Combine Poisson exposures without double counting shared events
Scale component rates separately and distinguish a total of independent observations from a sum that repeats an overlapping count.
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
- Convert event rates into window parameters.
- Distinguish disjoint observations from overlapping counts.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What parameter describes twenty minutes at six per hour?
What you will learn
- Scale and add independent exposure parameters.
- Use disjoint pieces to identify repeated overlap and its variance.
Scale each observation before adding
In a stipulated homogeneous process of six per hour, disjoint windows of twenty and ten minutes have Po(2) and Po(1) counts. Their sum is Po(3), so its zero probability is e⁻³ ≈ 0.0498. This also matches the thirty-minute union window.
For two separate independent observations, one at four per hour over thirty minutes and one at six per hour over twenty minutes, each parameter is two. Their total follows Po(4); its count-three mass is (64/6)e⁻⁴ ≈ 0.1954. The exposures need not have equal durations, but their component model assumptions and independence must be stated.
A summed overlap repeats the shared count
Let A observe 0–20 minutes and B observe 10–30 at six per hour. Split into independent Po(1) piece counts U,V,W. Then A + B = U + 2V + W, because the middle piece is counted twice. Its mean is four and variance 1 + 4 + 1 = 6.
A blind Po(4) conclusion would require variance four and therefore fails. The count of distinct events in the union is instead U + V + W ~ Po(3). Summing observation totals and counting unique events are different quantities. Draw the windows before choosing a distribution.
Pause and explain
For A + B = U + 2V + W with independent Po(1) pieces, what is its variance?
Worked example
At six per hour, combine disjoint twenty-minute and ten-minute counts.
Show the worked solution
- The component parameters are two and one.
- Disjoint counts are independent under the process.
- Their sum follows Po(3), with zero mass e⁻³ ≈ 0.0498.
Answer Po(3); P(total zero) ≈ 0.0498.
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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.
At six per hour, combine disjoint twenty- and ten-minute counts and calculate P(total zero).
Give me a hint
Scale both windows before using independence.
Compare my reasoning
- Parameters are two and one.
- The sum follows Po(3).
- P(0) = e⁻³ ≈ 0.0498.
Po(3); P(0) ≈ 0.0498.
Look for these in your work
- Convert both durations.
- Justify disjoint-count independence.
Separate independent observations use four per hour for thirty minutes and six per hour for twenty minutes. Find their sum distribution and P(total three).
Give me a hint
Each observation needs its own parameter.
Compare my reasoning
- The parameters are 4 × 0.5 = 2 and 6 × 1/3 = 2.
- The independent sum is Po(4).
- P(3) = (64/6)e⁻⁴ ≈ 0.1954.
Po(4); probability ≈ 0.1954.
Look for these in your work
- Scale the different exposures.
- Add parameters only after checking independence.
At six per hour, A observes 0–20 and B 10–30 minutes. Find moments of A + B and the distribution of unique union events.
Give me a hint
Split into three disjoint Po(1) pieces and identify the repeated middle count.
Compare my reasoning
- A + B = U + 2V + W.
- Its mean is four and variance six, so it is not Po(4).
- Unique union count is U + V + W ~ Po(3).
Summed observations: mean 4, variance 6; unique union events: Po(3).
Look for these in your work
- Identify the double count.
- Distinguish the two random quantities.
Common mistakes
- Scale each rate in consistent units.
- Do not add overlapping marginal counts as independent.
- Union events and summed observations can be different totals.
What can you explain now?
How often does the shared middle count enter A + B?
Compare with the explanation
Twice.
Its variance contribution is multiplied by four.
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Make it stick
Convert each exposure, draw any overlap and decide whether you are summing observations or counting unique events.
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.