Linear Combinations of Random Variables Revision Notes — 9709 Paper 6

E(aX + b) and Var(aX + b) — scaling and shifting one variable

Syllabus requirement · §6.2

“

use, in the course of solving problems, the results that E(aX + b) = aE(X) + b and Var(aX + b) = a²Var(X).

”

Plenty of quantities are just a fixed rule applied to a random variable you already understand: a total cost built from a fixed fee plus a rate per unit, a temperature converted from one scale to another, a price converted into another currency. If $X$ is a random variable and $Y=aX+b$ for constants $a$ and $b$ , the mean and variance of $Y$ follow directly from those of $X$ , with no new distribution to build:

E(aX+b) = aE(X)+b, \qquad Var(aX+b) = a^2\,Var(X)

The syllabus does not ask you to prove these, but the reason they take this shape is short — and knowing it means you can rebuild either formula if you forget it. Adding the constant $b$ shifts every possible value of $X$ up by $b$ — so the average shifts up by $b$ too, but the gaps between values are unchanged, so the spread (variance) is completely unaffected by $b$ . Scaling by $a$ stretches every value, and every gap between values, by a factor of $a$ — so the mean scales by $a$ , but since variance is built from squared deviations, the spread scales by $a^2$ , not $a$ .

The constant b vanishes from the variance

This is the single most common slip in the whole topic: $b$ shifts the mean but never appears in the variance at all — not as $+b$ , not as $+b^2$ . Only the multiplier $a$ affects the spread, and it does so as $a^2$ .

If X is normal, so is aX + b

The two rules above work for any random variable $X$ — they say nothing about its shape. But when $X$ happens to be normal, there is a bonus: $aX+b$ is normal too, with exactly the mean and variance those rules give.

X\sim N(\mu,\sigma^2) \quad\Longrightarrow\quad aX+b \sim N\big(a\mu+b,\ a^2\sigma^2\big)

That matters because a mean and a variance on their own cannot produce a probability — you need the shape as well. This is the one-variable version of §03's result, and the reasoning there is the same, applied to two variables at once.

Substituting into both formulas, piece by piece

A taxi charges a fixed $2.50 plus $3 per kilometre travelled. The distance travelled on a randomly chosen journey, $X$ km, has mean $10$ and variance $4$ . Find the mean and variance of the total fare, $Y = 3X + 2.5$ .

Show full working

1
Step 1 — identify $a$ and $b$ in $Y=aX+b$ . Here $a=3$ (the rate per km) and $b=2.5$ (the fixed charge).
Getting a and b the right way round is the whole question. The rate per kilometre multiplies the random distance, so it is a; the fixed charge is paid whatever happens, so it is b.
2
Step 2 — find $E(Y)$ , substituting $E(X)=10$ into the mean formula. $E(Y) = aE(X)+b = 3\times10+2.5$
3
Step 3 — evaluate. $E(Y) = 30+2.5 = 32.5$
4
Step 4 — find $Var(Y)$ , substituting $Var(X)=4$ into the variance formula. $Var(Y) = a^2\,Var(X) = 3^2\times4$
The fixed charge b = 2.5 plays no part in this step at all — only a is squared.
5
Step 5 — evaluate. $Var(Y) = 9\times4 = 36$

Answer

$E(Y) = 32.5$ , $Var(Y) = 36$ .

Find E(Y) and Var(Y) as two entirely separate calculations — the constant b enters only the mean formula, and only the multiplier a (squared) enters the variance formula.

A linear transformation of a normal variable

9709/61 O/N 2023 Q4(a)2 marks

The mass, in kilograms, of chemical $A$ produced per day by a factory is modelled by the random variable $X \sim N(10.3, 5.76)$ . The income generated by chemical $A$ is $2.50 per kilogram. Find the mean and variance of the daily income generated by chemical $A$ .

Show full working

1
Step 1 — write the income as a linear transformation of $X$ . Income $= 2.5X$ , so $a=2.5$ and $b=0$ (no fixed fee here).
There is no fixed fee here, so b = 0. Write it down anyway — it is what tells you nothing gets added on at the end.
2
Step 2 — find the mean income. $E(2.5X) = 2.5\times E(X) = 2.5\times10.3$
3
Step 3 — evaluate. $E(2.5X) = 25.75$
4
Step 4 — find the variance of the income, squaring the multiplier. $Var(2.5X) = 2.5^2\times Var(X) = 6.25\times5.76$
5
Step 5 — evaluate. $Var(2.5X) = 36$
6
Step 6 — name the distribution of the income. $X$ is normal, so $2.5X$ is normal too, carrying the mean and variance just found: $2.5X \sim N(25.75,\ 36)$
Worth writing even though this part asks only for a mean and a variance — it is what makes a follow-on part such as 'find the probability the income exceeds 30 dollars' answerable at all.

Answer

Mean income $=$ $25.75, variance $=36$ (in $ $^2$ ); the income is distributed $N(25.75, 36)$ .

Combining with a distribution's own mean/variance formula first

9709/62 M/J 2021 Q2(a)3 marks

The random variable $X$ has the distribution $B(400, 0.01)$ . Find $Var(4X+2)$ .

Show full working

1
Step 1 — find $Var(X)$ first, using the binomial variance formula (§5.4), since it isn't given directly. $Var(X) = np(1-p) = 400\times0.01\times0.99$
The question gives you a distribution, not a variance. Var(X) has to be produced from B(400, 0.01) first; the linear rule cannot start until it exists.
2
Step 2 — evaluate. $Var(X) = 3.96$
3
Step 3 — identify $a$ in $4X+2$ . Here $a=4$ ; the constant $2$ will not appear in the variance at all.
4
Step 4 — substitute into $Var(aX+b) = a^2\,Var(X)$ . $Var(4X+2) = 4^2\times3.96$
5
Step 5 — evaluate. $Var(4X+2) = 16\times3.96 = 63.36$

Answer

$Var(4X+2) = 63.36$ .

