Linear transformations and sums of random variables

A random variable in new units

Maple Ledger's follow-up cost for a batch is a fixed \$250 of clerk time plus \$40 per late invoice: $Y=250+40X$. $Y$ is a random variable because $X$ is; it takes the values 250, 290, 330 and 370 with the same probabilities as $x=0,1,2,3$. You could rebuild the table — but two rules make it unnecessary.

For constants $a$ and $b$,

$$E[aX+b]=a\mu +b,\mathrm{Var}(aX+b)=a^2{\sigma }^2,\mathrm{SD}(aX+b)=|a|\sigma$$

The mean transforms exactly as the values do. The variance ignores $b$ — adding a constant shifts every value and leaves the spread untouched — and scales by $a^2$, because variance is in squared units; the SD scales by $|a|$.

Worked example — the follow-up cost

With $\mu =0.95$ and ${\sigma }^2=0.9475$ from L6.2:

$$E[Y]=40\times 0.95+250=288,\mathrm{Var}(Y)=40^2\times 0.9475=1,600\times 0.9475=1,516,\mathrm{SD}(Y)=40\times 0.973=\$38.94$$

The expected follow-up cost is \$288 per batch with a standard deviation of about \$39. Rebuild the table if you doubt it: $250(0.40)+290(0.35)+330(0.15)+370(0.10)=288$.

Adding random variables

Five batches are processed in a day. Let $T=X_1+X_2+\cdots +X_5$ be the day's total of late invoices, where the five batches are independent and each has the L6.2 distribution. Then

$$E[T]=5\mu =4.75,\mathrm{Var}(T)=5{\sigma }^2=4.7375,\mathrm{SD}(T)=\sqrt{4.7375}=2.18$$

Means add, always. Variances add when the variables are independent — and only then. The standard deviation of the total is $\sqrt{5}\times 0.973=2.18$, not $5\times 0.973=4.87$. Standard deviations never add; the route to the SD of a sum is always through the variances.

In general, for independent $X$ and $Y$:

$$E[X+Y]={\mu }_X+{\mu }_Y,\mathrm{Var}(X+Y)={\sigma }_X^2+{\sigma }_Y^2$$

and — the surprise — $\mathrm{Var}(X-Y)={\sigma }_X^2+{\sigma }_Y^2$ as well. Subtracting an independent variable adds its uncertainty; the variance of a difference is a sum. Module 11's two-sample standard error is built on exactly that line.

Why this matters beyond the paper

$T$ counts five batches; $\bar{X}=T/5$ is the average per batch. Apply the transformation rule with $a = 1/5$: $E[\bar{X}]=\mu$ and $\mathrm{Var}(\bar{X})={\sigma }^2/5$. Averaging shrinks the variance by $n$ and the SD by $\sqrt{n}$. Module 8 states this as the sampling distribution of the mean; it is these two rules and nothing more.