Use exponential density areas and check divergent moments
Integrate a single-interval exponential density and contrast its finite moments with a valid heavy-tail model.
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
- Integrate exponentials and use integration by parts.
- Evaluate logarithmic limits.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which integral checks the mean of density 1/x² on x ≥ 1?
What you will learn
- Calculate exponential areas and moments with Paper 3 calculus.
- Recognise a valid density whose mean diverges.
An infinite interval can have finite moments
For f(x) = 2e^(−2x) on x ≥ 0, total area is one and P(X > t) = e^(−2t) for t ≥ 0. The median is ln(2)/2. A threshold containing 95% of values satisfies e^(−2t) = 0.05, so t = ln(20)/2 ≈ 1.4979.
Integration by parts gives E(X) = 1/2 and E(X²) = 1/2, hence variance 1/4. The exponential factor drives the boundary terms x e^(−2x) and x² e^(−2x) to zero. The formulae are derived for this stipulated density; recognising a named exponential distribution is not an extra syllabus requirement.
Unit area does not guarantee a finite mean
The density 1/x² on x ≥ 1 also has total area one. But its mean integrand is 1/x, whose integral from one to R is ln R and diverges. There is no finite mean or finite variance to obtain by ordinary moment subtraction.
This contrast is a convergence check for infinite-support densities, not a required classification of heavy-tail families. Do not subtract two divergent expressions or assign an artificial cutoff without stating a new truncated and re-normalised model.
Pause and explain
What follows when the first-moment integral diverges?
Worked example
For density 2e^(−2x), find P(X > 1) and its median.
Show the worked solution
- Upper area at one is e⁻².
- Set e^(−2m) = 1/2.
- m = ln(2)/2.
Answer Tail e⁻² ≈ 0.1353; median ln(2)/2.
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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.
For 2e^(−2x) on x ≥ 0, find P(X > 1) and median.
Give me a hint
Use tail area e^(−2t).
Compare my reasoning
- Integrate 2e^(−2x) from one to infinity to obtain e⁻².
- Median solves e^(−2m) = 1/2.
- Taking logarithms gives m = ln(2)/2, a non-negative supported location.
Tail e⁻² ≈ 0.1353; median ln(2)/2.
Look for these in your work
- Use an infinite-limit area.
- Solve a half-area equation.
Find mean, second moment, variance and a 95% upper threshold for this density.
Give me a hint
Use integration by parts and a 0.05 upper tail.
Compare my reasoning
- Mean = 1/2; second moment = 1/2.
- Variance = 1/4.
- Threshold is ln(20)/2.
Mean 1/2; second moment 1/2; variance 1/4; threshold ≈ 1.4979.
Look for these in your work
- Check moment limits.
- Use the correct tail probability.
Check whether density 1/x² on x ≥ 1 has unit area and a finite mean.
Give me a hint
Normalisation and first moment use different powers.
Compare my reasoning
- ∫₁∞x⁻² dx = 1.
- The mean is lim(R→∞) ln R.
- It diverges, so no finite mean or finite variance is available.
Valid density with area 1, but divergent mean; no finite moment-based variance.
Look for these in your work
- Separate validity from convergence.
- Do not subtract divergent quantities.
Common mistakes
- Do not substitute infinity into an unevaluated expression.
- Check moment boundary terms.
- Do not subtract divergent moments.
What can you explain now?
Does total density area one guarantee finite expectation?
Compare with the explanation
No.
The valid 1/x² density has a divergent mean integral.
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Make it stick
Check area, mean and second moment as separate improper integrals.
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.