Infer density parameters and reject invalid moment solutions
Solve normalisation and mean equations together, then check that the proposed density stays non-negative.
Section 6.3 · Continuous random variables
Exam scope and assessment notes
Probability & Statistics 2 (Paper 6, A Level) · Mathematics 9709, 2026–2027, section 6.3. Paper 6 builds on Paper 5 and assumes Paper 3 calculus. A probability density is non-negative and has total area one over a single interval; its support may be infinite. Density height is not probability, and individual points have probability zero. Calculate interval probabilities, median and other percentiles directly from density areas. Explicit cumulative distribution function knowledge is not required by this section. Calculate E(X) and E(X²) by integration, then Var(X) = E(X²) − E(X)² when the needed moments are finite. Infinite-support densities require convergent improper integrals; a valid density need not have finite moments. Only single-interval densities are used here, with zero density outside the stated support. Model conclusions are conditional on the stipulated distribution; examples do not establish a real population model. 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 simultaneous equations.
- Integrate a linear density and its moments.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which equation normalises ax + b on [0,1]?
What you will learn
- Determine two density coefficients from area and mean.
- Reject algebraic solutions that violate density conditions.
Two constraints determine two coefficients
For f(x) = ax + b on [0,1], normalisation gives a/2 + b = 1. A mean of 7/12 gives a/3 + b/2 = 7/12. Substituting b = 1 − a/2 leaves 1/2 + a/12 = 7/12, so a = 1 and b = 1/2.
The endpoint heights are 1/2 and 3/2, so this solution is valid. Its second moment is a/4 + b/3 = 5/12, giving variance 5/12 − 49/144 = 11/144.
Algebra alone is not enough
In this family endpoint non-negativity gives −2 ≤ a ≤ 2, since b = 1 − a/2. The feasible means are therefore 1/3 through 2/3. A requested mean of 5/6 would force a = 4 and b = −1, giving negative density near zero.
Changing the requested mean to 5/12 gives a = −1 and b = 3/2, a valid decreasing density. Its variance is also 11/144 by reflection, but its mean differs.
Pause and explain
What must follow a solution of density parameter equations?
Worked example
Find a,b for ax + b on [0,1] with mean 7/12.
Show the worked solution
- Use a/2 + b = 1.
- Use a/3 + b/2 = 7/12.
- Solve and check both endpoint heights.
Answer a = 1, b = 1/2.
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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.
Find a,b for ax + b on [0,1] with mean 7/12.
Give me a hint
Use area one and weighted area 7/12.
Compare my reasoning
- a/2 + b = 1.
- a/3 + b/2 = 7/12.
- a = 1, b = 1/2; both endpoint heights are positive.
a = 1, b = 1/2; valid density.
Look for these in your work
- Solve both equations.
- Check full-interval non-negativity.
Find variance for this solved density.
Give me a hint
Calculate a/4 + b/3 before subtracting the mean square.
Compare my reasoning
- Integrate x²(x + 1/2) from zero to one to obtain E(X²) = 1/4 + 1/6 = 5/12.
- Variance = 5/12 − (7/12)².
- Use a common denominator: 60/144 − 49/144 = 11/144.
11/144.
Look for these in your work
- Use exact moments.
- Retain a non-negative result.
Can a density ax + b on [0,1] have mean 5/6?
Give me a hint
Solve, then inspect the endpoint heights.
Compare my reasoning
- The equations force a = 4 and b = −1.
- Density is negative near zero.
- The admissible mean range is [1/3,2/3].
No valid density in this family; a = 4, b = −1 is inadmissible.
Look for these in your work
- Distinguish algebraic from admissible solutions.
- Explain the violated density property.
Common mistakes
- Use both normalisation and the given moment.
- Check density validity after solving.
- A feasible mean range is specific to this family.
What can you explain now?
Is every mean between zero and one possible in this linear family?
Compare with the explanation
No.
Non-negativity narrows the mean to [1/3,2/3].
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Make it stick
Density parameters must satisfy both equations and probability conditions.
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