Canonical Location-Scale Representation

Why Use a Baseline Representation?

Many familiar distributions come in families that share the same basic shape but differ in their position and spread. A normal model with one mean and standard deviation has the same standardized shape as a normal model with another mean and standard deviation. Writing a variable as a transformation of one fixed baseline variable makes that shared structure explicit.

The defining idea is simple: choose a random variable $Z$ whose distribution remains fixed, then set $X=\mu+\sigma Z$, where $\mu$ is a real location parameter and $\sigma$ is a positive scale parameter. The same construction can describe continuous or discrete distributions; it does not require a density.

Definition: Fix a real-valued random variable $Z$ with a specified distribution. The scalar location-scale family generated by $Z$ consists of the distributions of $$ X_{\mu,\sigma}=\mu+\sigma Z,\qquad \mu\in\mathbb R,\quad \sigma>0. $$ The variable $Z$ is the baseline variable. For a member $X_{\mu,\sigma}$, the corresponding standardized variable is $$ \frac{X_{\mu,\sigma}-\mu}{\sigma}. $$ It has the same distribution as $Z$. When variables are constructed on the same probability space, this standardized variable equals $Z$ exactly.

The family is determined by both the baseline distribution and the allowed values of $\mu$ and $\sigma$. The baseline does not change as the parameters change. For instance, varying $\mu$ shifts the baseline variable, while varying $\sigma$ multiplies its deviations from zero by a positive factor. Replacing $Z$ with a different baseline distribution generally gives a different family.

The word “standardized” here refers to undoing the specified location and scale transformation. It does not, by itself, mean that the resulting variable has mean zero or variance one. Those properties require suitable conditions on the baseline distribution and a compatible choice of $\mu$ and $\sigma$. Likewise, a location parameter need not be the mean, and a scale parameter need not be the standard deviation.

Standard and Parameterized Forms

A distribution may be described in either of two equivalent ways: by naming the distribution of $X$ together with its parameters, or by naming a fixed baseline variable $Z$ and writing $X=\mu+\sigma Z$. Translating between these descriptions means identifying the baseline law and the affine transformation; it does not require a density or a particular formula for a distribution function.

The basic algebra works in both directions. Starting from the baseline representation, subtract $\mu$ and divide by $\sigma$. Starting from a variable $X$ and proposed parameters $\mu,\sigma$, apply that same standardization and check whether the resulting law is the designated baseline law.

Worked Example: A Normal Distribution

Let $Z\sim N(0,1)$, and define $X=5+2Z$. Identify the parameterized normal distribution and standardize it again.

The baseline is standard normal. The transformation adds 5 and multiplies by 2, so the resulting normal variable has mean parameter 5 and standard deviation 2. Since the course notation writes the variance as the second argument, $$ X\sim N(5,2^2)=N(5,4). $$

Solving the transformation for $Z$ gives $$ Z=\frac{X-5}{2}. $$ Thus, if a realized value of $X$ is 1, its standardized value is $(1-5)/2=-2$. The scale in the representation is 2, not 4: 4 is the variance in the notation $N(5,4)$.

Worked Example: A Uniform Distribution

Let $Z\sim\operatorname{Uniform}(-1,1)$, and define $X=3+4Z$. Express the resulting distribution in a parameterized form and recover the baseline variable.

As $Z$ ranges from $-1$ to $1$, the corresponding values under the transformation range from $3+4(-1)=-1$ to $3+4(1)=7$. Therefore $X\sim\operatorname{Uniform}(-1,7)$. Solving for the baseline variable gives $$ Z=\frac{X-3}{4}. $$

For example, the value $X=5$ corresponds to $Z=(5-3)/4=1/2$. The location is 3 and the scale is 4 in the baseline representation. These numbers describe the affine transformation; they are not inferred by treating the endpoints as a mean and standard deviation.

Worked Example: A Shifted Exponential Distribution

Let $Z\sim\operatorname{Exp}(1)$, where the argument denotes the rate, and set $X=2+3Z$. Give the baseline representation and standardize a value of $X$.

The fixed baseline is a rate-one exponential variable, and the parameterized variable is its shift by 2 and positive rescaling by 3. In this construction, $X$ is a shifted exponential with location 2 and scale 3. The standardized variable is $$ \frac{X-2}{3}=Z\sim\operatorname{Exp}(1). $$

For a value $X=11$, standardization gives $Z=(11-2)/3=3$. In this example the location parameter is the shift in the construction, not the mean of the shifted variable. The example also illustrates why the definition does not require a symmetric baseline distribution.

Two Useful Facts, With Proofs

The representation is more than a convenient way to describe examples. It gives an exact test for whether a proposed standardized variable has the baseline law, and it shows how the parameters change when the variable is recoded by a positive affine transformation.

