Transform a random variable and track its moments
Separate the effect of a fixed shift from the squared effect of scaling on variance.
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
- Calculate expectation and variance.
- Read fixed coefficients and constants.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What happens to variance when every value is multiplied by two?
What you will learn
- Use affine mean and variance formulae.
- Distinguish units for mean, variance and standard deviation.
Transform centre and spread separately
If E(X) = 4 and Var(X) = 9, U = 2X + 5 has mean 2 × 4 + 5 = 13 and variance 2² × 9 = 36. Its standard deviation is six. The fixed five shifts every outcome equally, so it changes the centre but not deviations from that centre.
Scaling each deviation by two scales its square by four. This explains the squared coefficient in variance without requiring a formal proof. Mean and SD use the transformed variable’s units; variance uses squared units. A negative constant can be included in the mean in the same way.
Do not put a fixed fee into variance
For V = 3X − 2, mean is 3 × 4 − 2 = 10, variance is 3² × 9 = 81, and SD is nine. Neither subtracting two from variance nor squaring the constant is justified.
Changing U = 2X + 5 to 2X + 9 raises its mean from thirteen to seventeen and leaves variance thirty-six. The result holds for any variable with the stated finite moments; it does not establish that X or U is normal or Poisson. Distribution shape needs additional information.
Pause and explain
For mean 4 and variance 9, what are the moments of 2X + 5?
Worked example
E(X) = 4, Var(X) = 9. Find mean, variance and SD of 2X + 5.
Show the worked solution
- Mean = 2 × 4 + 5 = 13.
- Variance = 2² × 9 = 36.
- SD = √36 = 6.
Answer Mean 13; variance 36; SD 6.
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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.
E(X) = 4, Var(X) = 9. Find mean, variance and SD of U = 2X + 5.
Give me a hint
Apply the shift to mean and the squared scale to variance.
Compare my reasoning
- E(U) = 13.
- Var(U) = 4 × 9 = 36.
- SD(U) = 6.
Mean 13; variance 36; SD 6.
Look for these in your work
- Include the constant in mean.
- Keep variance and SD distinct.
For the same X, find moments of V = 3X − 2.
Give me a hint
A negative constant still shifts only the centre.
Compare my reasoning
- E(V) = 12 − 2 = 10.
- Var(V) = 9 × 9 = 81.
- SD(V) = 9.
Mean 10; variance 81; SD 9.
Look for these in your work
- Use the squared coefficient.
- Exclude the fixed constant from variance.
Compare 2X + 5 and 2X + 9 for this X. Does the higher fixed fee add uncertainty?
Give me a hint
Compare their centres and deviations separately.
Compare my reasoning
- Means are thirteen and seventeen.
- Both variances are thirty-six.
- The fixed fee adds four to every outcome but no random spread.
Mean increases by 4; both variances remain 36.
Look for these in your work
- Distinguish a fixed amount from a random term.
- Avoid inferring distribution shape from moments.
Common mistakes
- The constant belongs in mean, not variance.
- Square the scale coefficient for variance.
- Moments alone do not determine distribution shape.
What can you explain now?
Does adding five change the variance?
Compare with the explanation
No; a fixed shift leaves deviations unchanged.
The transformed mean shifts by the same five.
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Make it stick
Transform the mean with scale and shift; transform variance with the squared scale only.
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