Discrete Random Variables (Further)

This lesson extends your understanding of discrete random variables beyond what is covered in A-Level Mathematics. You will learn to compute expectations of functions of random variables, derive variance rigorously via the $E(X^2)$E(X2) method, manipulate the algebra of expectation and variance, and combine independent variables. These skills are the load-bearing foundation for the entire Further Statistics module: the Poisson, the continuous distributions, moment generating functions and the chi-squared tests all rest on a fluent command of $E$E and $\text{Var}$Var.

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1. Where this sits in AQA 7367

This topic belongs to the Paper 3 Statistics option (7367/3S). Paper 3 carries the more problem-solving-weighted assessment profile (AO1 40% / AO2 25% / AO3 35%), so although the mechanics of computing $E(X)$E(X) and $\text{Var}(X)$Var(X) are AO1, examiners deliberately wrap them in unfamiliar contexts and multi-step parameter-finding (AO3) and ask you to justify or interpret results (AO2). The A-Level Mathematics prerequisite is the discrete-distribution work in the applied paper (computing $E(X)$E(X) and $\text{Var}(X)$Var(X) from a table); Further Maths adds functions of a random variable, the linear-combination algebra, and sums of independent variables.

Students choose two of Mechanics (3M), Statistics (3S) and Discrete (3D). If you are reading this you are taking Statistics — this lesson is the gateway to the rest of 3S.

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2. Core theory: expectation and variance

A discrete random variable $X$X takes a countable set of values $x_1, x_2, \ldots$x1​,x2​,…, each with probability $P(X = x_i) = p_i$P(X=xi​)=pi​. For the distribution to be valid:

Property	Requirement
Non-negativity	$p_i \geq 0$pi​≥0 for all $i$i
Normalisation	$\displaystyle\sum_i p_i = 1$i∑​pi​=1

We use this running example throughout the lesson:

$x$x	1	2	3	4
$P(X = x)$P(X=x)	0.1	0.3	0.4	0.2

Check: $0.1 + 0.3 + 0.4 + 0.2 = 1$0.1+0.3+0.4+0.2=1. Valid.

Expectation

The expected value (mean) is the probability-weighted average of the values:

$E(X) = \mu = \sum_x x \, P(X = x).$E(X)=μ=∑x​xP(X=x).

It is the long-run mean of $X$X over many repetitions. For the running example,

$E(X) = 1(0.1) + 2(0.3) + 3(0.4) + 4(0.2) = 0.1 + 0.6 + 1.2 + 0.8 = 2.7.$E(X)=1(0.1)+2(0.3)+3(0.4)+4(0.2)=0.1+0.6+1.2+0.8=2.7.

$E(X)$E(X) need not be an attainable value of $X$X. It is a balance point, not a mode or median.

Expectation of a function of $X$X

For any function $g$g,

$E\big(g(X)\big) = \sum_x g(x) \, P(X = x).$E(g(X))=∑x​g(x)P(X=x).

This is the law of the unconscious statistician: you never need the distribution of $g(X)$g(X) — apply $g$g to each value and weight by the original probabilities. The most important case is $g(x) = x^2$g(x)=x2:

$E(X^2) = 1^2(0.1) + 2^2(0.3) + 3^2(0.4) + 4^2(0.2) = 0.1 + 1.2 + 3.6 + 3.2 = 8.1.$E(X2)=12(0.1)+22(0.3)+32(0.4)+42(0.2)=0.1+1.2+3.6+3.2=8.1.

Variance

The variance measures spread. By definition it is the expected squared deviation from the mean,

$\text{Var}(X) = E\big((X-\mu)^2\big) = \sum_x (x-\mu)^2 P(X=x),$Var(X)=E((X−μ)2)=∑x​(x−μ)2P(X=x),

but the computational form is far quicker. Expanding the square and using linearity:

$$\begin{aligned} \text{Var}(X) &= E\big(X^2 - 2\mu X + \mu^2\big) \\ &= E(X^2) - 2\mu E(X) + \mu^2 \\ &= E(X^2) - 2\mu^2 + \mu^2 \\ &= E(X^2) - \mu^2 = E(X^2) - \big(E(X)\big)^2. \end{aligned}$$Var(X)​=E(X2−2μX+μ2)=E(X2)−2μE(X)+μ2=E(X2)−2μ2+μ2=E(X2)−μ2=E(X2)−(E(X))2.​

For the running example,

$\text{Var}(X) = 8.1 - 2.7^2 = 8.1 - 7.29 = 0.81, \qquad \sigma = \sqrt{0.81} = 0.9.$Var(X)=8.1−2.72=8.1−7.29=0.81,σ=0.81​=0.9.

Because variance is an expected square, $\text{Var}(X) \geq 0$Var(X)≥0 always, and the identity forces $E(X^2) \geq \big(E(X)\big)^2$E(X2)≥(E(X))2 — with equality only when $X$X is constant.

