Check distribution claims for weighted counts and dependent inputs
Use support, moments and stated relationships to reject unsupported normal or Poisson combination claims.
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
- Know normal and Poisson closure conditions.
- Compare moments and possible values.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which quantity has exact Poisson closure for independent Poisson inputs?
What you will learn
- Reject a generic Poisson claim for scaled counts or differences.
- Use supplied dependence relations instead of unavailable independent theorems.
A weighted count is not automatically Poisson
For independent X ~ Po(2), Y ~ Po(3), D = X − Y has mean −1 and variance five. It can be negative, for example when X = 0,Y = 1, which has positive probability. Therefore D is not a Poisson variable, whose support is non-negative.
For X ~ Po(2), 2X has mean four but variance eight and only even values. It is not Po(4), which would have variance four and positive mass at one. Matching the mean does not establish either support or the entire distribution.
Dependent input needs the actual mechanism
Let X ~ N(5,4) and set Y = X. Both marginal models are N(5,4), but they are dependent. X + Y = 2X follows N(10,16) by the affine result, rather than the unjustified independent result N(10,8). A supplied relation can solve this particular dependent case.
With only two marginal normal models and no dependence information, the independent combination result cannot be assumed. Means can still be combined, but the variance and shape need appropriate joint information. This closes the section’s results; continuous densities, sampling and hypothesis tests remain separate later work, and self-checked practice is not teacher certification.
Pause and explain
If Y = X ~ N(5,4), what is X + Y’s distribution?
Worked example
X ~ Po(2). Is 2X distributed as Po(4)?
Show the worked solution
- Its mean is four but variance is eight.
- Its support contains only even counts.
- Po(4) has variance four and positive odd-count masses, so the models differ.
Answer No; 2X has mean 4, variance 8 and even support.
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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.
Independent X ~ Po(2), Y ~ Po(3). Find moments of X − Y and explain why it is not Poisson.
Give me a hint
Check support as well as the signed mean and variance.
Compare my reasoning
- Mean = 2 − 3 = −1.
- Variance = 2 + 3 = 5.
- The event X = 0,Y = 1 has positive probability and gives negative difference.
Mean −1; variance 5; not Poisson because negative values are possible.
Look for these in your work
- Add independent variances.
- Check non-negative-count support.
For X ~ Po(2), compare 2X with Po(4).
Give me a hint
Check variance and attainable count values, not only the mean.
Compare my reasoning
- Both means are four.
- 2X has variance eight and only even values.
- Po(4) has variance four and positive odd-count masses.
2X is not Po(4); its variance is 8 and support is even.
Look for these in your work
- Use squared scale.
- Reject matching-mean reasoning.
X ~ N(5,4) and Y = X. State X + Y’s distribution and explain why the independent-copy answer is wrong.
Give me a hint
Simplify the stated dependence before applying a theorem.
Compare my reasoning
- X + Y = 2X.
- Affine normal transformation gives N(10,16).
- N(10,8) would require fresh independent copies, which the equality excludes.
N(10,16); the variables repeat the same observation.
Look for these in your work
- Use the actual relation.
- Do not infer independence from matching marginal models.
Common mistakes
- A matching mean is not distribution closure.
- Differences can leave Poisson support.
- Equal marginal normal models do not establish independence.
What can you explain now?
Can X − Y take negative values for these Poisson inputs?
Compare with the explanation
Yes.
Positive probability at X = 0,Y = 1 rules out a non-negative Poisson support.
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Make it stick
Check the variable’s support, moments and dependence mechanism before claiming a distribution family.
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.