independent variables answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Mean $E(X)=G'(1)$
Examiner keyword
Differentiating $G_X(t)=\sum_x P(X=x)t^x$ gives $G_X'(1)=\sum_x xP(X=x)=E(X)$ . So the first derivative at $t=1$ is the mean.
Variance $G''(1)+G'(1)-[G'(1)]^2$
Examiner keyword
The variance of $X$ obtained from the PGF, where $G''(1)=E[X(X-1)]$ supplies $E(X^2)=G''(1)+G'(1)$ and then $\text{Var}(X)=E(X^2)-[E(X)]^2$ .
Factorial moment $E[X(X-1)]$
Examiner keyword
The value of the second derivative of the PGF at $t=1$ : $G_X''(1)=\sum_x x(x-1)P(X=x)=E[X(X-1)]$ . It is NOT $E(X^2)$ directly.
Product rule for sums
Examiner keyword
For independent $X$ and $Y$ , $G_{X+Y}(t)=G_X(t)G_Y(t)$ , because $E(t^{X+Y})=E(t^X)E(t^Y)$ factorises under independence.
Independence condition
Examiner keyword
The product rule $G_{X+Y}=G_XG_Y$ holds only when $X$ and $Y$ are independent. The factorisation of the expectation fails otherwise.
Convolution
The direct way to find the distribution of $X+Y$ by summing probabilities over all splits. Multiplying PGFs replaces this messy operation with a simple product.

Common Mistakes and Misconceptions — PGFs and applications to mean, variance & sums of independent variables

The traps other students keep falling into on pgfs and applications to mean, variance & sums of independent variables questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Omitting the $+G'(1)$ term, writing $\text{Var}(X)=G''(1)-[G'(1)]^2$
9231 Paper 4 Examiner Reports — most common error
Why it happens
Treating $G''(1)$ as $E(X^2)$ directly, mirroring the $\text{Var}=E(X^2)-[E(X)]^2$ shortcut.
How to avoid it
Remember $G''(1)=E[X(X-1)]$ , so $E(X^2)=G''(1)+G'(1)$ . The full formula is $G''(1)+G'(1)-[G'(1)]^2$ .
✕Assuming $G''(1)=E(X^2)$
9231 Paper 4 mark schemes
Why it happens
Expecting the second derivative to give the second moment by analogy with the mean.
How to avoid it
$G''(1)$ gives the second FACTORIAL moment $E[X(X-1)]$ . Add $E(X)=G'(1)$ to recover $E(X^2)$ .
✕Using $G_{X+Y}=G_XG_Y$ without stating (or checking) independence
9231 Paper 4 Examiner Reports
Why it happens
The product rule is so convenient that the condition is forgotten.
How to avoid it
State ' $X$ and $Y$ are independent' whenever you multiply PGFs — it is a marked condition of the theorem.
✕Claiming a sum of binomials is binomial when the values of $p$ differ
9231 Paper 4 Examiner Reports
Why it happens
Over-generalising $\text{B}(m,p)+\text{B}(n,p)=\text{B}(m+n,p)$ to unequal $p$ .
How to avoid it
The collapse needs the SAME $p$ so the bases $(q+pt)$ match. With different $p$ , the product is not a single binomial PGF.
✕Quoting a standard mean/variance without differentiating the PGF
9231 Paper 4 mark schemes
Why it happens
The results (e.g. Poisson mean $=\lambda$ ) are well known, so students skip the working.
How to avoid it
The skill examined is the PGF method: show $G'(t)$ and $G''(t)$ explicitly. An unsupported standard result earns little.
✕Chain-rule errors when differentiating $e^{\lambda(t-1)}$ or $(q+pt)^n$
9231 Paper 4 Examiner Reports
Why it happens
Forgetting the inner derivative ( $\lambda$ for the exponential, $p$ for the binomial base).
How to avoid it
Differentiate carefully: $\frac{d}{dt}e^{\lambda(t-1)}=\lambda e^{\lambda(t-1)}$ ; $\frac{d}{dt}(q+pt)^n=np(q+pt)^{n-1}$ .

PGFs and applications to mean, variance & sums of independent variables

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Mean of a Poisson from its PGF

2Mean of a geometric distribution

3Variance of a binomial from its PGF

4Sum of two independent Poissons

5Identifying the distribution of a binomial sum

6Mean and variance from a polynomial PGF

Mean from the PGF

Variance from the PGF

Second factorial moment

PGF of a sum (independent)

Sum of n identical independent variables

Sums of standard distributions

Mean E(X)=G′(1)E(X)=G'(1)E(X)=G′(1)

Variance G′′(1)+G′(1)−[G′(1)]2G''(1)+G'(1)-[G'(1)]^2G′′(1)+G′(1)−[G′(1)]2

Factorial moment E[X(X−1)]E[X(X-1)]E[X(X−1)]

Product rule for sums

Independence condition

Convolution

✕Omitting the +G′(1)+G'(1)+G′(1) term, writing Var(X)=G′′(1)−[G′(1)]2\text{Var}(X)=G''(1)-[G'(1)]^2Var(X)=G′′(1)−[G′(1)]2

✕Assuming G′′(1)=E(X2)G''(1)=E(X^2)G′′(1)=E(X2)

✕Using GX+Y=GXGYG_{X+Y}=G_XG_YGX+Y​=GX​GY​ without stating (or checking) independence

✕Claiming a sum of binomials is binomial when the values of ppp differ

✕Quoting a standard mean/variance without differentiating the PGF

✕Chain-rule errors when differentiating eλ(t−1)e^{\lambda(t-1)}eλ(t−1) or (q+pt)n(q+pt)^n(q+pt)n

Mean $E(X)=G'(1)$

Variance $G''(1)+G'(1)-[G'(1)]^2$

Factorial moment $E[X(X-1)]$

✕Omitting the $+G'(1)$ term, writing $\text{Var}(X)=G''(1)-[G'(1)]^2$

✕Assuming $G''(1)=E(X^2)$

✕Using $G_{X+Y}=G_XG_Y$ without stating (or checking) independence

✕Claiming a sum of binomials is binomial when the values of $p$ differ

✕Chain-rule errors when differentiating $e^{\lambda(t-1)}$ or $(q+pt)^n$