Identification as Uniqueness of an Observable Implication
A structural model connects parameters that describe an underlying mechanism to a probability distribution for the data that can actually be observed. The central identification question is whether that connection is one-to-one: could two different admissible structural parameter values produce exactly the same observable distribution? If they could, observations alone cannot distinguish those values, even with an arbitrarily large sample.
This is a population-level question about the model, not a question about whether a particular sample happens to estimate a parameter precisely. To ask it precisely, specify the admissible parameter space and the distribution implied by each parameter value.
\(\mathcal P(\theta_1)=\mathcal P(\theta_2)\ \Longrightarrow\ \theta_1=\theta_2.\)
Equivalently, each distribution in the model corresponds to at most one admissible parameter vector.
Equality here means equality of the full observable probability distributions, not merely similarity in a sample or equality of one selected statistic. The map \(\mathcal P\) is sometimes called the observable-distribution map. It includes whatever aspects of the data-generating setup are held fixed, such as the model specification and any specified design.
Identification can also be considered for a function of the parameters rather than for the entire vector. A function \(g(\theta)\) is identified if \(\mathcal P(\theta_1)=\mathcal P(\theta_2)\) implies \(g(\theta_1)=g(\theta_2)\). Thus a model may fail to identify every coordinate of \(\theta\) while still identifying a particular combination of them.
Worked Example: A One-Parameter Bernoulli Model
Suppose the observable variable \(X\) is Bernoulli, with \(\mathcal P(\theta)\) assigning probability \(p(\theta)=(1+\theta)/4\) to \(X=1\), where the structural parameter space is \(\Theta=[0,1]\). Is \(\theta\) globally identified?
For a Bernoulli distribution, the probability of \(X=1\) determines the full distribution: the probability of \(X=0\) is one minus that probability. If two parameter values produce the same distribution, they therefore satisfy
Multiplying both sides by 4 and subtracting 1 gives \(\theta_1=\theta_2\). This proves injectivity on the stated parameter space, so \(\theta\) is globally identified.
For illustration, suppose the exact observable success probability specified by a distribution is \(0.4\). Solving the model equation gives
This inversion illustrates uniqueness within the model; it does not claim that a finite sample reveals its exact population success probability.
Proving Injectivity and Finding Its Failure
A direct proof of global identification starts with two arbitrary admissible parameter values and assumes that they generate the same observable distribution. The proof then uses the model’s implications to show that the parameter values must be equal. The key is to establish the implication for every pair in the parameter space, not just for one example or for values near a particular point.
To demonstrate failure, it is enough to find a single pair of distinct admissible values that generate the same distribution. Such a pair is called observationally equivalent. Its existence disproves injectivity and shows that the full parameter vector is not globally identified on that parameter space.
Worked Example: A Sign Ambiguity in a Normal Model
Let \(Y\sim N(\theta^2,0.64)\), where \(\theta\in[-2,2]\). The second argument of \(N\) is the variance. Determine whether \(\theta\) is globally identified.
Take \(\theta_1=1.5\) and \(\theta_2=-1.5\). Both values belong to the parameter space, and their implied distributions are
The distributions are exactly equal, although \(1.5\ne-1.5\). The distribution map is therefore not injective, and \(\theta\) is not globally identified on \([-2,2]\).
The observable distribution does identify the function \(\theta^2\), because the mean of \(Y\) equals \(\theta^2\). But identifying that function does not identify the sign of \(\theta\). The distinction matters whenever a scientific or economic interpretation depends on the sign of the structural parameter.
Parameter Restrictions and the Domain of the Claim
Identification is always relative to the parameter space. A restriction can remove one or more observationally equivalent values, but it changes the set on which injectivity is being assessed. Consequently, a statement that a parameter is identified should specify the model and its admissible parameter space. A restriction may be justified by the substantive model or by a normalization that fixes an otherwise arbitrary representation; it should not be treated as evidence supplied by the observable distribution itself.
The following example makes the role of the domain explicit. The observable distribution reveals two features of the parameters, but without an ordering restriction those features do not distinguish which parameter is which.
