Did you know? If you multiply every value in a data set by three, its average becomes three times as large, but its variance becomes nine times as large. That is because variance measures squared distances from a mean. And if you add ten to every value instead, the mean rises by ten while the variance stays exactly the same.
For a learner preparing for JC1 H2 Mathematics in Bukit Timah, this is a particularly useful way to connect algebra with statistics. The 2027 Singapore–Cambridge H2 Mathematics syllabus 9758 includes discrete random variables, expectations, variances and transformations of random variables; individual junior colleges may teach the probability-and-statistics strands in different JC years. The fundamental distinction is stable regardless of the timetable: shifting changes location, while scaling changes spread.
This guide focuses on the pair of formulas E(aX + b) = aE(X) + b and Var(aX + b) = a²Var(X). We will explain why they differ, work through equal-probability outcomes and a Bernoulli model, show the role of negative multipliers and standard deviations, and diagnose the most common calculation errors. The aim is to help a student decide what happens to a distribution *before* calculating every transformed value.
Quick answer: the mean shifts, but variance only scales
Let X be a random variable with a finite mean and variance. Let a and b be real constants, and define Y = aX + b.
The expectation is
E(Y) = E(aX + b) = aE(X) + b.
The variance is
Var(Y) = Var(aX + b) = a²Var(X).
The standard deviation obeys the related rule
SD(Y) = |a|SD(X).
- Adding a fixed b shifts every value by b. It shifts the mean by b but does not change the variance.
- Multiplying every value by a multiplies the mean by a and the variance by a².
- The standard deviation is multiplied by |a|, not a².
- If a is negative, outcomes reverse their order, but variance remains non-negative.
- If a = 0, every transformed value equals b and the variance is zero.
- These rules do not require X to follow a normal distribution; they apply whenever the relevant moments exist.
A student can learn the two formulas by memory, but understanding distance from the mean explains why the constant b disappears in the variance calculation.
The ordinary meaning of expectation and variance
Expectation, often called the mean, is a probability-weighted average of the possible outcomes. For a finite discrete random variable with outcomes x₁, x₂, … and probabilities p₁, p₂, …, we calculate
E(X) = Σ xᵢpᵢ.
Variance measures how far values tend to be from that expected value, but it uses squared differences. It is defined by
Var(X) = E[(X − E(X))²].
For a finite discrete distribution, that becomes a weighted average of the squared distances of the possible outcomes from the mean.
This definition immediately explains why variance cannot be negative. Each squared distance is non-negative, and its probability weight is non-negative.
The square also explains why variance changes more dramatically under scaling than the mean does. When each distance is multiplied by three, its square becomes nine times larger.
Worked example 1: three equally likely outcomes
Let X take values 1, 2 and 3, each with probability 1/3.
Find the expectation:
E(X) = (1 + 2 + 3)/3 = 2.
The deviations from the mean are −1, 0 and +1. Their squares are 1, 0 and 1. Hence
Var(X) = (1 + 0 + 1)/3 = 2/3.
Now define Y = 3X + 5. The possible transformed values are 8, 11 and 14, still equally likely.
Their mean is (8 + 11 + 14)/3 = 11. The deviations from that mean are −3, 0 and +3. Squaring gives 9, 0 and 9.
Thus
Var(Y) = (9 + 0 + 9)/3 = 6.
Check the formula: a²Var(X) = 3²(2/3) = 9(2/3) = 6. It agrees exactly.
The mean formula also agrees: 3E(X) + 5 = 3(2) + 5 = 11.
Why the +5 vanished from the variance
Compare the original values 1, 2 and 3 with the transformed values 8, 11 and 14. The original mean is two; the transformed mean is eleven.
After accounting for the new mean, the distances are simply three times the old distances. The constant +5 moved *everyone* together and did not make values more spread out relative to one another.
To see this algebraically, subtract the transformed mean from any transformed observation:
(3X + 5) − (3E(X) + 5) = 3[X − E(X)].
The two +5 terms cancel. Squaring then produces a factor of 9. This is not a coincidence specific to the outcomes 1, 2 and 3; it works generally.
The derivation that makes the formula trustworthy
Take Y = aX + b. Its expectation is aE(X) + b.
By definition,
Var(Y) = E[(Y − E(Y))²].
Substitute both expressions:
Var(Y) = E[{(aX + b) − (aE(X) + b)}²].