The algebra of $E$E and $\text{Var}$Var

Expectation is linear; variance is not (it is quadratic in the scaling constant and blind to shifts):

Operation	Expectation	Variance
Scale by $a$a	$E(aX) = aE(X)$E(aX)=aE(X)	$\text{Var}(aX) = a^2\text{Var}(X)$Var(aX)=a2Var(X)
Shift by $b$b	$E(X+b) = E(X)+b$E(X+b)=E(X)+b	$\text{Var}(X+b) = \text{Var}(X)$Var(X+b)=Var(X)
Linear map	$E(aX+b) = aE(X)+b$E(aX+b)=aE(X)+b	$\text{Var}(aX+b) = a^2\text{Var}(X)$Var(aX+b)=a2Var(X)

A shift of $b$b slides the whole distribution along the axis without changing its shape, so the spread — and hence the variance — is untouched. A scaling by $a$a stretches deviations by $a$a, so squared deviations stretch by $a^2$a2.

We can confirm linearity numerically. Computing $E(3X+2)$E(3X+2) directly:

$E(3X+2) = 5(0.1) + 8(0.3) + 11(0.4) + 14(0.2) = 0.5 + 2.4 + 4.4 + 2.8 = 10.1,$E(3X+2)=5(0.1)+8(0.3)+11(0.4)+14(0.2)=0.5+2.4+4.4+2.8=10.1,

which matches $3E(X) + 2 = 3(2.7) + 2 = 10.1$3E(X)+2=3(2.7)+2=10.1. And $\text{Var}(3X+2) = 3^2(0.81) = 7.29$Var(3X+2)=32(0.81)=7.29.

Sums and combinations of independent variables

A great deal of Further Statistics — and almost every multi-source modelling problem — rests on how $E$E and $\text{Var}$Var behave when variables are combined. Expectation is always additive, whatever the dependence between the variables:

$E(X + Y) = E(X) + E(Y), \qquad E(aX + bY) = aE(X) + bE(Y).$E(X+Y)=E(X)+E(Y),E(aX+bY)=aE(X)+bE(Y).

This holds because expectation is a sum (an integral in the continuous case), and sums distribute over addition regardless of any relationship between $X$X and $Y$Y. Variance is more delicate. For independent $X$X and $Y$Y,

$\text{Var}(X + Y) = \text{Var}(X) + \text{Var}(Y), \qquad \text{Var}(X - Y) = \text{Var}(X) + \text{Var}(Y).$Var(X+Y)=Var(X)+Var(Y),Var(X−Y)=Var(X)+Var(Y).

Note the second identity carefully: even for a difference, the variances add. Intuitively, subtracting an uncertain quantity injects just as much uncertainty as adding it — the spread of $X - Y$X−Y is no smaller than the spread of $X + Y$X+Y. More generally, for independent variables and constants $a, b$a,b,

$\text{Var}(aX + bY) = a^2\text{Var}(X) + b^2\text{Var}(Y),$Var(aX+bY)=a2Var(X)+b2Var(Y),

where each coefficient is squared, exactly as in the single-variable rule $\text{Var}(aX) = a^2\text{Var}(X)$Var(aX)=a2Var(X). A frequent source of confusion deserves emphasis: the doubled variable $2X$2X and the sum $X_1 + X_2$X1​+X2​ of two independent copies of $X$X are not the same. The first scales a single outcome,

$\text{Var}(2X) = 2^2\,\text{Var}(X) = 4\,\text{Var}(X),$Var(2X)=22Var(X)=4Var(X),

while the second adds two independent outcomes,

$\text{Var}(X_1 + X_2) = \text{Var}(X) + \text{Var}(X) = 2\,\text{Var}(X).$Var(X1​+X2​)=Var(X)+Var(X)=2Var(X).

The means agree ($E(2X) = E(X_1 + X_2) = 2E(X)$E(2X)=E(X1​+X2​)=2E(X)), but the variances differ by a factor of two — averaging independent measurements reduces relative spread, which is precisely why repeating an experiment and taking a mean improves precision.

Worked illustration. Suppose $X$X (the running example) and an independent $Y$Y have $E(X) = 2.7,\ \text{Var}(X) = 0.81$E(X)=2.7, Var(X)=0.81 and $E(Y) = 4,\ \text{Var}(Y) = 2$E(Y)=4, Var(Y)=2. Then $E(X + Y) = 6.7$E(X+Y)=6.7 and, by independence, $\text{Var}(X + Y) = 0.81 + 2 = 2.81$Var(X+Y)=0.81+2=2.81. For $T = 2X - 3Y$T=2X−3Y: $E(T) = 2(2.7) - 3(4) = -6.6$E(T)=2(2.7)−3(4)=−6.6 and $\text{Var}(T) = 2^2(0.81) + 3^2(2) = 3.24 + 18 = 21.24$Var(T)=22(0.81)+32(2)=3.24+18=21.24 — note the minus sign in $T$T has no effect on the variance, which uses $(-3)^2 = 9$(−3)2=9.

