Why Identifiability Matters
A multinomial logit model describes category probabilities through a set of linear predictors. If several different parameter values produce exactly the same probabilities, the data cannot distinguish those parameter values. This is a problem of identifiability: before interpreting or estimating the parameters, we need to know whether the model assigns a unique parameter value to each probability distribution.
The ambiguity in an all-category logit model has a simple source. The softmax map uses relative, not absolute, values of the category logits. Adding the same quantity to every logit changes none of the probabilities. A baseline category removes this particular redundancy, though it does not correct other sources of non-identifiability, such as a rank-deficient predictor matrix.
All-Category Logits and Identifiability
Suppose a response has \(K\) categories, indexed by \(j=1,\ldots,K\). Let \(x\) be a \(q\)-dimensional predictor vector, with an intercept included as one of its components when the model has category-specific intercepts. Write the linear predictor for category \(j\) as \(\eta_j(x)=x^\mathsf{T}\beta_j\), where \(\beta_j\) is its category-specific coefficient vector. The softmax model assigns probabilities
The probabilities are positive and sum to one. The model is identifiable on a specified set of predictor values if no two distinct allowed parameter values give the same vector of category probabilities at every predictor value in that set. This definition concerns the probability model, not whether one particular sample happens to contain observations from every category.
For any real-valued function \(c(x)\), define new logits by \(\eta'_j(x)=\eta_j(x)+c(x)\) for every category. Then
Thus a shared shift does not change the probability vector. In the linear-predictor model, a common shift can itself be linear: if \(c(x)=x^\mathsf{T}\gamma\), then setting \(\beta'_j=\beta_j+\gamma\) for all \(j\) leaves all probabilities unchanged. With an intercept in \(x\), this includes adding a common constant to every category intercept and common amounts to every category’s slopes.
Worked Example: A Shared Shift Leaves Probabilities Unchanged
At a particular predictor value, suppose three category logits are \((0.2,-0.4,1.0)\). Their exponentials are approximately \((1.2214,0.6703,2.7183)\), with sum \(4.6100\). The probabilities are therefore approximately
Now add \(1\) to all three logits, giving \((1.2,0.6,2.0)\). Each exponential is multiplied by \(e\), so the new denominator is also \(e\) times the old denominator. In particular, the first probability is \(e^{1.2}/(e^{1.2}+e^{0.6}+e^2)=e\,e^{0.2}/[e(e^{0.2}+e^{-0.4}+e^1)]\), which equals the original first probability. The same cancellation applies to categories two and three. The logits differ, but the probability vectors are identical.
Exactly Which Logits Are Equivalent?
The common-shift argument gives one way to obtain equivalent logits. The next result shows that, at a fixed predictor value, it gives all such equivalences.
Proof. If \(\eta'_j=\eta_j+c\) for every \(j\), the common factor \(\exp(c)\) cancels from the softmax denominator, so \(p'_j=p_j\) for every category. Conversely, all softmax probabilities are positive, and for any category \(j\),
If \(p=p'\), the log probability ratios are equal, so \(\eta'_j-\eta'_K=\eta_j-\eta_K\) for every \(j\). Rearranging gives \(\eta'_j-\eta_j=\eta'_K-\eta_K\) for every \(j\). Taking \(c=\eta'_K-\eta_K\) proves that the difference is the same for every category. \(\square\)
This result explains the redundancy precisely: at each predictor value, only the differences among the logits are determined by the probabilities. In particular, the log odds of category \(j\) relative to category \(K\) are \(\eta_j(x)-\eta_K(x)\). The overall level shared by all logits cannot be recovered from the probability vector.
Fixing a Baseline Category
A standard solution is to select one category as the baseline and set its linear predictor to zero. Taking category \(K\) as the baseline gives
The baseline is a normalization, not a claim that category \(K\) has probability zero or has no relationship with the predictors. Its probability remains part of the model. The constraint chooses one representative from each set of logits that differ only by a shared shift.
Worked Example: Re-expressing Three Logits With a Baseline
At \(x=1\), suppose the three logits are \((1.0,0.2,-0.4)\). Use category 3 as the baseline by subtracting its logit, \(-0.4\), from every logit. The resulting logits are \((1.4,0.6,0)\). This subtraction is a shared shift, so it preserves probabilities.
The unnormalized weights in the baseline representation are \((e^{1.4},e^{0.6},1)\), approximately \((4.0552,1.8221,1)\). Their sum is \(6.8773\), giving
Computing from the original logits gives weights \((e^1,e^{0.2},e^{-0.4})\). These are the baseline weights multiplied by \(e^{-0.4}\), so normalization gives the same probabilities. The baseline logits are also the log probability ratios: \(\log(p_1/p_3)=1.4\) and \(\log(p_2/p_3)=0.6\).
