Affine Transformations of Normal Variables

Why Affine Transformations Matter

A normal model is often specified on one scale, while the question concerns a different one. A hypothetical cost may include a fixed charge as well as a percentage adjustment; a sensor may report temperature in one unit while a decision threshold is stated in another. In each case, the transformed variable is an affine function of the original variable: a constant plus a constant multiple of it.

The key result is simple but has two details that matter in calculations: adding a constant shifts the mean but does not change the variance, and multiplying by a constant multiplies the variance by the square of that constant. A negative multiplier also reverses inequalities when translating events back to the original variable.

Definition: Let $X\sim N(\mu,\sigma^2)$, where $\sigma>0$, and let $a$ and $b$ be fixed constants. The affine transformation of $X$ is $Y=a+bX$. If $b\ne 0$, then $Y$ is normal with mean $a+b\mu$ and variance $b^2\sigma^2$: $$ Y\sim N(a+b\mu,\ b^2\sigma^2). $$ If $b=0$, then $Y=a$ with probability 1; it is constant, not a nondegenerate normal variable.

The constants $a$ and $b$ must be fixed, not random. The result also assumes that $X$ really follows the stated normal distribution. It does not automatically apply to a truncated or censored normal variable, or to a variable whose distribution is only approximately normal.

Distribution Function and the Sign of the Multiplier

Write $F_X(x)=P(X\le x)$ for the distribution function of $X$, and let $\Phi$ be the standard normal distribution function. If $b>0$, solving $a+bX\le y$ for $X$ preserves the inequality. If $b<0$, dividing by $b$ reverses it. Since a normal variable is continuous, the probability of equality at a single point is zero.

F_Y(y)= \begin{cases} F_X\!\left(\dfrac{y-a}{b}\right), & b>0,\\[6pt] 1-F_X\!\left(\dfrac{y-a}{b}\right), & b<0. \end{cases}

For $X\sim N(\mu,\sigma^2)$, standardizing the argument in either case gives the same normal distribution function:

F_Y(y)=\Phi\!\left(\frac{y-(a+b\mu)}{|b|\sigma}\right),\qquad b\ne0.

This identifies the transformed mean as $a+b\mu$ and the transformed standard deviation as $|b|\sigma$. The variance is therefore $b^2\sigma^2$, not $b\sigma^2$ and not $|b|\sigma^2$.

Why the Mean and Variance Take These Forms

The distribution-function argument establishes that the transformed variable is normal. Its mean and variance can also be checked directly from the definitions of expectation and variance. Because $a$ and $b$ are constants,

E[Y]=E[a+bX]=a+bE[X]=a+b\mu.

For variance, subtract the transformed mean before squaring. The constant shift cancels, leaving $Y-E[Y]=b(X-\mu)$:

\operatorname{Var}(Y) =E\!\left[(Y-E[Y])^2\right] =E\!\left[b^2(X-\mu)^2\right] =b^2\operatorname{Var}(X) =b^2\sigma^2.

The absolute value appears in the standard deviation because standard deviations are nonnegative: $\sqrt{b^2\sigma^2}=|b|\sigma$. This also shows why a negative multiplier changes the direction of the scale but cannot produce a negative variance.

Calculating Probabilities After a Transformation

There are two equivalent approaches to a probability involving $Y=a+bX$. First, find the distribution of $Y$, then standardize using its transformed mean and standard deviation. Alternatively, translate the event into an event involving $X$, taking care to reverse inequalities if $b<0$. The second approach is often a useful check on the first.

Worked Example: A Fixed Charge and a Percentage Adjustment

Method: Transform the normal parameters, then standardize the requested threshold. Suppose a hypothetical claim amount $X$, measured in dollars, follows $X\sim N(1000,200^2)$. An adjusted amount is $C=50+1.1X$, also in dollars. Under this model, calculate $P(C>1590)$. The constants are a fixed addition of $\$50$ and a multiplier of $1.1$; no interest or mortality assumption is involved.

The transformed mean is $$ E[C]=50+1.1(1000)=50+1100=1150. $$ The transformed variance and standard deviation are $$ \operatorname{Var}(C)=1.1^2(200^2)=1.21(40000)=48400,\qquad \operatorname{SD}(C)=\sqrt{48400}=220. $$ Thus $C\sim N(1150,48400)$. Standardizing the threshold gives $$ z=\frac{1590-1150}{220}=\frac{440}{220}=2. $$ Therefore, $$ P(C>1590)=1-\Phi(2)=1-0.977249868\ldots=0.022750131\ldots. $$ To four decimal places, $P(C>1590)=0.0228$. As a check, the same event is $50+1.1X>1590$, or $X>1400$; its standardized threshold is $(1400-1000)/200=2$, giving the same probability. The method avoids treating the fixed charge as part of the random variation.

Worked Example: A Negative Multiplier and an Interval

Method: Use the transformed parameters, then verify the interval by solving its inequalities in the original variable. Suppose a hypothetical measurement $X$ follows $X\sim N(50,8^2)$, and define $Z=120-1.5X$. Find $P(33<Z<57)$. The measurement and transformed variable are on the same stated scale; the constants are fixed, and there is no interest or mortality assumption.