The b terms cancel:
Var(Y) = E[{a(X − E(X))}²].
Pull the constant through the square:
Var(Y) = E[a²(X − E(X))²] = a² E[(X − E(X))²].
Therefore
Var(aX + b) = a²Var(X).
This proof is short enough to remember as a chain of meaning. The transformed distance from its mean is a times the original distance. Variance squares those distances, so a² appears.
The proof also explains why replacing a² with a is wrong even for a simple positive multiplier. The square comes from the definition itself.
Worked example 2: adding a constant changes the mean only
Let X have values 1, 2 and 3 with equal probabilities, as before. Define Y = X + 10.
The outcomes become 11, 12 and 13, and their mean becomes 12.
The deviations from the new mean are −1, 0 and +1, exactly the same as the original deviations from two.
Thus Var(Y) = 2/3, unchanged from Var(X). The formula confirms it because a = 1, so a²Var(X) = Var(X).
If you have test marks and every student receives a ten-mark adjustment, the class mean increases by ten. The spread of scores does not change because the distances between students’ marks remain the same.
That story makes the constant term b easy to interpret: it changes the location of the distribution, not its variability.
Worked example 3: a negative multiplier does not make negative variance
Keep X equally likely to be 1, 2 or 3, but define
Y = −2X + 7.
The possible Y values are 5, 3 and 1. Their mean is three.
Notice that the order of outcomes has been reversed: a larger X leads to a smaller Y. But the deviations from the new mean are +2, 0 and −2. Their squares are 4, 0 and 4.
So Var(Y) = (4 + 0 + 4)/3 = 8/3.
Check using the formula: a²Var(X) = (−2)²(2/3) = 4(2/3) = 8/3.
The mean formula gives E(Y) = −2(2) + 7 = 3.
A student who writes Var(Y) = −2Var(X) obtains a negative number, which should immediately trigger a reasonableness check: a variance cannot be negative.
Standard deviation: why the absolute value appears
Variance is measured in squared units because deviations are squared. Standard deviation is the positive square root of variance, returning to the units of the original variable.
If Var(Y) = a²Var(X), then
SD(Y) = √[a²Var(X)] = |a|SD(X).
The square root of a² is |a|, not simply a. That matters when a is negative.
For Y = −2X + 7, the standard deviation is twice the standard deviation of X, not negative twice it. Changing the direction of a scale does not create negative spread.
A good learning comparison is:
Mean: multiply by a and add b.
Standard deviation: multiply by |a|.
Variance: multiply by a².
Three related measures, three slightly different transformation rules. The difference is mathematically meaningful.
Worked example 4: a simple probability model with two outcomes
Let X be one when a particular event occurs, and zero otherwise. Suppose the event has probability p = 0.3.
Then X is a Bernoulli random variable with possible values zero and one. Its mean is E(X) = 0.3.
Its variance is p(1 − p) = 0.3(0.7) = 0.21.
Now let Y = 100X + 20. The transformed outcomes are 20 when the event does not occur and 120 when it does.
The mean becomes E(Y) = 100(0.3) + 20 = 50.
The variance becomes Var(Y) = 100²(0.21) = 2100.
That number is in squared units. The standard deviation is √2100, approximately 45.83 units.
The fixed addition of twenty changes the average but contributes nothing to the variance; the factor of one hundred has a squared effect on the variance.
Why a “fixed bonus” does not increase uncertainty
Imagine a fictional simple payout model where a game awards either zero or one point, with a certain probability. If every outcome receives a guaranteed five-point bonus, the expected total rises by five points.
But the difference between the two possible outcomes remains one point. The bonus is identical in every case, so it does not introduce extra uncertainty about the outcome relative to the new mean.
Now double the point value of the random result. The distance between the possible scores doubles, and the variance becomes four times larger.
This provides a practical distinction between changing the baseline and changing the scale. Baseline shifts are controlled by b; scale changes are controlled by a.
In real modelling, additional uncertainty may arise from other random terms. The simple linear transformation aX + b assumes a and b are fixed constants, not additional random variables.
A unit-conversion example: centimetres to metres
Suppose a random length X is measured in centimetres. Define Y = X/100 as the same length measured in metres.
Here a = 1/100 and b = 0.
The mean in metres is E(Y) = E(X)/100. That matches ordinary unit conversion.
