Check density non-negativity and parameter bounds
Reject signed curves that have unit integral but are negative on part of the interval.
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
- Compare a linear function at its endpoints.
- Normalise an integral.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Can a probability density be negative on an interval?
What you will learn
- Check the full interval for non-negativity.
- Find admissible density parameter values.
Two conditions must hold together
The family f(x) = 1 + a(2x − 1) on [0,1] has integral one for every a, because the variable term has integral zero. This does not make every a valid.
Its endpoint heights are 1 − a and 1 + a. A linear function is non-negative throughout this interval exactly when both are non-negative, giving −1 ≤ a ≤ 1. The endpoint test works here because the density is linear; for other curves, check interior extrema too.
A negative region cannot be ignored
At a = 2 the height at zero is −1 and the curve remains negative on [0,0.25). Its integral over that interval is −1/8, an impossible probability. Clipping the negative part changes total area and creates a different model needing new normalisation.
At a = 0.6, the valid primitive is x + 0.6(x² − x). The probability below 0.5 is 0.35. Values at isolated endpoints do not alter probabilities, but negativity on an interval is fatal.
Pause and explain
Which a values make this linear family valid?
Worked example
Find admissible a for 1 + a(2x − 1) on [0,1].
Show the worked solution
- Its total area is always one.
- Endpoint heights are 1 − a and 1 + a.
- Require both to be non-negative.
Answer −1 ≤ a ≤ 1.
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.
Find all valid a for density 1 + a(2x − 1) on [0,1].
Give me a hint
Check integral and both endpoint heights.
Compare my reasoning
- The integral equals one.
- 1 − a ≥ 0 and 1 + a ≥ 0.
- Intersect these bounds.
−1 ≤ a ≤ 1.
Look for these in your work
- Check both properties.
- Retain endpoint cases.
For a = 0.6, calculate P(X < 0.5).
Give me a hint
Integrate the complete linear expression.
Compare my reasoning
- Primitive is x + 0.6(x² − x).
- At 0.5 it equals 0.5 − 0.15.
- Subtract its zero value at the lower support bound to obtain 0.35.
0.35.
Look for these in your work
- Use a valid parameter.
- Check probability range.
At a = 2, integrate over [0,0.25] and assess the model.
Give me a hint
A negative interval exposes an invalid density.
Compare my reasoning
- Density is 4x − 1.
- Integral is [2x² − x]₀⁰·²⁵ = −1/8.
- This cannot be a probability.
−1/8; the proposed density is invalid.
Look for these in your work
- Do not accept unit total area alone.
- Explain the negative region.
Common mistakes
- Unit signed area does not prove validity.
- Check interior minima for nonlinear curves.
- Do not silently clip a proposed density.
What can you explain now?
Can a negative interval represent probability?
Compare with the explanation
No.
Non-negativity and unit area are separate checks.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
A density needs both non-negative values and unit total area.
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.