The transformed mean and variance are $$ E[Z]=120-1.5(50)=120-75=45, $$ $$ \operatorname{Var}(Z)=(-1.5)^2(8^2)=2.25(64)=144. $$ Thus $Z\sim N(45,144)$, with standard deviation $12$. Standardizing both endpoints, $$ P(33<Z<57) =\Phi\!\left(\frac{57-45}{12}\right) -\Phi\!\left(\frac{33-45}{12}\right) =\Phi(1)-\Phi(-1). $$ Using unrounded distribution-function values, $$ \Phi(1)-\Phi(-1) \approx 0.841344746-0.158655254 =0.682689492. $$ The probability is $0.6827$ to four decimal places. For a check that accounts for the negative multiplier, solve the interval directly: $$ 33<120-1.5X<57 \quad\Longleftrightarrow\quad 42<X<58. $$ The bounds for $X$ are one standard deviation below and above its mean of 50, so this gives the same probability. The method avoids keeping the inequality directions unchanged when dividing by a negative number.

Worked Example: Converting a Normal Measurement From Fahrenheit to Celsius

Method: Express the unit conversion as an affine transformation and transform the variance as well as the mean. Suppose a hypothetical sensor's Fahrenheit reading $X$ follows $X\sim N(68,9^2)$. The Celsius reading is $C=\frac{5}{9}(X-32)$. Find $P(C>25)$. This is a unit conversion only; no interest or mortality assumption is involved.

First write the conversion in affine form: $$ C=\frac{5}{9}X-\frac{160}{9}. $$ Its mean is $$ E[C]=\frac{5}{9}(68-32)=\frac{5}{9}(36)=20. $$ Its variance and standard deviation are $$ \operatorname{Var}(C)=\left(\frac{5}{9}\right)^2(9^2) =\frac{25}{81}(81)=25,\qquad \operatorname{SD}(C)=5. $$ Thus $C\sim N(20,25)$. The standardized threshold is $$ z=\frac{25-20}{5}=1, $$ so $$ P(C>25)=1-\Phi(1) =1-0.841344746\ldots =0.158655254\ldots. $$ To four decimal places, the probability is $0.1587$. As a check, $C>25$ is equivalent to $X>77$ Fahrenheit; the original-scale standard score is $(77-68)/9=1$. The method avoids scaling the variance by $5/9$ instead of by $(5/9)^2$.

Common Exam Trap: The Negative Coefficient

A negative multiplier affects event inequalities, not just the mean. For example, if $Y=10-2X$, then $$ Y>4 \quad\Longleftrightarrow\quad 10-2X>4 \quad\Longleftrightarrow\quad -2X>-6 \quad\Longleftrightarrow\quad X<3. $$ A wrong calculation would divide by $-2$ but keep the inequality pointed the same way, concluding $X>3$. The correction is to reverse the inequality when dividing by a negative number. For a continuous normal $X$, the desired probability is $P(X<3)=F_X(3)$, not $P(X>3)=1-F_X(3)$.

A second frequent error is to transform a variance linearly. If $X$ has variance $\sigma^2$, the variance of $a+bX$ is $b^2\sigma^2$. The shift $a$ has no effect on variance, while a negative $b$ still gives a positive variance because it is squared.

Practical Checks Before Reporting a Result

Affine transformations are useful because they let a normal probability be calculated on the scale where the question is asked, without changing the underlying distribution family. They are especially valuable when a threshold combines units, fixed additions, or percentage adjustments. Before reporting a result, check the following:

State the model. Identify the original normal mean and variance, and confirm that the transformation constants are fixed.
Keep variance and standard deviation distinct. Transform variance using $b^2$, then take its square root if a standard deviation is needed.
Check the sign. A negative $b$ reverses inequalities when an event is translated to $X$.
Check the scale. Confirm that the threshold and transformed mean use the same units.
Check the direction. An upper-tail probability is $1-\Phi(z)$; a lower-tail probability is $\Phi(z)$.

A useful final check is to translate a threshold event back to $X$, if the algebra is manageable, and verify that its standardized value agrees with the one obtained from the transformed distribution. Small discrepancies usually point to an incorrect sign, a missed square in the variance, or a unit mismatch.

Key takeaway: If $X\sim N(\mu,\sigma^2)$ and $Y=a+bX$ with fixed $b\ne0$, then $Y\sim N(a+b\mu,b^2\sigma^2)$. Standardize with standard deviation $|b|\sigma$, and reverse an inequality whenever division by $b<0$ requires it.

Check Your Understanding

For each problem, treat the stated normal model and transformation as exact, and give probabilities to four decimal places unless a dollar threshold is requested.

A hypothetical instrument reading $X$ follows $X\sim N(30,4^2)$. Define $Y=2X+6$. Find the distribution of $Y$ and $P(Y>82)$. (Answer: $Y\sim N(66,64)$; $P(Y>82)=0.0228$.)
A hypothetical index $X\sim N(100,15^2)$ is transformed to $R=200-2X$. Find the distribution of $R$ and $P(-30<R<60)$. (Answer: $R\sim N(0,900)$; $P(-30<R<60)=0.8186$.)
A hypothetical processing-time measurement $X$, in hours, follows $X\sim N(12,3^2)$. Define $W=60-4X$. Find $P(W\le0)$, showing the equivalent event in terms of $X$. (Answer: $W\sim N(12,144)$; $W\le0$ is equivalent to $X\ge15$, so the probability is $0.1587$.)
A hypothetical sensor output $X\sim N(500,50^2)$ is converted to $Y=0.8X+20$. Find the distribution of $Y$ and $P(380\le Y\le460)$. (Answer: $Y\sim N(420,1600)$; $P(380\le Y\le460)=0.6827$.)
A hypothetical amount $X\sim N(40,6^2)$ is transformed to $Y=100-0.5X$. Find the value $t$ such that $P(Y>t)=0.10$, using the standard normal 90th percentile $1.2816$. (Answer: $Y\sim N(80,9)$, and $t=80+1.2816(3)=83.8448$, or $\$83.84$ if amounts are in dollars.)

What You'll Learn