But the variance in square metres is
Var(Y) = Var(X)/10000.
Why ten thousand? Because one metre contains one hundred centimetres, and variance measures squared length differences.
If the standard deviation in centimetres is ten, then the standard deviation in metres is 0.1. Squaring gives a variance of 0.01 square metres, consistent with the conversion.
This is a useful reasonableness check in scientific and data questions: variance should transform with the square of the measurement unit’s scaling factor.
The special case a = 0
Suppose Y = 0X + b. Then Y = b regardless of the value of X.
There is no randomness left in Y: every outcome produces the same constant value b.
Therefore E(Y) = b and Var(Y) = 0. The general formulas agree: aE(X) + b = b and a²Var(X) = 0.
This edge case is not merely a technical footnote. It makes the relationship between variance and spread exceptionally clear. A constant-valued random variable has no variation.
If a learner applies a formula and obtains non-zero variance for Y = b, the calculation is wrong.
Another formula: E(X²) minus the square of the mean
For a random variable with finite second moment, we can calculate variance as
Var(X) = E(X²) − [E(X)]².
This is often computationally useful, but the order of squaring matters. E(X²) is the average of the squared outcomes; [E(X)]² is the square of the average outcome.
For equally likely X = 1, 2, 3, we have E(X²) = (1 + 4 + 9)/3 = 14/3. The mean is two, so [E(X)]² = 4.
Thus Var(X) = 14/3 − 4 = 2/3.
A common wrong step is to assume E(X²) = [E(X)]². That would force variance to zero even when outcomes vary.
The two quantities differ precisely because there is spread around the mean.
Check the transformed variance with the second-moment formula
Take Y = 3X + 5, where X is again equally likely to be 1, 2 or 3.
The possible Y values are 8, 11 and 14. The mean is eleven.
Calculate E(Y²) = (8² + 11² + 14²)/3 = (64 + 121 + 196)/3 = 381/3 = 127.
Then Var(Y) = E(Y²) − [E(Y)]² = 127 − 121 = 6.
This agrees with a²Var(X) = 9(2/3) = 6. We have now checked the result by three routes: squared deviations, the transformation formula, and the second-moment formula.
Independent checks are useful in H2 Mathematics because different routes can reveal different careless errors. The method chosen should fit the information given in the question.
What changes when two random variables are added?
The formula Var(aX + b) applies to one random variable scaled and shifted by fixed constants.
When two random variables X and Z are added, the variance requires more care. If X and Z are independent, then
Var(X + Z) = Var(X) + Var(Z).
But independence is essential to that familiar simple addition formula. If the variables are dependent, the relationship can be different.
For example, if Z = X, then X + Z = 2X. Its variance is 4Var(X), not 2Var(X), unless Var(X) happens to be zero. The two terms are perfectly dependent because they are the same variable.
Do not confuse a fixed constant b with a second random variable. Adding a constant adds no new uncertainty; adding a random quantity may alter the spread.
For a fuller treatment of variance in independent sums and differences, use the JC2 H2 Mathematics guide to why variances add when values subtract.
A common exam error: forgetting which number is a constant
Consider Y = 4X − 7. A student correctly writes E(Y) = 4E(X) − 7, but then writes Var(Y) = 4Var(X) − 7.
There are two errors in the variance line. First, the coefficient four must be squared, giving sixteen. Second, the added fixed constant −7 must disappear from the variance.
The correct result is
Var(4X − 7) = 16Var(X).
An effective repair starts by asking why the −7 should disappear. If every possible Y is shifted down by seven, its mean shifts down by seven too, leaving each difference from its mean unchanged. Then ask why the four becomes sixteen: each difference is multiplied by four and subsequently squared.
The student now has an explanation rather than only a corrected formula.
What about a normal distribution?
Suppose X is normally distributed with mean μ and variance σ². A linear transformation Y = aX + b, with a ≠ 0, is also normally distributed, with mean aμ + b and variance a²σ².
This is consistent with the general expectation and variance formulas. The distribution family is preserved by a non-degenerate affine transformation, but its location and spread change.
For example, if X has mean 70 and variance 25, and Y = 2X + 10, then Y has mean 150 and variance 100. Its standard deviation is ten, double the original five.
If a = 0, Y collapses to a constant rather than an ordinary non-degenerate normal distribution. Keep that special case in mind.
