Combine expectations without assuming independence
Use linearity for sums and differences while separating it from the extra conditions needed for variance.
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
- Use signed coefficients.
- Calculate separate expectations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which term does a fixed +5 contribute to expectation?
What you will learn
- Calculate E(aX + bY + c) without independence.
- Avoid applying an independent variance rule to dependent quantities.
Keep coefficients and constants in the mean
If E(X) = 7 and E(Y) = 3, E(2X − 4Y + 5) = 14 − 12 + 5 = 7. Linearity of expectation does not require independence, provided the relevant expectations exist. Use each signed coefficient once.
For the same means, E(2X + Y) = 17 and E(3X − 2Y + 4) = 19. Expected values add algebraically; they need not be possible individual outcomes. This mean result alone does not supply a distribution or variance.
Do not transfer this permission to variance
If Y = X, the difference X − Y is identically zero and its expectation is zero. Linearity still works although the two quantities are completely dependent.
Its variance is also zero by the actual relation, rather than the sum of two positive variances. Independence is needed for the syllabus’s separate additive variance result. Check which result you are using before deciding whether dependence matters.
Pause and explain
Does linearity of expectation require independent variables?
Worked example
E(X) = 7, E(Y) = 3. Find E(2X − 4Y + 5) without assuming independence.
Show the worked solution
- Multiply each mean by its signed coefficient.
- Add 14 − 12 and the fixed five.
- The expectation is seven; no independence condition was needed.
Answer E(2X − 4Y + 5) = 7.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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.
E(X) = 7, E(Y) = 3. Find E(2X + Y).
Give me a hint
Multiply each separate mean by its own coefficient.
Compare my reasoning
- E(2X) = 14.
- E(Y) = 3.
- Add to get seventeen, without requiring independence.
17; independence is not required.
Look for these in your work
- Use both coefficients.
- State the correct condition.
For those means, calculate E(3X − 2Y + 4).
Give me a hint
Preserve the negative coefficient and fixed shift.
Compare my reasoning
- Three times seven is twenty-one.
- Subtract twice three and add four.
- The result is nineteen.
19.
Look for these in your work
- Retain signed coefficients.
- Include the constant.
If Y = X, calculate expectation and variance of X − Y. Explain why the independent variance rule is inappropriate.
Give me a hint
Simplify the random variable using its known relation.
Compare my reasoning
- X − Y = X − X = 0 for every outcome.
- Its expectation and variance are zero.
- The independent additive variance rule is unavailable because Y repeats X.
Mean 0; variance 0; the variables are dependent.
Look for these in your work
- Use the actual relation.
- Distinguish mean and variance conditions.
Common mistakes
- Expectation linearity does not need independence.
- Mean alone does not specify variance or shape.
- Dependent repeated variables need their actual relation.
What can you explain now?
Does E(X + Y) = E(X) + E(Y) require independence?
Compare with the explanation
No, when the expectations exist.
The additive variance rule has a different condition.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Add signed expected values whether independent or dependent, then check variance assumptions separately.
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.