Worked Example: Two Parameters Known Only Through Their Sum and Product
Let \(X_1\) and \(X_2\) be independent, with \(X_1\sim N(a+b,1)\) and \(X_2\sim N(ab,1)\). The observable distribution is the joint distribution of \((X_1,X_2)\). First take \((a,b)\in\mathbb R^2\). Are \(a\) and \(b\) globally identified?
Consider the distinct parameter values \((a,b)=(1,3)\) and \((a,b)=(3,1)\). Under both,
Thus both parameter values imply independent normal variables with the same marginal distributions: \(X_1\sim N(4,1)\) and \(X_2\sim N(3,1)\). Independence is specified under both parameter values, so the full joint distributions agree as well. The parameter vector is not globally identified on \(\mathbb R^2\).
Now restrict the parameter space to \(0<a<b\). Suppose an observable distribution is generated by some parameter pair in this restricted space, and write its two means as \(s=E[X_1]=a+b\) and \(q=E[X_2]=ab\). The two parameters must be the roots of
because \((t-a)(t-b)=t^2-(a+b)t+ab\). The observable distribution fixes \(s\) and \(q\), so it fixes this quadratic equation and its set of roots. The restriction \(a<b\) then fixes which root is \(a\) and which is \(b\). Hence no two distinct ordered pairs in the restricted space generate the same distribution, and the parameter vector is globally identified there.
For example, if the observable means are \(s=4\) and \(q=3\), then
The roots are 1 and 3. Without the ordering restriction either assignment is admissible; under \(0<a<b\), the unique assignment is \(a=1\), \(b=3\). This conclusion depends on the restricted parameter space, not on a difference between the distributions generated by \((1,3)\) and \((3,1)\).
A General Proof Strategy
The examples suggest a reusable proof structure. It does not require a particular distributional family, but it does require precise definitions of the parameter space and the observable distribution.
Write the admissible parameter space \(\Theta\) and specify the full observable distribution \(\mathcal P(\theta)\) for each \(\theta\in\Theta\).
Take arbitrary \(\theta_1,\theta_2\in\Theta\) and suppose \(\mathcal P(\theta_1)=\mathcal P(\theta_2)\).
Use properties that must agree under equal distributions, such as probabilities of events or well-defined moments, to derive restrictions on \(\theta_1\) and \(\theta_2\).
If the restrictions force \(\theta_1=\theta_2\) for every pair, the map is injective. If a distinct pair satisfies distributional equality, global identification fails.
Moments, Full Distributions, and Common Pitfalls
Moments can simplify a proof, but their role needs care. If two distributions are equal and their relevant moments exist, those moments must be equal. Therefore, if a vector of moments uniquely determines \(\theta\) over \(\Theta\), it provides a valid route to prove global identification. The logic is: equal observable distributions imply equal moments, and equal moments then imply equal parameters.
The converse is not automatic. Equal selected moments do not generally imply equal distributions. For example, two distributions can share a mean and variance while differing in other features. So a collision in a moment map alone proves observational equivalence only when those moments determine the full distribution in the specified model. A proof based on moments must establish the implication it needs.
A related mistake is to confuse a sample calculation with identification. A large sample may estimate features of the observable distribution very accurately, but it cannot separate parameters that imply exactly the same distribution. Conversely, a globally injective model does not guarantee that a finite-sample estimate will be precise. Identification concerns uniqueness in the population model; estimation concerns learning from realized data.
Check Your Understanding
Answer each question by referring to the map from parameters to observable distributions.
- In the Bernoulli example, which feature of the observable distribution determines the full Bernoulli law?
- Why do \(\theta=1.5\) and \(\theta=-1.5\) disprove global identification in the normal example?
- What is the difference between identifying \(\theta^2\) and identifying \(\theta\) in that example?
- In the sum-and-product example, why does the restriction \(a<b\) resolve the parameter assignment without changing the observable distribution?
- When does injectivity of a moment map provide a valid proof of global identification, and why is a collision in selected moments not always enough to show nonidentification?