Variance Under Shifts and Scaling

Why Shifts and Scales Matter

A random variable may be reported in different units or measured relative to a different reference point without changing the underlying uncertainty in the same way. Converting a temperature from Celsius to Fahrenheit adds an offset and changes the unit size. A hypothetical cost model might add a fixed charge to a random amount. To compare variability correctly, it is important to know which part of a linear transformation changes the variance and which part does not.

The rule is simple, but two errors are common: treating an added constant as if it affects variance, and multiplying variance by the scale factor rather than by its square. Both errors can be avoided by returning to the definition of variance: it measures squared deviations from the mean.

Definition: Let $X$ be a real-valued random variable with finite variance, and let $a$ and $b$ be fixed constants. For $Y=aX+b$, $$ \operatorname{Var}(Y)=a^2\operatorname{Var}(X). $$ Thus, adding $b$ does not change variance, while multiplying $X$ by $a$ multiplies variance by $a^2$. The condition that $X$ has finite variance is needed for this rule to give a finite variance.

The constants $a$ and $b$ are not random variables. If they were random, or if they depended on $X$ in a nonlinear way, the stated rule would not automatically apply. The rule does not require $X$ to have a particular distribution, nor does it require independence from another variable: it concerns one random variable transformed using fixed constants.

Why the Rule Holds

The mean changes under the transformation as $E[Y]=aE[X]+b$. Variance is the expected squared difference between a variable and its own mean. Subtracting the transformed mean therefore removes the added constant.

\begin{aligned} \operatorname{Var}(aX+b) &=E\left[\left((aX+b)-E[aX+b]\right)^2\right]\\ &=E\left[\left((aX+b)-(aE[X]+b)\right)^2\right]\\ &=E\left[\left(a(X-E[X])\right)^2\right]\\ &=E\left[a^2(X-E[X])^2\right]\\ &=a^2E\left[(X-E[X])^2\right]\\ &=a^2\operatorname{Var}(X). \end{aligned}

The cancellation of $b$ is the reason a shift does not affect variance: every possible value and the mean move by the same amount, leaving each deviation from the mean unchanged. The factor $a$, by contrast, multiplies every deviation by $a$. Since variance squares those deviations, the resulting factor is $a^2$. A negative $a$ reverses the direction of deviations but still produces a nonnegative variance.

Useful special cases: For any fixed $b$, $\operatorname{Var}(X+b)=\operatorname{Var}(X)$. For any fixed $a$, $\operatorname{Var}(aX)=a^2\operatorname{Var}(X)$. If $a=0$, the transformed variable is the constant $b$, so its variance is zero.

Worked Applications

In the examples below, each random variable has a finite probability distribution, so its mean and variance exist. No interest rate or mortality assumption is relevant to these hypothetical measurement and cost models; no life-contingent payment or investment accumulation is being modeled. For each example, first determine the variance of the original variable, then apply the transformation rule and check the result where useful.

Worked Example: A Fixed Charge Plus a Random Cost

A hypothetical service operation has a daily event count $X$ with the following probability distribution. Its daily cost, in dollars, is $Y=80X+25$: a fixed charge of $25 plus $80 for each event. Find the mean and variance of $Y$. The distribution is specified by the table; there are no interest or mortality assumptions.

$x$	$P(X=x)$
0	0.2
1	0.5
3	0.3

Method. Calculate $E[X]$ and $\operatorname{Var}(X)$, then use the mean and variance transformation rules with $a=80$ and $b=25$.

First, the mean event count is $E[X]=(0)(0.2)+(1)(0.5)+(3)(0.3)=0+0.5+0.9=1.4$. The second raw moment is $E[X^2]=(0^2)(0.2)+(1^2)(0.5)+(3^2)(0.3)=0+0.5+2.7=3.2$. Therefore, $\operatorname{Var}(X)=E[X^2]-(E[X])^2=3.2-(1.4)^2=3.2-1.96=1.24$.

