Scaling and shifting a single variable
Suppose every value of X is doubled and then 5 is added. What happens to the mean and to the spread? The mean follows along exactly as you would expect. The spread does not.
Adding a constant slides the whole distribution sideways without changing its shape, so the variance is untouched. Multiplying stretches it, and because variance is built from squared distances, it is multiplied by the square of the scale factor. The a² is the single most important detail in this topic.
- a
- the scale factorsquared in the variance, and never negative once squared
- b
- the shiftaffects the mean only, never the variance
- Var
- the variancebuilt from squared deviations, hence the a²
Var(3X) is 9Var(X), not 3Var(X)
And Var(X − 4) is just Var(X), unchanged. If you find yourself subtracting a constant from a variance, something has gone wrong. A useful check: variance can never come out negative, and a negative scale factor squares to a positive one — Var(−2X) = 4Var(X).
Combining independent variables
Means always add, whether or not the variables are independent. Variances add too, but only when the variables are independent, and the result that surprises everyone is that they add even when the variables are subtracted.
The reason is worth understanding rather than memorising. Subtracting makes the difference more variable, not less: if X is high and Y is low, X − Y is unusually large; if the reverse, it is unusually small. Both sources of variability contribute, so the uncertainties accumulate whichever way the variables are combined.
- independent
- the essential conditionwithout it the variance rule fails
- a², b²
- squared coefficientsso a minus sign disappears
- normal
- closed under linear combinationwhich is what makes these questions solvable
X ~ N(50, 16) and Y ~ N(30, 9) are independent. Find the distribution of X − Y, and hence P(X − Y > 25).
- E(X − Y) = 50 − 30 = 20.Means subtract when the variables are subtracted.
- Var(X − Y) = Var(X) + Var(Y) = 16 + 9 = 25.Variances ADD despite the subtraction. Writing 16 − 9 = 7 here is the standard error.
- So X − Y ~ N(20, 25), with standard deviation 5.A linear combination of independent normals is normal, which is what allows the next step.
- z = (25 − 20)/5 = 1.Standardising the value of interest using the new mean and standard deviation.
- P(X − Y > 25) = 1 − Φ(1) = 1 − 0.8413 = 0.1587.Upper tail, so subtract from 1.
X − Y ~ N(20, 25); P = 0.159
n copies is not the same as n times one
This is the distinction that separates a correct answer from a confident wrong one. 2X means take one observation and double it. X₁ + X₂ means take two separate observations and add them. Their means are the same; their variances are not.
Doubling one observation doubles the deviation from the mean, so the variance quadruples. Adding two independent observations lets their deviations partly cancel — one may be above average while the other is below — so the variance only doubles.
The wording decides which applies. "Twice the mass of one item" is 2X. "The total mass of two items" is X₁ + X₂.
| Expression | Mean | Variance | In words |
|---|---|---|---|
| 2X | 2μ | 4σ² | one observation, doubled |
| X₁ + X₂ | 2μ | 2σ² | two separate observations, added |
| X₁ + … + Xₙ | nμ | nσ² | the total of n observations |
| X̄ = (ΣX)/n | μ | σ²/n | the mean of n observations |
Why the sample mean is more reliable
The last row of that table is the foundation of all statistical inference. The mean of n observations has variance σ²/n, so its standard deviation is σ/√n — it shrinks as the sample grows. That is the precise sense in which a larger sample gives a more reliable estimate, and it is why quadrupling the sample size only halves the uncertainty.
The rules on one page
E(aX + b) = aE(X) + b— means behave exactly as expected.Var(aX + b) = a²Var(X)— the shift drops out, the scale factor squares.Var(X ± Y) = Var(X) + Var(Y)for independent variables — always a plus.2Xhas variance4σ²;X₁ + X₂has variance2σ².- Any linear combination of independent normal variables is itself normal.
- Independence is required for every variance rule here, and questions expect you to say so.
Working through a worded combination
The difficulty in these questions is almost never the algebra — it is deciding which variable the words describe before any formula is applied.
Read for two things. First, is this one observation scaled or several observations added? "Three times the weight of one box" is 3X; "the weight of three boxes" is X₁ + X₂ + X₃, and their variances differ by a factor of three. Second, is a constant being added, which leaves the variance untouched entirely?
| The words | The variable | Variance |
|---|---|---|
| "twice the length of a rod" | 2X | 4σ² |
| "the total length of two rods" | X₁ + X₂ | 2σ² |
| "a rod plus a 5 cm cap" | X + 5 | σ² unchanged |
| "the difference in length of two rods" | X₁ − X₂ | 2σ² — still added |
| "the mean length of ten rods" | X̄ | σ²/10 |
A packing question is the classic trap
A box containing six items, where the box itself also has weight, is X₁ + … + X₆ + B — six separate observations plus one more variable, not 6X + B. The two have the same mean and very different variances, and questions are written specifically to see whether the distinction has been understood.