For a linear-predictor model, the baseline constraint can be applied to coefficient vectors. Given unrestricted vectors \(\beta_1,\ldots,\beta_K\), define \(\widetilde{\beta}_j=\beta_j-\beta_K\) for \(j<K\), and \(\widetilde{\beta}_K=0\). Then \(x^\mathsf{T}\widetilde{\beta}_j=x^\mathsf{T}\beta_j-x^\mathsf{T}\beta_K\), which subtracts the baseline logit from every category. The probabilities are unchanged, and the resulting \(K-1\) coefficient vectors specify the log odds relative to the baseline.
The following result states when this normalization removes the shared-shift ambiguity from the coefficient parameters. Its rank condition matters: the model must have enough independent predictor information to distinguish the coefficients.
Proof. Suppose first that the probability vectors agree at every design row. By the preceding theorem, for every row \(i\) and category \(j\),
Stacking these equalities over rows gives \(X[(\beta'_j-\beta'_K)-(\beta_j-\beta_K)]=0\). Since \(X\) has full column rank, its null space contains only the zero vector, and hence \(\beta'_j-\beta'_K=\beta_j-\beta_K\) for each \(j\). Set \(\gamma=\beta'_K-\beta_K\). Rearranging yields \(\beta'_j=\beta_j+\gamma\) for every \(j\). Conversely, if all coefficient vectors are shifted by the same \(\gamma\), every row’s logits receive the same shift \(x_i^\mathsf{T}\gamma\); the softmax probabilities therefore remain unchanged. Finally, if both arrays satisfy \(\beta_K=\beta'_K=0\), then the shared-shift relation forces \(\gamma=0\), so the arrays are equal. \(\square\)
Worked Example: A Baseline Does Not Fix Rank Deficiency
Suppose the predictor vector is \(x=(1,z,2z)^\mathsf{T}\), so the three design columns are \(1\), \(z\), and \(2z\). These columns are linearly dependent because the third is twice the second. For category 1, compare coefficient vectors \(\beta_1=(0.4,0.3,0.1)^\mathsf{T}\) and \(\beta'_1=(0.4,0.5,0)^\mathsf{T}\). At any value of \(z\),
and
Thus the two vectors give exactly the same category 1 predictor at every design value. This can occur even when category 3 is fixed as the baseline and all other category coefficients are held fixed: the probabilities still cannot distinguish these two category 1 coefficient vectors. The shared-shift ambiguity has been addressed, but the predictors themselves do not contain enough independent information to identify all three coefficients.
What the Baseline Solves—and What It Does Not
With \(K\) categories and \(q\) coefficients per category, an unrestricted all-category model lists \(Kq\) coefficients. The common-shift invariance accounts for \(q\) redundant directions: each component of the shared vector \(\gamma\) can be added to every category’s coefficient vector. A baseline parameterization has \((K-1)q\) free coefficients, matching the number of independent category contrasts.
That count assumes the predictor columns are linearly independent on the design being used to identify the coefficients. If they are not, distinct coefficient vectors can produce the same linear predictors, as in the worked example. Thus, fixing a baseline removes the category-wide shift but does not by itself guarantee identifiability under every design. It is useful to check both the category normalization and the rank of the predictor matrix.
Common Pitfall
- Wrong reasoning: “The probabilities sum to one, so setting one category’s probability or logit to zero solves the redundancy.” The probabilities already sum to one by the softmax definition, and the baseline category generally has positive probability.
- Correction: Fix the baseline linear predictor or coefficient vector to zero. This sets a reference for the logits without removing the baseline category from the response.
- Another incomplete conclusion: “Once a baseline is fixed, every coefficient must be identifiable.” This is only true when the predictor information is sufficiently rich; rank-deficient columns can still yield multiple coefficient vectors with the same predictions.
Check Your Understanding
Use the definitions and results above to check your understanding.
- For a three-category softmax model, explain algebraically why adding the same constant to all three logits leaves the probabilities unchanged.
- If two sets of finite logits give the same softmax probability vector, what must be true about their differences category by category?
- With category \(K\) as baseline, what is the model meaning of setting \(\eta_K(x)=0\)? Does it imply \(p_K(x)=0\)?
- Why does a full-column-rank design matrix matter when translating identifiable category contrasts into unique coefficient vectors?
- A model has a baseline constraint, but two predictor columns are identical. What additional identifiability problem can remain?