The expected cost is $E[Y]=80E[X]+25=(80)(1.4)+25=112+25=\$137$. Its variance is $\operatorname{Var}(Y)=80^2\operatorname{Var}(X)=6{,}400(1.24)=7{,}936$ dollars squared.

Check. The possible costs are $25, $105, and $265, with mean $137. Their deviations from the mean are $-112$, $-32$, and $128$. The weighted squared deviations are $(0.2)(112^2)+(0.5)(32^2)+(0.3)(128^2)=2{,}508.8+512+4{,}915.2=7{,}936$, as required. The fixed charge shifts each cost but does not alter its variance.

Common mistake avoided: Multiplying the original variance by 80 would give $99.2$, not $7{,}936$. Variance scales by $80^2$, because it is based on squared deviations.

Worked Example: Changing Temperature Units

For a hypothetical weather study, a daily temperature $X$, measured in degrees Celsius, takes values $-2$, $0$, and $5$ with probabilities $0.25$, $0.5$, and $0.25$, respectively. Convert to Fahrenheit using $F=1.8X+32$. Find the mean and variance in Fahrenheit units. This is a finite discrete distribution with the probabilities given; no interest rate or mortality assumption applies.

Method. Find the original mean and variance, then use $a=1.8$ and $b=32$. The offset changes the mean but not the variance.

The Celsius mean is $E[X]=(-2)(0.25)+(0)(0.5)+(5)(0.25)=-0.5+0+1.25=0.75$ degrees. The second raw moment is $E[X^2]=(4)(0.25)+(0)(0.5)+(25)(0.25)=1+0+6.25=7.25$. Hence $\operatorname{Var}(X)=7.25-(0.75)^2=7.25-0.5625=6.6875$ degrees Celsius squared.

The Fahrenheit mean is $E[F]=1.8(0.75)+32=1.35+32=33.35$ degrees Fahrenheit. The Fahrenheit variance is $\operatorname{Var}(F)=(1.8)^2(6.6875)=3.24(6.6875)=21.6675$ degrees Fahrenheit squared.

Check. The Fahrenheit values are $28.4$, $32$, and $41$. Their mean is $(0.25)(28.4)+(0.5)(32)+(0.25)(41)=7.1+16+10.25=33.35$. Their weighted squared deviations from 33.35 are $(0.25)(4.95^2)+(0.5)(1.35^2)+(0.25)(7.65^2)=6.125625+0.91125+14.630625=21.6675$. The direct calculation agrees with the transformation rule.

Common mistake avoided: Adding 32 to the variance would be incorrect. The 32-degree offset changes the center of the distribution, not the deviations around that center.

Worked Example: A Shifted Reserve Margin With Negative Scaling

A hypothetical reserve margin $M$, measured in thousands of dollars, is defined by $M=9-2X$, where $X$ is a loss index. The index has the finite distribution $P(X=2)=0.6$ and $P(X=5)=0.4$. Find the mean and variance of the margin. The stated distribution and probabilities fully specify the model; no interest rate or mortality assumption is used.

Method. Write the transformation as $M=(-2)X+9$, so the scale factor is negative. Calculate the original moments and then use the square of $-2$.

The mean of $X$ is $E[X]=(2)(0.6)+(5)(0.4)=1.2+2=3.2$. Its second raw moment is $E[X^2]=(2^2)(0.6)+(5^2)(0.4)=2.4+10=12.4$. Thus $\operatorname{Var}(X)=12.4-(3.2)^2=12.4-10.24=2.16$.

The expected margin is $E[M]=9-2E[X]=9-(2)(3.2)=9-6.4=2.6$ thousand dollars. Its variance is $\operatorname{Var}(M)=(-2)^2(2.16)=4(2.16)=8.64$ thousand-dollars squared.

Check. If $X=2$, then $M=9-4=5$, with probability 0.6. If $X=5$, then $M=9-10=-1$, with probability 0.4. The mean is $(5)(0.6)+(-1)(0.4)=3-0.4=2.6$. The variance is $(0.6)(5-2.6)^2+(0.4)(-1-2.6)^2=(0.6)(5.76)+(0.4)(12.96)=3.456+5.184=8.64$.

