Infer linear coefficients from moment conditions
Solve the squared scale equation, retain both signs and use independent mean information to test candidate transformations.
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
- Solve quadratic equations and simultaneous conditions.
- Use affine and independent normal moments.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which solutions follow from a² = 4?
What you will learn
- Retain signed branches when inferring a coefficient from variance.
- Check every candidate against original moment conditions.
Variance alone does not choose the scale sign
For X ~ N(4,9), U = aX + b with mean ten and variance thirty-six gives 9a² = 36, so a = 2 or a = −2. Its mean equation is 4a + b = 10, producing (a,b) = (2,2) or (−2,18). Both transforms follow N(10,36), though one preserves order and one reverses it.
If the problem separately states a > 0, choose (2,2). Do not silently discard the negative branch because variance must be non-negative: negative scale is perfectly valid. A zero target variance would instead force a = 0 for this positive-variance input.
Use all independent-combination conditions
Now let independent X ~ N(4,9), Y ~ N(2,4), with W = aX + Y having mean ten and variance forty. The mean equation 4a + 2 = 10 gives a = 2; the variance equation 9a² + 4 = 40 verifies it.
The variance equation alone also admits a = −2, but its mean would be −6, not ten. Check the original pair of conditions. The resulting model is N(10,40) because independent normal inputs were supplied, not because two moments by themselves force normality.
Pause and explain
For X ~ N(4,9), which negative-scale transform has mean 10 and variance 36?
Worked example
For X ~ N(4,9), find all (a,b) making aX + b have mean 10 and variance 36.
Show the worked solution
- 9a² = 36 gives a = ±2.
- 4a + b = 10 gives b = 10 − 4a.
- Thus (2,2) and (−2,18) both satisfy the conditions.
Answer (a,b) = (2,2) or (−2,18).
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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.
Find all a,b for X ~ N(4,9) and aX + b with mean 10, variance 36.
Give me a hint
Solve the squared coefficient before substituting into the mean equation.
Compare my reasoning
- a² = 4 gives a = ±2.
- b = 10 − 4a.
- The pairs are (2,2) and (−2,18).
(2,2) or (−2,18).
Look for these in your work
- Keep both variance roots.
- Verify both moment equations.
For that problem, add the condition a > 0. State the unique transform.
Give me a hint
Use the extra restriction after retaining both algebraic roots.
Compare my reasoning
- The candidates have a = 2 and a = −2.
- Only a = 2 satisfies positivity.
- Then b = 2, giving 2X + 2 ~ N(10,36).
a = 2, b = 2; transform 2X + 2.
Look for these in your work
- Apply the explicit sign restriction.
- Check the resulting moments.
Independent X ~ N(4,9), Y ~ N(2,4). Infer a if aX + Y has mean 10 and variance 40.
Give me a hint
Use the mean equation and test it in the independent variance equation.
Compare my reasoning
- 4a + 2 = 10 gives a = 2.
- 9a² + 4 = 40 verifies this value.
- The variance-only root −2 fails the mean; the result is N(10,40).
a = 2; W ~ N(10,40).
Look for these in your work
- Use every supplied condition.
- Justify distribution shape from the input models.
Common mistakes
- Variance gives a squared scale, often two signs.
- Check the original mean condition.
- Moments alone do not establish normality.
What can you explain now?
Does a² = 4 force a = 2?
Compare with the explanation
No; a = −2 is also a solution.
Use any separately supplied sign restriction or mean condition.
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Make it stick
Solve squared scales with both signs, apply extra conditions and verify every candidate in the original equations.
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.