FS2 4.1 - Linear combinations of normal variables

Keep the signs in the mean and square them in the variance

Four independent readings and four times one reading have the same mean. Why can their variances differ by a factor of $4$? If each reading has distribution $\mathrm N(10,4)$, the two totals both have mean $40$, but their variances are $16$ and $64$. The expression “four readings” needs a careful model.

A coefficient can represent a change of scale, a subtraction or an average. A fixed added constant shifts every outcome by the same amount, so it changes the mean without adding spread.

Means combine with their signed coefficients. For independent variables, variances combine with squared coefficients: a negative coefficient changes a deviation's sign before it is squared, so variance contributions still add.

The formula booklet supplies the independent variance algebra $\operatorname{Var}(aX\pm bY)=a^2\operatorname{Var}(X)+b^2\operatorname{Var}(Y)$. The following full normal-distribution result also specifies the distribution's shape.

The same rule extends to any finite collection of independent normal variables. Write each distinct independent variable once, then sum its coefficient times its mean and its squared coefficient times its variance.

Means still combine linearly without independence, but the simple variance addition needs independence. Readings sharing a calibration error, or expressions sharing the same observation, cannot automatically be treated as independent. If all random terms cancel, the result is a constant, rather than a normal variable with positive spread.

Here is a diagnostic about the conditions, rather than the arithmetic.

Turn a comparison into one normal variable

Move every random term to one side of the inequality. The remaining fixed threshold tells you which tail or interval to use.

In the next pair, $X\sim\mathrm N(20,9)$ and $Y\sim\mathrm N(5,4)$ are independent. The second parameters are variances, so their standard deviations are $3$ and $2$.

Worked example 2 changes only the coefficient of the same reading $Y$, from $2$ to $3$.

A condition such as $|W|<d$ becomes $-d<W<d$, so it uses two normal CDF values. A non-zero mean must still be subtracted at both endpoints.

Pearson rewards the model, its parameters and the correctly translated event, so show them before the calculator output.

Change only that coefficient again.

A sum of observations is different from scaling one observation

Write subscripts when they mean different readings. $X_1+X_2+X_3+X_4$ has four independent sources of variation; $4X$ has one source multiplied by $4$.

[DIAGRAM: asset_name: Equal-mean normal totals with different dependence; asset_slug: 9fm0-fs2-127-sum-versus-scale; caption: Four independent readings have less spread than four times a single reading; description: A single coordinate plot compares two normal density curves centred at forty. The blue narrower density represents the sum of four independent observations each normal with mean ten and variance four, giving a normal total with mean forty and variance sixteen, standard deviation four. The clay wider density represents four times one observation, giving mean forty and variance sixty-four, standard deviation eight. Grey solid and dashed curve outlines distinguish the models, a vertical grey guide marks the common mean, and a clear legend explains both accents.]

For the average $A_n=S_n/n$, scaling the total gives mean $n\mu/n=\mu$ and variance $n\sigma^2/n^2=\sigma^2/n$. The same average is commonly written $\bar X$.