These combination rules reappear throughout 3S: the additivity of the Poisson (Lesson 2), the variance of a sample mean, and the chained approximations of Lesson 3 are all consequences of them.

When variables are not independent: covariance

The independence assumption is doing real work in the variance rules, and it is worth seeing what happens without it. For any two variables,

$\text{Var}(X + Y) = \text{Var}(X) + \text{Var}(Y) + 2\,\text{Cov}(X, Y), \qquad \text{Cov}(X, Y) = E(XY) - E(X)E(Y).$Var(X+Y)=Var(X)+Var(Y)+2Cov(X,Y),Cov(X,Y)=E(XY)−E(X)E(Y).

The covariance measures how the two variables move together: it is positive when high $X$X tends to accompany high $Y$Y, negative when high $X$X accompanies low $Y$Y, and zero when there is no linear association. Independence forces $E(XY) = E(X)E(Y)$E(XY)=E(X)E(Y), hence $\text{Cov}(X, Y) = 0$Cov(X,Y)=0, which is exactly why the cross-term vanishes and the variances simply add for independent variables.

A concrete computation makes this tangible. Suppose $X$X and $Y$Y each take the values $0$0 and $1$1, with the joint distribution

	$Y = 0$Y=0	$Y = 1$Y=1
$X = 0$X=0	$0.1$0.1	$0.3$0.3
$X = 1$X=1	$0.4$0.4	$0.2$0.2

The marginal for $X$X is $P(X=0) = 0.4,\ P(X=1) = 0.6$P(X=0)=0.4, P(X=1)=0.6, so $E(X) = 0.6$E(X)=0.6; the marginal for $Y$Y is $P(Y=0) = 0.5,\ P(Y=1) = 0.5$P(Y=0)=0.5, P(Y=1)=0.5, so $E(Y) = 0.5$E(Y)=0.5. The only term contributing to $E(XY)$E(XY) is the cell where both equal 1: $E(XY) = 1\cdot 1\cdot 0.2 = 0.2$E(XY)=1⋅1⋅0.2=0.2. Therefore

$\text{Cov}(X, Y) = E(XY) - E(X)E(Y) = 0.2 - (0.6)(0.5) = 0.2 - 0.3 = -0.1.$Cov(X,Y)=E(XY)−E(X)E(Y)=0.2−(0.6)(0.5)=0.2−0.3=−0.1.

The negative covariance says that, in this joint distribution, $X = 1$X=1 tends to coincide with $Y = 0$Y=0 — and indeed the largest probability, $0.4$0.4, sits in exactly that cell. Because the covariance is non-zero, $X$X and $Y$Y are not independent, and you could not add their variances directly: you would need the $2\,\text{Cov}(X,Y) = -0.2$2Cov(X,Y)=−0.2 correction. (A cleaner check of independence: under independence the top-left cell would be $P(X=0)P(Y=0) = 0.4\times 0.5 = 0.2$P(X=0)P(Y=0)=0.4×0.5=0.2, but the table gives $0.1$0.1 — they differ, confirming dependence.) Although full two-variable distributions are at the edge of the A-Level Further specification, recognising when the additive variance rule applies — and being able to articulate that it requires independence — is squarely examinable and is precisely the reasoning the variance algebra is built on.

Mode, median and the discrete CDF

Two further summaries round out the description of a discrete distribution. The mode is the value of $x$x carrying the largest probability; a distribution is bimodal (or multimodal) when two or more values tie for the maximum. The median is a value $m$m with $P(X \leq m) \geq 0.5$P(X≤m)≥0.5 and $P(X \geq m) \geq 0.5$P(X≥m)≥0.5 — the "middle" value once the probability is accumulated. For the parameter example (with $k = 0.1$k=0.1, values $0,1,2,3$0,1,2,3 carrying $0.1, 0.2, 0.3, 0.4$0.1,0.2,0.3,0.4), the mode is $x = 3$x=3 (probability $0.4$0.4). Accumulating probability gives $P(X \leq 0) = 0.1,\ P(X \leq 1) = 0.3,\ P(X \leq 2) = 0.6$P(X≤0)=0.1, P(X≤1)=0.3, P(X≤2)=0.6, so the median is $m = 2$m=2, the first value at which the running total reaches or passes $0.5$0.5.