Theorem: Let $Z$ be a fixed baseline variable, let $\mu\in\mathbb R$, and let $\sigma>0$. If $X=\mu+\sigma Z$, then $(X-\mu)/\sigma=Z$. More generally, if $X$ is defined only through its distribution and $X\overset{d}{=}\mu+\sigma Z$, then $(X-\mu)/\sigma\overset{d}{=}Z$.

Use the representation.
When $X=\mu+\sigma Z$ on one probability space, subtract $\mu$ from both sides to obtain $X-\mu=\sigma Z$.

Divide by the positive scale.
Since $\sigma>0$, division is valid and gives $(X-\mu)/\sigma=Z$. Equality of the variables implies equality of their distributions.

Handle equality in distribution.
If $X\overset{d}{=}\mu+\sigma Z$, applying the same measurable function $x\mapsto(x-\mu)/\sigma$ to both sides preserves equality in distribution. The transformed right-hand side is exactly $Z$, so $(X-\mu)/\sigma\overset{d}{=}Z$. $\square$

This result gives a direct verification procedure. For a proposed location-scale description, standardize $X$ using its proposed parameters; the description is correct precisely when the resulting variable has the proposed baseline distribution. This statement concerns the specified baseline and parameter pair; it does not assert that no other pair could describe the same distribution.

Theorem: Suppose $X=\mu+\sigma Z$, where $\sigma>0$, and recode it as $Y=a+bX$, where $a\in\mathbb R$ and $b>0$. Then $$ Y=(a+b\mu)+(b\sigma)Z. $$ Thus the recoded variable has the same baseline $Z$, with location $a+b\mu$ and scale $b\sigma$.

Proof. Substitute the representation of $X$ into the definition of $Y$: $$ Y=a+bX=a+b(\mu+\sigma Z)=a+b\mu+b\sigma Z. $$ Regrouping gives $Y=(a+b\mu)+(b\sigma)Z$. The new scale is positive because $b>0$ and $\sigma>0$, so this is again a location-scale representation using the same baseline variable. $\square$

The restriction $b>0$ matters: the definition uses a positive scale. A negative multiplier reverses the baseline values and generally changes the baseline shape as well as its orientation. The theorem makes no claim that such a reversal can always be represented using the original baseline and a positive scale.

Recognizing the Structure

When checking whether a model is a location-scale family, separate three questions:

What is fixed? Identify one baseline distribution that does not vary with the parameters.
What is transformed? Confirm that the model can be written as that baseline variable multiplied by a positive number and then shifted.
What are the parameters? Read $\mu$ and $\sigma$ from the transformation, not automatically from familiar summaries such as the mean or standard deviation.

A simple discrete illustration shows that no continuous-law assumption is hidden in the definition. If $Z$ is Bernoulli with success probability $1/3$, then $X=10+2Z$ takes the values 10 and 12 with the same probabilities as $Z$ assigns to 0 and 1. Its standardized variable $(X-10)/2$ is Bernoulli with success probability $1/3$. The transformation changes the numerical values but preserves the fixed baseline law.

Common Mistake: Treating Scale as Standard Deviation

A frequent error is to read $\sigma$ in $X=\mu+\sigma Z$ as the standard deviation of $X$, regardless of the baseline. That conclusion does not follow from the representation alone. The multiplier $\sigma$ is the scale parameter relative to $Z$; its relationship to the standard deviation depends on the baseline distribution.

For example, in the normal representation $X=\mu+\sigma Z$ with $Z\sim N(0,1)$, $\sigma$ is the standard deviation, so the variance is $\sigma^2$. But the definition also permits an exponential or discrete baseline, for which the same interpretation need not hold. The correction is to identify the baseline first and use “location” and “scale” for the parameters in the affine representation unless additional distributional facts justify a more specific interpretation.

A related mistake is to use a different baseline distribution for different parameter values and still call the result one location-scale family. If the baseline law changes, the common-shape construction has not been established. Check that every member can be standardized to the same fixed baseline law.

Key takeaway: A scalar location-scale family is generated by one fixed baseline variable through $X=\mu+\sigma Z$, with $\mu\in\mathbb R$ and $\sigma>0$. Standardizing reverses this transformation and recovers the baseline distribution.

Check Your Understanding

For each question, distinguish the baseline distribution from the location and scale parameters.

If $Z\sim N(0,1)$ and $X=4+3Z$, what is the distribution of $X$, using the convention that the second normal argument is the variance?
If $X=7+2Z$, write the standardized variable in terms of $X$. What condition on its distribution would verify a proposed baseline law for $Z$?
Let $Z\sim\operatorname{Uniform}(-1,1)$ and $X=1+5Z$. Find the inverse standardization and the interval of possible values of $X$.
Why does a positive scale multiplier in the definition not imply that the scale is the standard deviation?
If $Y=2+4X$ and $X=\mu+\sigma Z$, with $\sigma>0$, express $Y$ using the same baseline variable and identify its new location and scale.