When Var(X) isn't given directly, find it first using whatever distribution X actually has — the linear-transformation rule only ever needs Var(X) as an input, however that number was obtained.

Common mistakes

Writing $Var(aX+b) = a^2\,Var(X) + b^2$ , or $+b$
$Var(aX+b) = a^2\,Var(X)$ exactly — the additive constant never appears in the variance
A constant shift moves every value by the same amount, so it can't change how spread out those values are relative to each other.
Using $Var(aX+b) = a\,Var(X)$ , forgetting to square the multiplier
Square $a$ before multiplying: $Var(aX+b)=a^2\,Var(X)$
Variance is built from squared deviations, so any linear scaling of X carries through as the square of that scale factor.

The same idea, drawn as a graph

Once a random variable's probability density function has a graph (§6.3), these two rules have a picture: adding $b$ slides the whole graph sideways without changing its shape or height; scaling by $a$ stretches it horizontally by a factor of $a$ and, to keep the total area under the curve equal to $1$ , shrinks its height by the same factor. Cambridge sets exactly this as a sketching question — "sketch the density of $aX+b$ " — once §6.3 has introduced what that density graph actually is. Keep the E/Var rules from this section in mind; §6.3 comes back to them for the picture.

Your turn

1
A random variable $X$ has mean $8$ and variance $5$ . Find $E(2X-3)$ and $Var(2X-3)$ .
Stuck? Show hint
E(aX+b) = aE(X)+b uses both a and b; Var(aX+b) = a²Var(X) uses only a, squared.
Show solution
1. 1
  Step 1 — identify $a$ and $b$ . Comparing $2X-3$ with $aX+b$ gives $a=2$ and $b=-3$ .
  The minus sign belongs to b. Writing b = −3, rather than thinking 'subtract 3', is what keeps the sign right when it is substituted.
2. 2
  Step 2 — substitute into the mean formula. $E(2X-3) = aE(X)+b = 2\times8-3$
3. 3
  Step 3 — evaluate. $E(2X-3) = 16-3 = 13$
4. 4
  Step 4 — substitute into the variance formula, which uses only $a$ . $Var(2X-3) = a^2\,Var(X) = 2^2\times5$
  The −3 is not written anywhere in this line. That is the rule, not an oversight.
5. 5
  Step 5 — evaluate. $Var(2X-3) = 4\times5 = 20$
Answer
$E(2X-3)=13$ , $Var(2X-3)=20$ .
23 marks
A company converts a measured temperature $X$ (in an internal sensor unit, mean $50$ , variance $16$ ) to degrees Celsius using $C = 0.5X - 10$ . Find the mean and standard deviation of $C$ .
Stuck? Show hint
Find E(C) and Var(C) separately, then take a square root for the standard deviation.
Show solution
1. 1
  Step 1 — identify $a$ and $b$ in $C = 0.5X - 10$ . Here $a=0.5$ and $b=-10$ .
2. 2
  Step 2 — find the mean. $E(C) = 0.5\times50-10 = 25-10 = 15$
3. 3
  Step 3 — find the variance, squaring $a$ . $Var(C) = 0.5^2\times16 = 0.25\times16 = 4$
  0.5 squared is 0.25, not 0.5. A multiplier below 1 shrinks the spread by more than it shrinks the mean.
4. 4
  Step 4 — the question asks for the standard deviation, so square-root the variance. $\text{s.d.}(C) = \sqrt{4} = 2$
  Read the last line of the question again before answering. Quoting the variance where a standard deviation was asked for throws away the final mark.
Answer
$E(C)=15$ , standard deviation of $C = 2$ .

Practise E(aX + b) and Var(aX + b) from real Paper 6 papersReal past-paper questions · E(aX + b) and Var(aX + b)

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Can you do all of these?

The additive constant b never appears in Var(aX+b) — only the multiplier a, squared
Var(aX+bY) = a²Var(X) + b²Var(Y) needs independence — and adds variances even for a difference
Distinguish 'n independent copies summed' (variance × n) from 'one copy scaled by n' (variance × n²) — the means agree, the variances never do
Convert a standard deviation to a variance before combining — mark schemes reject any SD/variance mix
Rearrange any inequality between two variables (e.g. L < 3S) onto one side before reading off a and b
A linear combination is normal only when every variable in it is independent AND individually normal
The sum of independent Poisson variables is Poisson, with parameter equal to the sum of the individual parameters
State the new distribution N(mean, variance) or Po(λ) as its own explicit step before standardising or substituting further
When a question says 'stating a necessary assumption', write the independence assumption out in words — it carries its own mark
Put every Poisson rate on the interval the question asks about BEFORE adding the parameters — never scale the combined λ afterwards
'Differ by more than k' is P(D > k) + P(D < −k) added — not twice one tail, unless E(D) = 0
The total of n independent copies has standard deviation σ√n, never σn

Now do the questions

95 real Paper 6 parts from 2021–2025, sorted by difficulty, with mark schemes

The Poisson Distribution

Continuous Random Variables

Linear Combinations of Random Variables

E(aX + b) and Var(aX + b) — scaling and shifting one variable

E(aX + bY) and Var(aX + bY) — combining two independent variables

A linear combination of normal variables is normal

Can you do all of these?

E(aX + b) and Var(aX + b) — scaling and shifting one variable

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