That running total is the cumulative distribution function (CDF),

$F(x) = P(X \leq x) = \sum_{x_i \leq x} P(X = x_i),$F(x)=P(X≤x)=∑xi​≤x​P(X=xi​),

a step function that climbs from 0 to 1, jumping at each value of $X$X by the size of its probability. The CDF and the probability mass function carry the same information: you recover an individual probability as the size of the jump,

$P(X = x) = F(x) - F(x^-),$P(X=x)=F(x)−F(x−),

where $F(x^-)$F(x−) is the value of $F$F just to the left of $x$x. For the example, $F$F jumps by $0.1, 0.2, 0.3, 0.4$0.1,0.2,0.3,0.4 at $x = 0, 1, 2, 3$x=0,1,2,3 respectively, reaching exactly $1$1 at $x = 3$x=3. The CDF is the natural tool for "at most" and "more than" questions — $P(X > 2) = 1 - F(2) = 1 - 0.6 = 0.4$P(X>2)=1−F(2)=1−0.6=0.4 — and it is the discrete shadow of the continuous CDF that dominates the second half of this course.

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3. Worked examples with M1/A1 mark schemes

Example 1 — full table analysis

A random variable $X$X has the distribution below, where $k$k is constant.

$x$x	0	1	2	3
$P(X = x)$P(X=x)	$k$k	$2k$2k	$3k$3k	$4k$4k

Find $k$k, $E(X)$E(X), $\text{Var}(X)$Var(X) and $E(2X^2 - 3X + 1)$E(2X2−3X+1).

$k + 2k + 3k + 4k = 1 \implies 10k = 1 \implies k = 0.1.$k+2k+3k+4k=1⟹10k=1⟹k=0.1. (M1 set sum of probabilities $=1$=1; A1 $k = 0.1$k=0.1.)

$E(X) = 0(0.1) + 1(0.2) + 2(0.3) + 3(0.4) = 2.0.$E(X)=0(0.1)+1(0.2)+2(0.3)+3(0.4)=2.0. (M1 $\sum x\,P(X=x)$∑xP(X=x); A1 $E(X)=2.0$E(X)=2.0.)

$E(X^2) = 0(0.1) + 1(0.2) + 4(0.3) + 9(0.4) = 0 + 0.2 + 1.2 + 3.6 = 5.0.$E(X2)=0(0.1)+1(0.2)+4(0.3)+9(0.4)=0+0.2+1.2+3.6=5.0. (M1 $\sum x^2 P(X=x)$∑x2P(X=x); A1 $E(X^2)=5.0$E(X2)=5.0.)

$\text{Var}(X) = 5.0 - 2.0^2 = 1.0.$Var(X)=5.0−2.02=1.0. (M1 apply $E(X^2)-(E(X))^2$E(X2)−(E(X))2; A1 $\text{Var}(X)=1.0$Var(X)=1.0.)

$E(2X^2 - 3X + 1) = 2E(X^2) - 3E(X) + 1 = 2(5.0) - 3(2.0) + 1 = 5.$E(2X2−3X+1)=2E(X2)−3E(X)+1=2(5.0)−3(2.0)+1=5. (M1 expand using linearity; A1 $=5$=5.)

Example 2 — the variance algebra under transformation

The number of faults $X$X on a metre of cable has $E(X) = 1.5$E(X)=1.5 and $\text{Var}(X) = 0.75$Var(X)=0.75. The inspection cost in pounds is $C = 4X + 10$C=4X+10. Find $E(C)$E(C), $\text{Var}(C)$Var(C) and the standard deviation of $C$C.

$E(C) = 4E(X) + 10 = 4(1.5) + 10 = 16.$E(C)=4E(X)+10=4(1.5)+10=16. (M1 $E(aX+b)=aE(X)+b$E(aX+b)=aE(X)+b; A1 $\pounds 16$£16.)

$\text{Var}(C) = 4^2 \, \text{Var}(X) = 16(0.75) = 12.$Var(C)=42Var(X)=16(0.75)=12. (M1 $\text{Var}(aX+b)=a^2\text{Var}(X)$Var(aX+b)=a2Var(X), with the $+10$+10 ignored; A1 $12$12.)

$\text{SD}(C) = \sqrt{12} = 2\sqrt{3} \approx 3.46.$SD(C)=12​=23​≈3.46. (A1 $\approx \pounds 3.46$≈£3.46.)

Example 3 — recovering a distribution from its moments

A variable $X$X takes the values $0, 1, 2$0,1,2 with $P(X=1) = 0.5$P(X=1)=0.5. Given $E(X) = 1$E(X)=1, find the full distribution and $\text{Var}(X)$Var(X).

Let $P(X=0) = a$P(X=0)=a and $P(X=2) = c$P(X=2)=c. Then

$a + 0.5 + c = 1 \implies a + c = 0.5.$a+0.5+c=1⟹a+c=0.5. (M1 normalisation equation.)

$E(X) = 0\cdot a + 1(0.5) + 2c = 1 \implies 0.5 + 2c = 1 \implies c = 0.25.$E(X)=0⋅a+1(0.5)+2c=1⟹0.5+2c=1⟹c=0.25. (M1 expectation equation; A1 $c = 0.25$c=0.25, hence $a = 0.25$a=0.25.)

