Why Existence and Uniqueness Need Separate Checks
A likelihood can have a score equation with a solution, but that alone does not say whether the solution is unique. Nor does the shape of a likelihood guarantee that its maximum occurs at a finite parameter value: the likelihood may increase toward the edge of the natural parameter space. In an exponential family, these questions can be organized around two objects: the curvature of the log-partition function and the observed sufficient statistic.
The central distinction is between uniqueness, which follows from strict concavity under minimality, and existence, which requires the observed statistic to match the model mean at some finite natural parameter. The condition is exact for the regular minimal family described below; membership in the interior of a convex support is an equivalent shortcut only under additional assumptions.
Likelihood in Canonical Form
Let \(X_1,\ldots,X_n\) be independent and identically distributed observations from a \(d\)-dimensional canonical exponential family with density or mass function, relative to a fixed base measure,
Here \(\eta\in\mathbb{R}^d\) is the natural parameter, \(T(X)\in\mathbb{R}^d\) is the sufficient statistic for one observation, \(A(\eta)\) is the log-partition function, and \(\Omega\) is the natural parameter space. Assume \(\Omega\) is open and convex, and \(A(\eta)\) is finite on \(\Omega\). Write
Apart from terms not involving \(\eta\), the sample log-likelihood is
The usual regularity for differentiating the normalizing integral is assumed: around each \(\eta\in\Omega\), the defining integral for \(A\) remains finite in a neighborhood. This permits differentiation under the integral and yields the familiar moment identities below.
Log-Partition Derivatives and Strict Concavity
The log-partition function is not merely a normalizing constant. Its first derivative gives the mean of the sufficient statistic, and its second derivative gives its covariance matrix. These identities make the curvature of the likelihood directly interpretable.
Proof. Let \(Z(\eta)\) be the integral of \(h(x)\exp\{\eta^T T(x)\}\) against the base measure, so \(A(\eta)=\log Z(\eta)\). The neighborhood finiteness assumption justifies differentiating this integral: locally, factors that are polynomial in \(T(x)\) can be bounded by a finite sum of exponential integrands with natural parameters in that neighborhood. Thus
Differentiating once more gives
For any \(a\in\mathbb{R}^d\), \(a^T\nabla^2A(\eta)a=\operatorname{Var}_\eta(a^TT(X))\). If this variance is zero, then \(a^TT(X)\) is constant almost surely. Minimality says this can happen only when \(a=0\). Hence the Hessian is positive definite. Finally, \(\nabla^2\ell(\eta)=-n\nabla^2 A(\eta)\), which is negative definite, so \(\ell\) is strictly concave on the convex set \(\Omega\). \(\square\)
The Exact Criterion for a Finite, Unique MLE
Proof. Differentiating the sample log-likelihood gives
If a finite maximum is attained at \(\hat\eta\in\Omega\), the openness of \(\Omega\) means that \(\hat\eta\) is an interior point of the parameter domain. A differentiable function at an interior maximum has zero gradient, so \(\nabla A(\hat\eta)=\bar T\). This proves necessity.
Conversely, suppose that \(\nabla A(\hat\eta)=\bar T\) for some \(\hat\eta\in\Omega\). Then \(\nabla\ell(\hat\eta)=0\). Concavity implies that a stationary point is a global maximum: for any \(\eta\in\Omega\), concavity gives \(\ell(\eta)\leq\ell(\hat\eta)+\nabla\ell(\hat\eta)^T(\eta-\hat\eta)=\ell(\hat\eta)\). Strict concavity, proved above, makes this maximum unique. Therefore the MLE exists, is finite, and is unique. \(\square\)
The set \(\{\nabla A(\eta):\eta\in\Omega\}\) is the family’s mean-parameter space. It is the precise target for the observed average. In regular models it lies inside the relative interior of the convex support of \(T(X)\), the closed convex hull of values that \(T(X)\) can take. But interior convex-support membership is not, without further conditions such as steepness, a universal substitute for membership in the mean-parameter space.
Worked Examples
Worked Example: Bernoulli Trials
Suppose \(X_1,\ldots,X_8\) are Bernoulli observations and there are \(s=5\) successes. In canonical form, \(T(X)=X\), \(\eta=\log\{p/(1-p)\}\), and \(A(\eta)=\log(1+e^\eta)\). The observed average is \(\bar T=5/8\).
The mean map is \(\nabla A(\eta)=e^\eta/(1+e^\eta)\). Solving the score equation, \(e^{\hat\eta}/(1+e^{\hat\eta})=5/8\), gives \(e^{\hat\eta}=5/3\), hence \(\hat\eta=\log(5/3)\). Transforming back, \(\hat p=e^{\hat\eta}/(1+e^{\hat\eta})=5/8\).
