Add independent Poisson counts with exact distribution closure
Use the exact Poisson-sum result and distinguish probabilities of a total from probabilities of one particular contributing pair.
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
- Use exact Poisson masses.
- Multiply independent events and condition on a positive event.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which parameter matches the independent sum of Po(2) and Po(3)?
What you will learn
- State Po(λ₁ + λ₂) for independent Poisson counts.
- Calculate total-count and conditional component probabilities.
Add the parameters for independent unweighted counts
Let independent X ~ Po(2), Y ~ Po(3). Their sum S follows exactly Po(5). Mean and variance are each five, matching the separate independent moment calculations. This is distribution closure, not the rare-binomial approximation from section 6.1.
P(S = 2) = e⁻⁵5²/2! = 12.5e⁻⁵ ≈ 0.0842. The event includes component pairs (0,2),(1,1),(2,0). Their independent masses have coefficients 4.5,6,2 times e⁻⁵, adding to 12.5e⁻⁵. Multiplying P(X = 2)P(Y = 2) would instead describe total four.
Keep conditions and parameter limits
P(S ≥ 1) = 1 − e⁻⁵ ≈ 0.9933. Given S = 1, P(X = 1 | S = 1) has numerator P(X = 1,Y = 0) = 2e⁻⁵ and denominator P(S = 1) = 5e⁻⁵, giving 2/5.
The counts must be independent and their parameters finite and non-negative. A zero-parameter component is a constant zero, so adding it leaves the other count unchanged. If the sum parameter is zero, conditioning on a positive sum is undefined. Weighted sums and differences do not have this general Poisson closure.
Pause and explain
Which pairs contribute to X + Y = 2?
Worked example
For independent Po(2), Po(3), find the sum model and its probability of two.
Show the worked solution
- Add parameters to get Po(5).
- The count-two mass is 25e⁻⁵/2.
- Evaluate ≈ 0.0842.
Answer Po(5); P(S = 2) ≈ 0.0842.
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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.
Independent X ~ Po(2), Y ~ Po(3). State S = X + Y’s model and P(S = 2).
Give me a hint
Use exact sum closure before the mass formula.
Compare my reasoning
- S ~ Po(5).
- P(S = 2) = 12.5e⁻⁵.
- The value is approximately 0.0842.
Po(5); probability ≈ 0.0842.
Look for these in your work
- State independence.
- Label exact model closure.
For the same sum, find the probability of at least one event.
Give me a hint
Use its zero-count complement.
Compare my reasoning
- S has parameter five.
- P(S = 0) = e⁻⁵.
- P(S ≥ 1) = 1 − e⁻⁵ ≈ 0.9933.
1 − e⁻⁵ ≈ 0.9933.
Look for these in your work
- Use the sum parameter.
- Preserve the whole upper event.
For those independent counts, find P(X = 1 | S = 1).
Give me a hint
The numerator requires X = 1 and Y = 0.
Compare my reasoning
- The joint numerator is 2e⁻⁵.
- The sum-one mass is 5e⁻⁵ > 0.
- The ratio is 2/5.
2/5.
Look for these in your work
- Intersect the condition correctly.
- Divide by the sum-event probability.
Common mistakes
- Independence is essential to the stated sum theorem.
- A total event includes all contributing pairs.
- Poisson-sum closure does not cover every weighted difference.
What can you explain now?
Does S = 2 require both components equal two?
Compare with the explanation
No; that would total four.
The total-two pairs are (0,2),(1,1),(2,0).
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Make it stick
Check independent Poisson inputs, add their parameters, and distinguish the total event from a particular contributing pair.
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