Identifiability of Location and Scale Parameters

Why Identifiability Matters

A location-scale model describes a random variable as $X=\mu+\sigma Z$, where the distribution of $Z$ is fixed, $\mu$ shifts the distribution, and $\sigma$ changes its scale. The central question here is whether the distribution of $X$ determines $\mu$ and $\sigma$ uniquely. This is a property of the family of distributions, not a claim that a particular finite sample will estimate the parameters accurately.

We take the scale parameter to be strictly positive. This usual convention matters: a negative multiplier reflects the baseline distribution, and if the baseline is symmetric that reflection changes nothing. We will see that this is a parameterization ambiguity rather than a failure of identifiability under the positive-scale convention.

Definition: Let $Z$ have a fixed probability distribution $G$ on $\mathbb{R}$, and consider the family $X_{\mu,\sigma}=\mu+\sigma Z$, with parameter space $\mu\in\mathbb{R}$ and $\sigma>0$. The family is identifiable if, for any $(\mu_1,\sigma_1)$ and $(\mu_2,\sigma_2)$, $$ \mu_1+\sigma_1 Z \overset{d}{=} \mu_2+\sigma_2 Z \quad\Longrightarrow\quad (\mu_1,\sigma_1)=(\mu_2,\sigma_2). $$ The equality is equality of probability distributions; the copies of $Z$ on the two sides need not be the same random object.

The Role of Affine Self-Symmetry

If two parameter pairs give the same law, an affine transformation of that law produces a corresponding symmetry of the baseline distribution. Specifically, applying $x\mapsto (x-\mu_2)/\sigma_2$ to both sides of a distributional equality gives $$ Z \overset{d}{=} \frac{\mu_1-\mu_2}{\sigma_2}+\frac{\sigma_1}{\sigma_2}Z. $$ The multiplier $\sigma_1/\sigma_2$ is positive. Conversely, if $Z\overset{d}{=}a+bZ$ for some $a\in\mathbb{R}$ and $b>0$, then $$ \mu+\sigma Z \overset{d}{=} (\mu+\sigma a)+(\sigma b)Z. $$ Thus a nontrivial positive affine symmetry of $G$ produces two different parameter pairs representing the same law.

Theorem: The scalar location-scale family with $\sigma>0$ is identifiable if and only if the only $a\in\mathbb{R}$ and $b>0$ satisfying $Z\overset{d}{=}a+bZ$ are $a=0$ and $b=1$.

This criterion is useful because it reduces parameter identifiability to a question about the fixed baseline law. For probability distributions on the real line, there is a sharper conclusion: every nondegenerate baseline has no nontrivial positive affine self-symmetry.

Theorem: If $Z$ has a nondegenerate probability distribution on $\mathbb{R}$, then the family $X_{\mu,\sigma}=\mu+\sigma Z$, with $\mu\in\mathbb{R}$ and $\sigma>0$, is identifiable.

Proof. For a distribution function $F$, define its generalized quantile by $Q(u)=\inf\{x:F(x)\geq u\}$, for $0<u<1$. Quantiles transform under positive affine changes as $Q_{\mu+\sigma Z}(u)=\mu+\sigma Q_Z(u)$. This follows by transforming the event $\{\mu+\sigma Z\leq x\}$, or equivalently by applying the increasing map $z\mapsto\mu+\sigma z$ to the quantiles.

A nondegenerate distribution has two quantile levels $0<p<r<1$ for which $Q_Z(p)<Q_Z(r)$. Indeed, its generalized quantile function is nondecreasing; if it were constant throughout $(0,1)$, the distribution would be concentrated at that one value. Suppose now that $\mu_1+\sigma_1Z\overset{d}{=}\mu_2+\sigma_2Z$. Equality of distributions gives equality of their quantiles at $p$ and $r$: $$ \mu_1+\sigma_1 Q_Z(p)=\mu_2+\sigma_2 Q_Z(p), \qquad \mu_1+\sigma_1 Q_Z(r)=\mu_2+\sigma_2 Q_Z(r). $$ Subtracting the first equation from the second yields $$ (\sigma_1-\sigma_2)\bigl(Q_Z(r)-Q_Z(p)\bigr)=0. $$ The quantile difference is strictly positive, so $\sigma_1=\sigma_2$. Substitution into either equality gives $\mu_1=\mu_2$. Therefore the family is identifiable. $\square$

The proof requires neither a density nor finite moments. It works for discrete, continuous, and mixed distributions alike. Quantiles also give a practical way to recover the parameters when the chosen baseline has two known, distinct quantiles.

Worked Examples

Worked Example: Recovering Parameters From Exponential Quantiles

Suppose $Z\sim\operatorname{Exp}(1)$, with distribution function $F_Z(z)=1-e^{-z}$ for $z\geq 0$. A location-scale variable $X=\mu+\sigma Z$, where $\sigma>0$, is known to have 25th and 75th percentiles $x_{.25}=2+3\ln(4/3)$ and $x_{.75}=2+3\ln 4$. Recover its parameters.