The key point for this guide is that the variance formula does not depend on normality. It is a more general property of linear transformations.
Ten short practice questions with answers
Question 1. E(X) = 4 and Var(X) = 3. Find E(2X + 5). Answer: 2(4) + 5 = 13.
Question 2. For the same X, find Var(2X + 5). Answer: 2²(3) = 12.
Question 3. Find Var(X − 9). Answer: Var(X), unchanged.
Question 4. If Var(X) = 7, find Var(−3X + 2). Answer: (−3)²(7) = 63.
Question 5. If SD(X) = 4, find SD(−2X + 8). Answer: |−2|×4 = 8.
Question 6. If E(X) = 10, find E(−X + 3). Answer: −10 + 3 = −7.
Question 7. If Y = 5 is constant, what is Var(Y)? Answer: 0.
Question 8. For equal-probability outcomes X = 1, 2, 3, find E(X²). Answer: 14/3.
Question 9. For those outcomes, find Var(X). Answer: 14/3 − 2² = 2/3.
Question 10. X is measured in centimetres with variance 400 cm². What is the variance of X/100 measured in metres? Answer: 400/10000 = 0.04 m².
The questions deliberately alternate between mean, variance and standard deviation. A student who gives the same scaling rule for all three has not yet distinguished their meanings.
A five-minute tutoring diagnostic
I would start with three simple equally likely values: 1, 2 and 3. Ask the student for the mean and the three deviations. No formula beyond ordinary averaging is needed.
Then add ten to each value. Ask whether the deviations from the new mean have changed. They have not, so the variance has not changed.
Next multiply the original values by three. The deviations triple, so their squares become nine times larger. The variance must be nine times larger.
Finally change the multiplier to negative three. The order of outcomes reverses, but the squared deviations still multiply by nine. A learner who explains this correctly has built the idea that negative spread is impossible.
Only after these examples do we write E(aX + b) and Var(aX + b) symbolically.
An independent three-stage learning route
Stage one: meaning of the measures
Calculate mean, variance and standard deviation for a three-outcome distribution. Explain in words what each measures and why variance involves squares.
Stage two: linear transformations
Apply shifts, positive scales and negative scales. Predict each result first, then calculate. Check the a = 0 edge case.
Stage three: model interpretation
Use a simple two-outcome payout or a unit conversion. Require the learner to state whether the transformed value is a random variable, which constants are fixed, and how the unit of variance changes.
The final goal is not to memorise a formula for every numerical question. It is to understand which changes move a distribution and which make it more or less spread out.
Frequently asked questions
Why is Var(aX + b) equal to a²Var(X)?
Because deviations from the transformed mean equal a times the original deviations, and variance squares those deviations.
Why doesn’t b appear in the variance formula?
Adding b shifts every outcome and the mean by the same amount, leaving deviations from the mean unchanged.
Can variance ever be negative?
No. It is an expectation of squared deviations and therefore non-negative.
What if a is negative?
The outcome order reverses, but variance is multiplied by a², which is non-negative. Standard deviation scales by |a|.
Do mean and standard deviation obey the same rule?
No. Mean obeys E(aX + b) = aE(X) + b; standard deviation obeys SD(aX + b) = |a|SD(X).
Must X be normally distributed?
No. The linear-transformation formulas for expectation and variance hold for any random variable with the required finite moments.
Are these rules part of H2 Mathematics?
Yes. The 2027 H2 Mathematics syllabus 9758 includes expectations and variances of discrete random variables and the transformation rules for mean and variance in probability-and-statistics problems. Specific teaching order varies between colleges.
Related reading and an honest enquiry route
For the full distinction between one random variable and adding separate variables, read why independent variances add even when values subtract. To revisit the meaning of independence, use mutually exclusive versus independent events. The official 2027 H2 Mathematics syllabus 9758 provides the examination scope, and the Bukit Timah H2 Mathematics statistics overview offers broader context.
The immutable eduKateSG Secondary 1 Mathematics small-group reference illustrates the practice of diagnosing, explaining and checking independent understanding. Its Secondary 1 class arrangements are not automatically JC1 arrangements. For current enquiries, see Bukit Timah JC1 tuition.
The central idea is easy to retain: moving all values changes where the distribution sits; multiplying them changes the distances between values; squaring those distances changes variance by the square of the multiplier.