So $P(X=0) = 0.25,\ P(X=1) = 0.5,\ P(X=2) = 0.25$P(X=0)=0.25, P(X=1)=0.5, P(X=2)=0.25. Then

$E(X^2) = 0 + 1(0.5) + 4(0.25) = 1.5, \qquad \text{Var}(X) = 1.5 - 1^2 = 0.5.$E(X2)=0+1(0.5)+4(0.25)=1.5,Var(X)=1.5−12=0.5. (M1 $E(X^2)$E(X2); A1 $\text{Var}(X) = 0.5$Var(X)=0.5.)

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4. Specimen-style exam question

(Specimen-style — not from any past paper.) The discrete random variable $X$X has probability distribution

$x$x	$-1$−1	0	1	2
$P(X=x)$P(X=x)	$0.2$0.2	$a$a	$0.3$0.3	$b$b

Given that $E(X) = 0.6$E(X)=0.6: (a) show that $a + b = 0.5$a+b=0.5 and find a second equation in $a$a and $b$b; hence find $a$a and $b$b; (b) find $\text{Var}(X)$Var(X); (c) find $\text{Var}(5 - 2X)$Var(5−2X).

Model solution.

(a) Probabilities sum to 1: $0.2 + a + 0.3 + b = 1 \Rightarrow a + b = 0.5$0.2+a+0.3+b=1⇒a+b=0.5. Expectation:

$E(X) = (-1)(0.2) + 0\cdot a + 1(0.3) + 2b = 0.1 + 2b = 0.6 \implies b = 0.25,$E(X)=(−1)(0.2)+0⋅a+1(0.3)+2b=0.1+2b=0.6⟹b=0.25,

so $a = 0.25$a=0.25. Both lie in $[0,1]$[0,1], so the distribution is valid.

(b) $E(X^2) = (-1)^2(0.2) + 0 + 1^2(0.3) + 2^2(0.25) = 0.2 + 0.3 + 1.0 = 1.5$E(X2)=(−1)2(0.2)+0+12(0.3)+22(0.25)=0.2+0.3+1.0=1.5, so

$\text{Var}(X) = 1.5 - 0.6^2 = 1.5 - 0.36 = 1.14.$Var(X)=1.5−0.62=1.5−0.36=1.14.

(c) Variance ignores the shift and squares the scale: $\text{Var}(5 - 2X) = (-2)^2 \text{Var}(X) = 4(1.14) = 4.56.$Var(5−2X)=(−2)2Var(X)=4(1.14)=4.56.

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5. Synoptic links

• Poisson and binomial (this module): both are discrete random variables; everything here — $E$E, $\text{Var}$Var, the $E(X^2)$E(X2) identity — specialises to them. For $\text{Po}(\lambda)$Po(λ) the headline fact is $E(X) = \text{Var}(X) = \lambda$E(X)=Var(X)=λ.
• Continuous random variables (later lessons): every formula has a continuous analogue with $\sum$∑ replaced by $\int$∫: $E(X) = \int x f(x)\,dx$E(X)=∫xf(x)dx, $\text{Var}(X) = E(X^2) - (E(X))^2$Var(X)=E(X2)−(E(X))2 unchanged.
• Moment generating functions (later): $M_X(t) = E(e^{tX})$MX​(t)=E(etX) packages every moment; $E(X) = M_X'(0)$E(X)=MX′​(0) and $E(X^2) = M_X''(0)$E(X2)=MX′′​(0) regenerate the quantities above.
• A-Level Maths — algebra of series and binomial expansion: the $\sum$∑ manipulations and $\sum p_i = 1$∑pi​=1 reasoning mirror summing a probability series.
• Linear coding (A-Level statistics): the coding $Y = \frac{X - a}{b}$Y=bX−a​ used to simplify mean/variance calculations is exactly $E(aX+b)$E(aX+b)/$\text{Var}(aX+b)$Var(aX+b) in disguise.

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6. Mark-scheme literacy

• "Show that": the target value is printed, so you are paid for the method and communication, not the number. Display the full sum $\sum x P(X=x)$∑xP(X=x) and the substitution — landing on the given value with no working scores little.
• "Hence": you must use the previous part. If part (a) gave $E(X^2)$E(X2), a "hence find $\text{Var}(X)$Var(X)" expects $E(X^2) - (E(X))^2$E(X2)−(E(X))2, not a fresh calculation.
• Follow-through (ft): if you find a wrong $E(X)$E(X) but then apply $\text{Var}(X) = E(X^2) - (E(X))^2$Var(X)=E(X2)−(E(X))2 correctly, the variance A-mark is usually awarded "ft" on your value. So always finish the method, even after a slip.
• Exact vs decimal: give exact fractions or full decimals; a variance such as $0.81$0.81 should not be rounded to $0.8$0.8 unless asked. Premature rounding of $E(X)$E(X) before squaring is a classic accuracy-mark loss.
• Units and interpretation: AO2 marks attach to a sentence — "the expected cost is £16" — not a bare number.

