Distinguish independent copies from repeating one observation
Compare two new observations with doubling a single realized value and explain their different spread.
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 squared affine variance.
- Use independent additive variance.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What distinguishes X₁ + X₂ from 2X in this example?
What you will learn
- Distinguish X₁ + X₂ from 2X.
- Calculate moments and normal distributions for each mechanism.
Same expectation, different randomness
If X has mean five and variance four, two independent copies X₁,X₂ give S = X₁ + X₂ with mean ten and variance eight. Doubling one observation gives D = 2X with mean ten and variance sixteen. Both have the same expectation but the second repeats the same deviation twice.
Independent does not mean identical observed values. Identically distributed does not mean independent. A shared measurement, reused input or common deterministic equality can make observations dependent; the description of the sampling mechanism matters.
Apply the actual relation
X₁ − X₂ has mean zero and variance eight under independence, while X − X is identically zero. Treating two labels for the same observation as independent would give a fictitious spread.
If X ~ N(5,4), independent copies sum to N(10,8), whereas 2X follows N(10,16). Their SDs are √8 and four. This exact normal result follows from the stipulated distributions and independence, not merely from the equal moments.
Pause and explain
If Var(X) = 4, which variance belongs to 2X?
Worked example
Compare X₁ + X₂ and 2X when X has mean 5 and variance 4 and the copies are independent.
Show the worked solution
- Both means are ten.
- Independent-copy sum has variance eight.
- Repeated-value doubling has variance sixteen.
Answer Same mean 10; variances 8 and 16 respectively.
Try a topic test
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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.
Compare moments of X₁ + X₂ and 2X for mean 5, variance 4 and independent copies.
Give me a hint
Identify whether the same realization is reused.
Compare my reasoning
- Both means equal ten.
- Independent-copy sum has variance eight.
- 2X has variance sixteen.
Means 10; variances 8 and 16 respectively.
Look for these in your work
- Distinguish the mechanisms.
- Apply the correct variance rule.
Compare X₁ − X₂ with X − X for the same setup.
Give me a hint
Simplify the repeated-value difference separately.
Compare my reasoning
- Independent difference has mean zero.
- Its variance is eight.
- The repeated difference is always zero with variance zero.
Independent difference: mean 0, variance 8; repeated difference: constant 0.
Look for these in your work
- Use independent squared contributions.
- Recognise deterministic cancellation.
Now X ~ N(5,4). State distributions for the independent-copy sum and 2X.
Give me a hint
Normal combination and affine transformation use different variance calculations.
Compare my reasoning
- Independent sum has mean ten and variance eight.
- Thus X₁ + X₂ ~ N(10,8).
- 2X has mean ten and variance sixteen, giving N(10,16).
Independent sum N(10,8); doubled value N(10,16).
Look for these in your work
- Justify normality from the stated models.
- Keep the two spreads distinct.
Common mistakes
- A repeated observation is not a fresh independent copy.
- Equal marginal distributions do not imply independence.
- Same mean does not imply same variability.
What can you explain now?
Does 2X have variance 2Var(X)?
Compare with the explanation
No; it has 4Var(X).
The same random deviation is multiplied twice, not independently resampled.
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Make it stick
Read the observation mechanism: fresh independent copies add variance, while a repeated value uses a squared scale.
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.