Common mistake avoided: The negative sign does not make variance negative. Squaring the scale factor gives $(-2)^2=4$, while the shift of 9 has no effect on variance.

How to Apply the Rule Reliably

When a problem gives a transformed random variable, identify its fixed scale factor and fixed shift before calculating. Then check that the original variable has finite variance. If its variance is known, the transformed variance follows immediately; if not, calculate it from the distribution or other information supplied. The variance formula does not require recalculating the transformed distribution.

Write the transformation in standard form.
Express the quantity as $Y=aX+b$, with fixed constants $a$ and $b$. The sign of $a$ matters for direction and the mean, but its square determines the variance factor.

Find or confirm the original variance.
Use the distribution or given information to obtain $\operatorname{Var}(X)$. The transformation rule presumes that this variance exists and is finite.

Square the scale factor.
Calculate $a^2\operatorname{Var}(X)$. Do not include $b$ in the variance calculation.

Check units and special cases.
Variance is expressed in squared units. If $a=1$ or $a=-1$, the variance is unchanged; if $a=0$, the transformed variance is zero.

The corresponding standard deviation, when it is useful to report it, scales by $|a|$, not by $a^2$: taking the square root of $a^2\operatorname{Var}(X)$ gives $|a|\sqrt{\operatorname{Var}(X)}$. This distinction matters because standard deviation has the same units as the variable, while variance has squared units. Do not substitute the standard-deviation rule when a question asks for variance.

Common Exam Traps

Scaling variance only once. For $Y=5X+b$, the correct result is $25\operatorname{Var}(X)$, not $5\operatorname{Var}(X)$.
Adding the shift to the variance. For $Y=X+12$, variance is unchanged. The mean increases by 12, but deviations from the mean do not change.
Leaving a negative variance factor. If $Y=-3X+b$, the variance factor is $(-3)^2=9$, not $-3$.
Ignoring units. If $X$ is measured in dollars, its variance is in dollars squared. Multiplying $X$ by a conversion factor also changes the squared unit scale of its variance.
Applying the result without its conditions. The rule requires fixed constants and a finite variance for $X$. A nonlinear transformation or a random multiplier is not covered by this formula alone.

Key takeaway: For $Y=aX+b$, the mean is $aE[X]+b$, but the variance is $a^2\operatorname{Var}(X)$. A constant shift moves the distribution without changing its spread; a scale changes each deviation, and variance squares that change.

Check Your Understanding

For each finite distribution, use the transformation rule and check which part of the transformation affects variance.

A hypothetical daily count $X$ has probabilities $0.5, 0.25, 0.25$ at $1,2,4$, respectively. Let $Y=3X-7$. Find $\operatorname{Var}(X)$ and $\operatorname{Var}(Y)$. (Answer: $\operatorname{Var}(X)=1.5$, $\operatorname{Var}(Y)=13.5$.)
A hypothetical index $W$ takes values $-1,1,3$ with probabilities $0.2,0.5,0.3$, respectively. Let $Q=10-0.5W$. Find the variance of $Q$. (Answer: $\operatorname{Var}(Q)=0.49$.)
A hypothetical service time $X$, in minutes, takes values $20,40,60$ with probabilities $0.25,0.5,0.25$, respectively. A payment is $Y=12+0.4X$, in dollars. Find the variance of the payment. (Answer: $\operatorname{Var}(X)=200$ minutes squared and $\operatorname{Var}(Y)=32$ dollars squared.)
A hypothetical loss index $X$ is 2 with probability 0.6 and 5 with probability 0.4. A margin is $R=7-3X$. Find its mean and variance. (Answer: $E[R]=-2.6$, $\operatorname{Var}(R)=19.44$.)
A random variable $X$ has finite variance $16$. For $T=-2.5X+4$, determine the variance and explain whether the shift affects it. (Answer: $\operatorname{Var}(T)=100$; the shift of 4 does not affect variance.)

What You'll Learn