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7. Grade-band model answers

Question. The random variable $X$X has $E(X) = 4$E(X)=4 and $\text{Var}(X) = 9$Var(X)=9. Find $E(X^2)$E(X2) and $\text{Var}(2X - 1)$Var(2X−1), and explain why $\text{Var}(2X-1) \neq 2\text{Var}(X)$Var(2X−1)=2Var(X).

Mid-band response. $E(X^2) = 9 + 16 = 25$E(X2)=9+16=25. $\text{Var}(2X-1) = 4 \times 9 = 36$Var(2X−1)=4×9=36. It is not $2 \times 9$2×9 because you square the 2.

Examiner-style commentary: The two numerical answers are correct and would earn the method and accuracy marks. The explanation is too thin for the AO2 mark — "you square the 2" states the rule rather than the reason, and the $-1$−1 is not addressed.

Stronger response. Rearranging $\text{Var}(X) = E(X^2) - (E(X))^2$Var(X)=E(X2)−(E(X))2 gives $E(X^2) = \text{Var}(X) + (E(X))^2 = 9 + 16 = 25$E(X2)=Var(X)+(E(X))2=9+16=25. For the transformation, $\text{Var}(2X - 1) = 2^2\,\text{Var}(X) = 4(9) = 36$Var(2X−1)=22Var(X)=4(9)=36; the $-1$−1 is a shift and does not affect spread. It is not $2\text{Var}(X)$2Var(X) because variance scales with the square of the multiplier.

Examiner-style commentary: Fully correct with the identity quoted and rearranged, and the shift correctly dismissed. The final sentence earns the reasoning mark. To reach the top band the explanation could connect the $a^2$a2 factor to squared deviations.

Top-band response. From $\text{Var}(X) = E(X^2) - (E(X))^2$Var(X)=E(X2)−(E(X))2, $E(X^2) = 9 + 4^2 = 25$E(X2)=9+42=25. Writing $Y = 2X - 1$Y=2X−1, $\text{Var}(Y) = E((Y - E(Y))^2)$Var(Y)=E((Y−E(Y))2); since $Y - E(Y) = 2(X - E(X))$Y−E(Y)=2(X−E(X)), the deviation is scaled by 2 and the squared deviation by $2^2 = 4$22=4, giving $\text{Var}(Y) = 4\,\text{Var}(X) = 36$Var(Y)=4Var(X)=36. The additive constant $-1$−1 cancels in $Y - E(Y)$Y−E(Y), so it cannot change the variance. Hence $\text{Var}(2X-1) = 36 \neq 18 = 2\text{Var}(X)$Var(2X−1)=36=18=2Var(X): variance is quadratic in the scale factor, not linear.

Examiner-style commentary: This is exemplary. The candidate derives the $a^2$a2 factor from first principles via $Y - E(Y) = 2(X - E(X))$Y−E(Y)=2(X−E(X)), explains why the shift cancels, and states the general principle. Every available AO1 and AO2 mark is secured.

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8. Common misconceptions

1. "$E(X^2) = (E(X))^2$E(X2)=(E(X))2." False except for a constant. The gap between them is the variance: $E(X^2) - (E(X))^2 = \text{Var}(X) \geq 0$E(X2)−(E(X))2=Var(X)≥0.
2. "$\text{Var}(aX+b) = a\,\text{Var}(X) + b$Var(aX+b)=aVar(X)+b." No — variance squares the scale and ignores the shift: $\text{Var}(aX+b) = a^2\text{Var}(X)$Var(aX+b)=a2Var(X).
3. "$\text{Var}(2X) = \text{Var}(X+X)$Var(2X)=Var(X+X)." These are different. $2X$2X is one variable doubled, giving $4\text{Var}(X)$4Var(X); $X_1 + X_2$X1​+X2​ of two independent copies gives $2\text{Var}(X)$2Var(X).
4. "Expectation must be a possible value." $E(X) = 2.7$E(X)=2.7 is fine even though $X$X only takes integer values.
5. "$E(1/X) = 1/E(X)$E(1/X)=1/E(X)" or "$E(\sqrt{X}) = \sqrt{E(X)}$E(X​)=E(X)​." Only linear functions pass through $E$E. For non-linear $g$g, compute $E(g(X)) = \sum g(x)P(X=x)$E(g(X))=∑g(x)P(X=x).
6. "Variance can be negative." Impossible — it is an expected square. A negative answer signals an arithmetic slip (often $\big(E(X)\big)^2$(E(X))2 computed as $E(X^2)$E(X2) by mistake).
7. "$\text{Var}(X+Y) = \text{Var}(X)+\text{Var}(Y)$Var(X+Y)=Var(X)+Var(Y) always." Only when $X, Y$X,Y are independent; otherwise add $2\,\text{Cov}(X,Y)$2Cov(X,Y).