The solution is finite because \(5/8\) is attained by the mean map at the finite value \(\log(5/3)\). Also, \(A''(\eta)=e^\eta/(1+e^\eta)^2>0\) for finite \(\eta\), so the family is strictly convex in its log-partition function and the log-likelihood is strictly concave. The estimate is therefore unique.
Worked Example: Poisson Counts
Suppose five independent Poisson observations have total count \(S=12\). For a Poisson mean \(\lambda=e^\eta\), the canonical representation has \(T(X)=X\) and \(A(\eta)=e^\eta\), with terms independent of \(\eta\) absorbed into the carrier. Thus \(\bar T=12/5=2.4\).
The mean map is \(\nabla A(\eta)=e^\eta\). The score equation is \(e^{\hat\eta}=2.4\), so \(\hat\eta=\log(2.4)\), a finite value. On the mean scale, \(\hat\lambda=e^{\hat\eta}=2.4\). The Hessian of \(A\) is \(A''(\eta)=e^\eta>0\); therefore the likelihood is strictly concave in \(\eta\), and this finite MLE is unique.
This calculation checks both parts of the criterion: the observed average is a model mean at a finite parameter, and minimality gives the strict curvature needed for uniqueness.
Worked Example: Three-Category Multinomial Sampling
Consider ten independent categorical observations with category counts \((4,3,3)\). Take category 3 as the reference category and define \(T(X)=(\mathbf{1}\{X=1\},\mathbf{1}\{X=2\})\). The natural parameters are \(\eta_1=\log(p_1/p_3)\) and \(\eta_2=\log(p_2/p_3)\), and \(A(\eta)=\log(1+e^{\eta_1}+e^{\eta_2})\). The observed average is \(\bar T=(4/10,3/10)\).
The gradient components are \[ \frac{\partial A}{\partial\eta_1} =\frac{e^{\eta_1}}{1+e^{\eta_1}+e^{\eta_2}},\qquad \frac{\partial A}{\partial\eta_2} =\frac{e^{\eta_2}}{1+e^{\eta_1}+e^{\eta_2}}. \] Matching these to \(0.4\) and \(0.3\) gives fitted probabilities \((\hat p_1,\hat p_2,\hat p_3)=(0.4,0.3,0.3)\). Hence \(\hat\eta_1=\log(0.4/0.3)=\log(4/3)\) and \(\hat\eta_2=\log(0.3/0.3)=0\).
Both natural parameters are finite because all three fitted probabilities are positive. The two indicator statistics are not related by a nonzero linear combination that is constant across all three categories, so this representation is minimal. The theorem therefore gives a unique finite estimate.
What Convexity Does—and Does Not—Guarantee
Strict concavity means there cannot be two distinct finite maximizers in a minimal representation. It does not by itself ensure that a maximizer exists inside \(\Omega\). For existence, the observed average must be achieved by the gradient of \(A\) at a parameter in the allowed natural parameter space.
A tempting shortcut is to check only whether the observed average lies in the interior of the convex support of \(T(X)\). This is useful when additional assumptions ensure that the gradient map covers that interior, as in many steep full exponential families. Without those assumptions, the safe test is the exact one: solve \(\nabla A(\eta)=\bar T\) and verify that a solution lies in \(\Omega\).
Another important qualification is minimality. If a representation contains redundant statistics, the likelihood can be flat along some parameter directions. A maximum may then exist without its parameter value being unique. Removing redundant directions, or working in a minimal representation, is what makes the strict-concavity conclusion apply.
Common Mistake
- Wrong reasoning: “The likelihood is concave, so it has a unique MLE.” Concavity alone allows a flat set of maxima, and it does not guarantee that the supremum is attained at a finite parameter.
- Correction: Establish strict concavity using minimality to get uniqueness if a maximum exists. Then check existence separately by asking whether \(\bar T\) equals \(\nabla A(\eta)\) for some \(\eta\in\Omega\).
Check Your Understanding
Use the score equation and the mean-parameter criterion to answer these questions.
- For an iid canonical exponential family, write the sample log-likelihood in terms of \(S=\sum_i T(X_i)\) and \(A(\eta)\).
- Why does minimality make the log-likelihood strictly concave?
- In the Bernoulli example, what observed average would correspond to the finite natural parameter \(\hat\eta=0\)?
- What is the exact necessary and sufficient condition for a finite MLE in the regular minimal setting?
- Why is convex-support interior membership not automatically the exact existence test for every regular exponential family?