Add squared variance contributions for independent variables
Use independence to add variance terms, including for differences, and identify when that rule is unavailable.
Section 6.2 · Linear combinations of random variables
Exam scope and assessment notes
Probability & Statistics 2 (Paper 6, A Level) · Mathematics 9709, 2026–2027, section 6.2. Paper 6 builds on Paper 5 and assumes Paper 3 calculus. Use E(aX + b) = aE(X) + b and Var(aX + b) = a²Var(X). Expectation is linear without independence, when the relevant moments exist. For independent X and Y, Var(aX + bY) = a²Var(X) + b²Var(Y); negative coefficients still give positive squared contributions. A constant changes the mean but not variance. SD scales by |a|, not a². Affine transforms of normal variables and linear combinations of independent normal variables are normal when the resulting variance is positive; zero variance is a constant, so handle it directly rather than standardising by zero. Independent Poisson counts have an exact Poisson sum with the parameters added; a scaled count or difference is not generally Poisson. Equal marginal distributions do not establish independence. Proofs are not required. Normal probabilities here use quoted cumulative values when supplied and otherwise unrounded standardisation with numerical areas. Exact distribution results are conditional on the stipulated models; real applications need evidence. Continuous density functions, sampling, estimation and hypothesis testing belong to later sections. Written practice is self-checked; cumulative assessment and human subject-teacher review remain pending.
Before you begin
- Square negative and fractional coefficients.
- Distinguish independent variables from repeated values.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For independent X,Y, how do variance terms combine in X − Y?
What you will learn
- Calculate a²Var(X) + b²Var(Y) under independence.
- Explain why subtracting variables does not subtract independent variances.
Square before adding
For independent X,Y with variances four and nine, Var(2X − 3Y + 7) = 2² × 4 + (−3)² × 9 = 97. The fixed seven contributes no variance. The standard deviation is √97, not the sum of the separate SDs.
For X − Y, the coefficients are one and minus one; both squares are one, so variance is thirteen. For 3X + 2Y it is 9 × 4 + 4 × 9 = 72. Mean uses signed coefficients, but independent variance contributions add after squaring.
Check the dependence condition first
The additive variance rule here assumes independence. When Y = X and Var(X) = 4, Var(X − Y) = 0, while blindly adding would give eight. A known deterministic relation must be used before calculation.
Do not infer independence from equal marginal means, different names, or a claim that two observations are separate. The mechanism or question must justify it. Covariance formulae are not required for this section; where the independent rule is unavailable, say what further information is needed or use a supplied relationship.
Pause and explain
Why does Y = X prevent using the independent variance sum for X − Y?
Worked example
Independent X,Y have variances 4 and 9. Find Var(2X − 3Y + 7).
Show the worked solution
- Independence permits additive variance contributions.
- Square coefficients to obtain 4 × 4 and 9 × 9.
- Add to get 97; the constant contributes zero.
Answer Variance 97; SD √97.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Independent variables have variances 4 and 9. Find variance and SD of X − Y.
Give me a hint
Both coefficients square to one.
Compare my reasoning
- Independence is stipulated.
- Variance = 4 + 9 = 13.
- SD = √13 ≈ 3.6056.
Variance 13; SD √13.
Look for these in your work
- State independence.
- Add variances before taking the square root.
For those independent variables, find variance and SD of 3X + 2Y.
Give me a hint
Square each coefficient separately.
Compare my reasoning
- The X contribution is 9 × 4 = 36.
- The Y contribution is 4 × 9 = 36.
- Total variance is 72 and SD √72.
Variance 72; SD √72.
Look for these in your work
- Use both squared scales.
- Do not add separate SDs.
Instead Y = X with Var(X) = 4. Compare actual Var(X − Y) with a blind independent calculation.
Give me a hint
Simplify using the stated equality.
Compare my reasoning
- The difference is always zero.
- Its actual variance is zero.
- The unjustified independent sum would give eight; independence is false.
Actual variance 0; blind independent result 8 is invalid.
Look for these in your work
- Use the dependence relationship.
- Explain which assumption fails.
Common mistakes
- Differences still add independent variances.
- Square whole signed coefficients.
- Do not apply the rule without independence.
What can you explain now?
Why is Var(X − Y) = 13 here, not −5?
Compare with the explanation
Both independent variance coefficients are squared.
Variance cannot be obtained by subtracting these two variances.
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Make it stick
Check independence, square each coefficient, add variance contributions and take one final square root for SD.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.