Handle negative scales and constant random variables
Reflect values with negative coefficients and handle zero variance without a zero-denominator standardisation.
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 affine moment rules.
- Reverse inequalities when dividing by a negative value.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is (−3)² in the variance formula?
What you will learn
- Use |a| for SD under negative scaling.
- Recognise zero-scale transforms as constants.
A reflection still has non-negative variance
For E(X) = 10 and Var(X) = 4, U = 30 − 3X has mean zero and variance (−3)² × 4 = 36, so SD is six. The coefficient reverses the order of values but cannot make variance or SD negative.
The aligned scales show X values 8,10,12 mapping to U values 6,0,−6. For V = 5 − 0.5X, mean is zero, variance one and SD one. In general, SD(aX + b) = |a|SD(X), whereas variance uses a².
A zero scale removes randomness
W = 0X + 7 equals seven for every outcome. Mean is seven and variance and SD are zero. P(W ≤ 7) = 1 but P(W < 7) = 0, since its whole probability lies at seven.
Do not standardise a constant by dividing by SD zero or call it an ordinary positive-variance normal distribution. The normal transformation result applies to a non-degenerate result; when the variance vanishes, calculate constant events directly.
Pause and explain
How should probabilities for W = 0X + 7 be calculated?
Worked example
E(X) = 10, Var(X) = 4. Find moments of 30 − 3X.
Show the worked solution
- Mean = 30 − 30 = 0.
- Variance = (−3)² × 4 = 36.
- SD = |−3| × 2 = 6.
Answer Mean 0; variance 36; SD 6.
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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.
E(X) = 10, Var(X) = 4. Find moments of 30 − 3X.
Give me a hint
Square the negative coefficient before calculating variance.
Compare my reasoning
- Mean is zero.
- Variance is thirty-six.
- SD is six, not minus six.
Mean 0; variance 36; SD 6.
Look for these in your work
- Retain the signed mean coefficient.
- Use positive spread.
For this X, find moments of 5 − 0.5X.
Give me a hint
Use the square for variance and absolute scale for SD.
Compare my reasoning
- Mean = 5 − 5 = 0.
- Variance = 0.25 × 4 = 1.
- SD = 0.5 × 2 = 1.
Mean 0; variance 1; SD 1.
Look for these in your work
- Square the fractional scale.
- Separate variance from SD.
Find mean, variance, P(W ≤ 7) and P(W < 7) for W = 0X + 7.
Give me a hint
The zero scale makes W exactly seven in every outcome.
Compare my reasoning
- Mean is seven and variance zero.
- Seven satisfies W ≤ 7, so that probability is one.
- Seven does not satisfy W < 7, so that probability is zero.
Mean 7; variance 0; P(W ≤ 7) = 1; P(W < 7) = 0.
Look for these in your work
- Recognise the constant case.
- Avoid dividing by zero.
Common mistakes
- Negative scaling does not give negative variance.
- Use absolute scale for SD.
- Handle zero variance directly.
What can you explain now?
Is SD(−3X) negative?
Compare with the explanation
No; it is three times SD(X).
Spread is non-negative even when order reverses.
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Make it stick
A negative scale reflects order but keeps variance positive; a zero scale gives a constant to handle directly.
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