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9. Common errors

• Forgetting to square inside $E(X^2)$E(X2): computing $\sum xP$∑xP again instead of $\sum x^2 P$∑x2P.
• Rounding $E(X)$E(X) before squaring it in $\text{Var}(X) = E(X^2) - (E(X))^2$Var(X)=E(X2)−(E(X))2, introducing rounding error into the variance.
• Dropping a term when a value is $0$0 or negative — e.g. omitting the $x=0$x=0 row (it contributes 0 to $E(X)$E(X) but its probability still matters for normalisation) or mishandling $(-1)^2 = 1$(−1)2=1.
• Keeping the additive constant in a variance: writing $\text{Var}(3X+2) = 9\text{Var}(X) + 2$Var(3X+2)=9Var(X)+2.
• Sign slip in $\text{Var}(a - bX)$Var(a−bX): the answer uses $b^2$b2, so the minus sign disappears.
• Probabilities not summing to 1 because an unknown $k$k was found but never substituted back to check.

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10. Going further (STEP / MAT / Oxbridge)

The identity $\text{Var}(X) = E(X^2) - (E(X))^2$Var(X)=E(X2)−(E(X))2 is the second instance of a deeper pattern. Define the moments $\mu_k = E(X^k)$μk​=E(Xk) and the central moments $\nu_k = E\big((X-\mu)^k\big)$νk​=E((X−μ)k). Then $\nu_2$ν2​ is the variance, $\nu_3$ν3​ governs skewness and $\nu_4$ν4​ governs kurtosis, and each $\nu_k$νk​ expands into the raw moments by the binomial theorem:

$\nu_k = E\big((X-\mu)^k\big) = \sum_{j=0}^{k} \binom{k}{j} (-\mu)^{k-j} \mu_j.$νk​=E((X−μ)k)=∑j=0k​(jk​)(−μ)k−jμj​.

Setting $k=2$k=2 recovers exactly $\nu_2 = \mu_2 - \mu^2$ν2​=μ2​−μ2. A short STEP-flavoured challenge: show that the central third moment is

$E\big((X-\mu)^3\big) = E(X^3) - 3\mu E(X^2) + 2\mu^3,$E((X−μ)3)=E(X3)−3μE(X2)+2μ3,

and verify it on the running example (where $\mu = 2.7$μ=2.7). A second, genuinely useful result is the variance of a sum: for any $X, Y$X,Y,

$\text{Var}(X+Y) = \text{Var}(X) + \text{Var}(Y) + 2\,\text{Cov}(X,Y), \quad \text{Cov}(X,Y) = E(XY) - E(X)E(Y),$Var(X+Y)=Var(X)+Var(Y)+2Cov(X,Y),Cov(X,Y)=E(XY)−E(X)E(Y),

which collapses to additive variances precisely when $\text{Cov}(X,Y) = 0$Cov(X,Y)=0 — guaranteed by independence, since then $E(XY) = E(X)E(Y)$E(XY)=E(X)E(Y). The converse is false: uncorrelated does not imply independent, a favourite Oxbridge interview trap.

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11. Additional A* practice (with worked answers)

P1. $X$X has $P(X=x) = \dfrac{x}{15}$P(X=x)=15x​ for $x = 1,2,3,4,5$x=1,2,3,4,5. Find $E(X)$E(X) and $\text{Var}(X)$Var(X).

$E(X) = \frac{1}{15}(1+4+9+16+25) = \frac{55}{15} = \frac{11}{3} \approx 3.667.$E(X)=151​(1+4+9+16+25)=1555​=311​≈3.667. $E(X^2) = \frac{1}{15}(1+8+27+64+125) = \frac{225}{15} = 15.$E(X2)=151​(1+8+27+64+125)=15225​=15. $\text{Var}(X) = 15 - \left(\frac{11}{3}\right)^2 = 15 - \frac{121}{9} = \frac{14}{9} \approx 1.556.$Var(X)=15−(311​)2=15−9121​=914​≈1.556.

P2. Given $E(X) = 3$E(X)=3 and $E(X^2) = 13$E(X2)=13, find (a) $\text{Var}(X)$Var(X); (b) $E\big((X-3)^2\big)$E((X−3)2); (c) $\text{Var}(4 - 3X)$Var(4−3X).

(a) $13 - 9 = 4.$13−9=4. (b) $E((X-\mu)^2) = \text{Var}(X) = 4.$E((X−μ)2)=Var(X)=4. (c) $(-3)^2(4) = 36.$(−3)2(4)=36.

P3. A fair four-sided die shows $1,2,3,4$1,2,3,4. Let $X$X be the score and $Y = X^2$Y=X2. Find $E(Y)$E(Y) and $\text{Var}(Y)$Var(Y).