Solving $F_Z(Q_Z(u))=u$ gives $Q_Z(u)=-\ln(1-u)$. Therefore $Q_Z(.25)=\ln(4/3)$ and $Q_Z(.75)=\ln 4$. The quantile transformation gives $$ x_{.75}-x_{.25} =\sigma\bigl(\ln 4-\ln(4/3)\bigr) =\sigma\ln 3. $$ The specified quantile difference is $3\ln 4-3\ln(4/3)=3\ln 3$, so $\sigma=3$. Then $$ \mu=x_{.25}-\sigma Q_Z(.25) =2+3\ln(4/3)-3\ln(4/3) =2. $$ Both parameters are determined uniquely. The calculation also illustrates why two distinct baseline quantiles suffice: their difference determines scale, and either quantile then determines location.

Worked Example: Symmetry and the Sign of Scale

Let $Z$ take the values $-1$ and $1$, each with probability $1/2$. Under the positive-scale convention, $X=\mu+\sigma Z$ has support $\{\mu-\sigma,\mu+\sigma\}$. With $(\mu,\sigma)=(3,2)$, the support is $\{1,5\}$, with probability $1/2$ at each point.

If one instead allowed a signed scale, the pair $(3,-2)$ would also give values $5$ and $1$, each with probability $1/2$. Thus $(3,2)$ and $(3,-2)$ would describe the same law. The reason is symmetry: $Z\overset{d}{=}-Z$. Restricting scale to $\sigma>0$ removes this sign ambiguity. With that restriction, the two support points identify the parameters: their midpoint is $\mu=(1+5)/2=3$, and half their separation is $\sigma=(5-1)/2=2$.

Worked Example: A Degenerate Baseline Destroys Scale Identifiability

Suppose $Z=5$ with probability one. Then $X=\mu+\sigma Z$ is the constant $\mu+5\sigma$. The distinct parameter pairs $(\mu,\sigma)=(1,2)$ and $(\mu,\sigma)=(3,1.6)$ both give the constant $11$: $$ 1+5(2)=11, \qquad 3+5(1.6)=11. $$ Their distributions are identical, so this location-scale family is not identifiable. In affine-symmetry terms, a point mass at $5$ is unchanged by any positive affine map $z\mapsto a+bz$ that fixes $5$, meaning $a+5b=5$. The nondegeneracy condition in the theorem is therefore necessary.

Worked Example: Invariant Support Does Not Mean Invariant Law

The support of $Z\sim\operatorname{Exp}(1)$ is $[0,\infty)$, which remains $[0,\infty)$ after multiplication by any positive constant. But equal supports do not imply equal distributions. For example, $Z$ and $2Z$ have different probabilities above $1$: $$ P(Z>1)=e^{-1}, \qquad P(2Z>1)=P(Z>1/2)=e^{-1/2}. $$ These probabilities differ, so the distributions differ even though their supports are identical. It is equality of laws—not equality of supports alone—that defines an affine self-symmetry and determines identifiability.

What the Result Does and Does Not Say

For a fixed, nondegenerate baseline distribution and a positive scale, location and scale are identifiable. This is a population-level uniqueness statement: it says that two parameter settings cannot produce exactly the same distribution. It does not say that parameters can be estimated precisely from a small sample, or that a likelihood has a convenient form.

The choice of parameter space is part of the statement. Allowing negative scales can introduce reflection ambiguities when the baseline is symmetric. A degenerate baseline can make different positive scales yield exactly the same law. Neither issue contradicts the theorem: the first changes the parameter convention, and the second removes its nondegeneracy hypothesis.

Common Mistake

A tempting but incorrect argument is: “The support is unchanged by scaling, so the scale parameter cannot be identified.” The exponential example shows the gap: support records which values are possible, but not how probability is distributed among them. To establish non-identifiability, one must show equality of the probability laws for distinct parameter pairs, not merely equality of their supports. Conversely, when proving identifiability, equality of distributions provides the full quantile information used in the proof.

Key takeaway: A scalar location-scale family with a fixed, nondegenerate baseline law is identifiable when scale is restricted to be positive. The relevant exception is a degenerate baseline; sign ambiguities arise only if the parameterization also allows negative scales.

Check Your Understanding

Use the definitions and arguments above to check each conclusion.

What does identifiability require when two parameter pairs produce equal distributions?
Why does the quantile proof not require $E[Z]$ or $\operatorname{Var}(Z)$ to exist?
If a baseline is symmetric about zero, why can allowing negative scales create two representations of the same law?
For the constant baseline $Z=5$, give a condition relating two pairs $(\mu_1,\sigma_1)$ and $(\mu_2,\sigma_2)$ that makes their resulting distributions equal.
Why is equality of supports alone insufficient to establish equality of probability distributions?