Scaling and shifting one random variable

By the end of this lesson

By the end you can find the mean, variance and standard deviation of $aX + b$ from those of $X$ , explain why adding $b$ never changes the spread, and use the fact that a normal variable stays normal.

Key to the symbols

Symbol	Meaning
$X$	a random variable, such as the length of a taxi journey in miles
$a$ , $b$	fixed numbers (constants), such as $2$ or $-5$
$aX + b$	the new random variable you get by multiplying every value of $X$ by $a$ , then adding $b$
$E(X)$	the mean (expected value) of $X$
$\text{Var}(X)$	the variance of $X$ : the average squared distance from the mean
$\text{sd}(X)$	the standard deviation of $X$ , which is $\sqrt{\text{Var}(X)}$
$\lvert a \rvert$	the size of $a$ without its sign, so $\lvert -3 \rvert = 3$
$N(\mu, \sigma^2)$	the normal distribution with mean $\mu$ and variance $\sigma^2$

A change of the form $aX + b$ is called a linear change, because a graph of $aX + b$ against $X$ is a straight line.

A taxi fare

A taxi charges a fixed £3 plus £2 per mile. If a journey is $X$ miles, the fare in pounds is

$F = 2X + 3$

The journey length $X$ is random, so the fare $F$ is random too. You do not need a new table for $F$ . Its mean and spread follow straight from those of $X$ .

If journeys average 4 miles, fares average $2 \times 4 + 3 = £11$ .
The fixed £3 is added to every fare, so it moves all fares up together. It cannot make fares more or less spread out.
The £2 per mile doubles every gap between journeys. A journey 1 mile longer than average costs £2 more than average.

Definition

Rules for aX + b

For any random variable $X$ and constants $a$ and $b$ :

$E(aX + b) = aE(X) + b$

$\text{Var}(aX + b) = a^2\,\text{Var}(X)$

$\text{sd}(aX + b) = \lvert a \rvert\,\text{sd}(X)$

In words:

the mean does exactly what each value does: multiply by $a$ , add $b$ ;
the variance ignores $b$ and is multiplied by $a^2$ ;
the standard deviation ignores $b$ and is multiplied by the size of $a$ .

You may use these results without proving them.

Why b disappears and a is squared

Take a tiny example. $X$ is $1$ , $2$ or $3$ , each with probability $\frac{1}{3}$ , so $E(X) = 2$ .

Now let $Y = 2X + 3$ . The values become $5$ , $7$ and $9$ , so $E(Y) = 7 = 2 \times 2 + 3$ .

Look at the distances from the mean:

	values	distances from the mean	squared distances
$X$	$1, 2, 3$	$-1, 0, 1$	$1, 0, 1$
$Y$	$5, 7, 9$	$-2, 0, 2$	$4, 0, 4$

The $+3$ moved every value and the mean by 3, so the distances did not change at all.
The $\times 2$ doubled every distance.
Variance uses squared distances, so doubling each distance makes the variance $2^2 = 4$ times bigger.

Here $\text{Var}(X) = \frac{1 + 0 + 1}{3} = \frac{2}{3}$ and $\text{Var}(Y) = \frac{4 + 0 + 4}{3} = \frac{8}{3}$ , which is exactly $4 \times \frac{2}{3}$ .

Worked example: Example 1: shifting and scaling

$X$ has $E(X) = 20$ and $\text{Var}(X) = 9$ . Find the mean, variance and standard deviation of $Y = 3X - 5$ .

Have a go first?

Check 1

CalculateQuestion 1 · Recall

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