$E(Y) = E(X^2) = \frac14(1+4+9+16) = \frac{30}{4} = 7.5.$E(Y)=E(X2)=41​(1+4+9+16)=430​=7.5. $E(Y^2) = E(X^4) = \frac14(1+16+81+256) = \frac{354}{4} = 88.5.$E(Y2)=E(X4)=41​(1+16+81+256)=4354​=88.5. $\text{Var}(Y) = 88.5 - 7.5^2 = 88.5 - 56.25 = 32.25.$Var(Y)=88.5−7.52=88.5−56.25=32.25.

P4. $X$X takes values $1, 2, 3$1,2,3 with probabilities $p, q, p$p,q,p (symmetric). Given $\text{Var}(X) = 0.5$Var(X)=0.5, find $p$p and $q$q.

By symmetry $E(X) = 2$E(X)=2. Normalisation: $2p + q = 1$2p+q=1. $E(X^2) = p + 4q + 9p = 10p + 4q$E(X2)=p+4q+9p=10p+4q. Then $\text{Var}(X) = 10p + 4q - 4 = 0.5$Var(X)=10p+4q−4=0.5, so $10p + 4q = 4.5$10p+4q=4.5. Subtract $4(2p+q)=4$4(2p+q)=4: $2p = 0.5 \Rightarrow p = 0.25$2p=0.5⇒p=0.25, $q = 0.5$q=0.5.

P5. The number of heads $X$X in two tosses of a biased coin ($P(\text{head}) = 0.6$P(head)=0.6) has distribution $P(0)=0.16,\ P(1)=0.48,\ P(2)=0.36$P(0)=0.16, P(1)=0.48, P(2)=0.36. Verify $E(X) = 1.2$E(X)=1.2 and find $\text{Var}(X)$Var(X); compare with $np$np and $np(1-p)$np(1−p).

$E(X) = 0(0.16) + 1(0.48) + 2(0.36) = 1.2 = np = 2(0.6).$E(X)=0(0.16)+1(0.48)+2(0.36)=1.2=np=2(0.6). $E(X^2) = 0 + 0.48 + 4(0.36) = 1.92.$E(X2)=0+0.48+4(0.36)=1.92. $\text{Var}(X) = 1.92 - 1.44 = 0.48 = np(1-p) = 2(0.6)(0.4).$Var(X)=1.92−1.44=0.48=np(1−p)=2(0.6)(0.4). The table reproduces the binomial moments exactly.

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12. Board-alignment footer

Aligned to AQA A-Level Further Mathematics 7367, Paper 3 Statistics (7367/3S): discrete random variables, $E(X)$E(X), $E(g(X))$E(g(X)), $\text{Var}(X) = E(X^2) - (E(X))^2$Var(X)=E(X2)−(E(X))2, and the linear-transformation algebra. The same content appears in Edexcel Further Statistics 1 (discrete random variables, $E(X)$E(X)/$\text{Var}(X)$Var(X) of linear functions) and OCR(A)/OCR(MEI) Statistics modules; the formulae and conventions are identical across boards.

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13. Visual summary

$$\boxed{\;E(X) = \sum_x x\,P(X=x) \quad\bullet\quad \text{Var}(X) = E(X^2) - \big(E(X)\big)^2 \quad\bullet\quad \text{Var}(aX+b) = a^2\text{Var}(X)\;}$$E(X)=x∑​xP(X=x)∙Var(X)=E(X2)−(E(X))2∙Var(aX+b)=a2Var(X)​

Quantity	Discrete formula	Key behaviour under $aX+b$aX+b
$E(X)$E(X)	$\sum x\,P(X=x)$∑xP(X=x)	$aE(X)+b$aE(X)+b (linear)
$E(g(X))$E(g(X))	$\sum g(x)\,P(X=x)$∑g(x)P(X=x)	apply $g$g then weight
$\text{Var}(X)$Var(X)	$E(X^2)-(E(X))^2$E(X2)−(E(X))2	$a^2\text{Var}(X)$a2Var(X) (shift ignored)
$\sigma$σ	$\sqrt{\text{Var}(X)}$Var(X)​	$\lvert a\rvert\,\sigma$∣a∣σ

graph LR
  A["Distribution table x, P(X=x)"] --> B["E(X) = sum x P"]
  A --> C["E(X^2) = sum x^2 P"]
  B --> D["Var(X) = E(X^2) - (E(X))^2"]
  C --> D
  D --> E["SD = sqrt(Var)"]
  B --> F["Transform: E(aX+b)=aE(X)+b"]
  D --> G["Transform: Var(aX+b)=a^2 Var(X)"]

Recap: build a clean table, compute $E(X)$E(X) and $E(X^2)$E(X2), then $\text{Var}(X) = E(X^2) - (E(X))^2$Var(X)=E(X2)−(E(X))2. Expectation is linear; variance squares the scale and ignores shifts. Master this and every distribution in 7367/3S becomes a special case.