How Does a Linear Transformation Affect the Expected Value and Variance of a Random Variable?

For a linear transformation $Y=aX+b$, the expected value becomes $E(Y)=aE(X)+b$, while the variance becomes $\operatorname{Var}(Y)=a^2\operatorname{Var}(X)$. The constant $b$ shifts the centre of the distribution but does not change its spread.

Why the Formulas Work

Expected value represents the long-run mean. Multiplying every value of $X$ by $a$ multiplies the mean by $a$. Adding $b$ then shifts every value, including the mean, by the same amount:

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

Variance measures the average squared deviation from the mean. Adding $b$ changes each observation and the mean equally, so every deviation remains unchanged. Multiplying by $a$ multiplies each deviation by $a$; squaring those deviations produces the factor $a^2$:

$$\operatorname{Var}(aX+b)=a^2\operatorname{Var}(X).$$

More formally, write $\mu=E(X)$. The deviation of $Y$ from its mean is

$$Y-E(Y)=a(X-\mu).$$

Thus, the transformed deviation is $a$ times the original deviation. Variance averages the squares of these deviations, so the multiplier becomes $a^2$. This also proves why $b$ disappears: it moves every value and the mean by exactly the same amount.

The corresponding standard deviation is multiplied by the absolute value of the scale factor:

$$\sigma_Y=|a|\sigma_X.$$

These formulas also show that a negative value of $a$ reflects the distribution while changing its spread by (|a|). If $a=0$, every transformed value equals $b$, so $E(Y)=b$, $\mathrm{Var}$, and , because the resulting distribution has no spread.

Worked Example

Suppose $E(X)=12$, $\operatorname{Var}(X)=9$, and $Y=-2X+5$. Then

$$E(Y)=-2(12)+5=-19,$$

and

$$\operatorname{Var}(Y)=(-2)^2(9)=36.$$

The original standard deviation is $\sqrt{9}=3$, while the transformed standard deviation is $\sqrt{36}=6$. This agrees with . The negative multiplier reverses the ordering of values, but spread cannot be negative.

A common misconception is $\operatorname{Var}(aX+b)=a\operatorname{Var}(X)+b$. This is incorrect: translation by $b$ does not affect variance, and scaling by $a$ changes variance by , not .

IB Exam Technique

This is HL-only content from AHL 4.14, involving linear transformations of random variables. In an IB Mathematics AI response, state the general formulas before substituting, distinguish variance from standard deviation, and retain full GDC precision until the final answer. Remember that $a^2$ is positive even when $a<0$.