Fractional Logit QMLE and Boundary Outcomes

Why a Bernoulli Criterion Can Fit Fractional Outcomes

A response measured as a fraction can take any value from zero to one, including the endpoints. Examples include a share of a budget allocated to a category, a proportion of eligible days on which a service was used, or a fraction of a group meeting a criterion. The response is not necessarily a binary indicator, so a Bernoulli distribution is not an appropriate description of its observed values.

Fractional logit uses the Bernoulli log-likelihood formula as an estimation criterion, not as a claim that each observed response is a Bernoulli draw. This criterion is well-defined for fractional responses and has a score that targets a logit conditional mean. Its use therefore does not require the outcome to be only zero or one.

Definition: For observations $i=1,\ldots,n$, let $Y_i\in[0,1]$ be the response and $X_i$ a vector of regressors, including an intercept when appropriate. The fractional logit conditional-mean model is $$ E[Y_i\mid X_i]=\mu_i=\Lambda(X_i'\beta), \qquad \Lambda(\eta)=\frac{\exp(\eta)}{1+\exp(\eta)}. $$ Here $\beta$ is the parameter vector, and $\Lambda$ is the logistic function. The fractional logit quasi-maximum likelihood estimator (QMLE) maximizes $$ Q_n(\beta)=\sum_{i=1}^n\left[Y_i\log(\mu_i)+(1-Y_i)\log(1-\mu_i)\right]. $$

For every finite $X_i'\beta$, $0<\mu_i<1$, so both logarithms are defined. The response $Y_i$, however, need only lie in the closed interval $[0,1]$. In particular, the criterion is finite for $Y_i=0$, when its contribution is $\log(1-\mu_i)$, and for $Y_i=1$, when its contribution is $\log(\mu_i)$.

Deriving the Score and Curvature

The logistic function satisfies $\frac{d\mu_i}{d(X_i'\beta)}=\mu_i(1-\mu_i)$. Differentiating one observation’s contribution $q_i(\beta)=Y_i\log(\mu_i)+(1-Y_i)\log(1-\mu_i)$ gives

\begin{aligned} \frac{\partial q_i}{\partial \beta} &=\left(\frac{Y_i}{\mu_i}-\frac{1-Y_i}{1-\mu_i}\right) \frac{\partial\mu_i}{\partial\beta}\\ &=\left(\frac{Y_i-\mu_i}{\mu_i(1-\mu_i)}\right) \mu_i(1-\mu_i)X_i\\ &=X_i(Y_i-\mu_i). \end{aligned}

Summing the contributions gives the score. Differentiating again gives the Hessian. These expressions hold for fractional as well as binary $Y_i$.

S_n(\beta)=\sum_{i=1}^n X_i(Y_i-\mu_i), \qquad H_n(\beta)=-\sum_{i=1}^n X_iX_i'\mu_i(1-\mu_i).

The Hessian is negative semidefinite, since for any vector $a$, $a'H_n(\beta)a=-\sum_i(a'X_i)^2\mu_i(1-\mu_i)\leq 0$. Thus the criterion is concave in $\beta$; with a full-rank design it is strictly concave when the observations give positive curvature in every parameter direction. Concavity helps with optimization, but does not by itself guarantee a finite maximizer. For example, complete separation can send some coefficients toward infinity.

Worked Example: A Fractional Observation’s Criterion and Score

Suppose a hypothetical observation has $Y_i=0.40$, $X_i=1$, and a candidate intercept $\beta=0$. The fitted mean is $\mu_i=\Lambda(0)=0.50$. Calculate the observation’s criterion contribution and score contribution.

The criterion contribution is $$ q_i=0.40\log(0.50)+0.60\log(0.50)=\log(0.50)=-0.6931. $$ The score contribution is $$ X_i(Y_i-\mu_i)=1(0.40-0.50)=-0.1000. $$ The negative score indicates that, at this candidate intercept, increasing the intercept would decrease this observation’s contribution locally. The calculation uses the actual fractional response directly; it does not convert $0.40$ into a binary outcome.

Why the Criterion Targets the Conditional Mean

The central justification is that the expected criterion is maximized by setting the fitted mean equal to the true conditional mean. This is a statement about the criterion’s expectation given $X_i$, not an assumption that the outcome has a Bernoulli distribution.

Result: Suppose $m_i=E[Y_i\mid X_i]$ lies strictly between zero and one. Conditional on $X_i$, the expected contribution at a fitted mean $\mu_i$ is $$ E[q_i\mid X_i]=m_i\log(\mu_i)+(1-m_i)\log(1-\mu_i). $$ As a function of $\mu_i$, this is uniquely maximized at $\mu_i=m_i$.

To verify the result, differentiate the expected contribution with respect to $\mu_i$:

\frac{d}{d\mu_i}E[q_i\mid X_i] =\frac{m_i}{\mu_i}-\frac{1-m_i}{1-\mu_i} =\frac{m_i-\mu_i}{\mu_i(1-\mu_i)}.

The derivative is positive if $\mu_i<m_i$, zero at $\mu_i=m_i$, and negative if $\mu_i>m_i$. The second derivative is $-m_i/\mu_i^2-(1-m_i)/(1-\mu_i)^2<0$, so the maximum is unique. Consequently, if the conditional mean is correctly specified as $\Lambda(X_i'\beta_0)$, maximizing the population criterion targets $\beta_0$, subject to identification. No further feature of the conditional distribution of $Y_i$ is needed for this mean-targeting argument.

Worked Example: Boundary Outcomes in an Intercept-Only Model

Consider four hypothetical fractional responses $0,\ 1,\ 0.25,\ 0.75$ and an intercept-only fractional logit model. Their sample mean is $$ \bar{Y}=\frac{0+1+0.25+0.75}{4}=\frac{2}{4}=0.50. $$ At an intercept of $\beta=0$, the fitted mean is $\Lambda(0)=0.50$.

Each of the four observations contributes $\log(0.50)=-0.6931$ to the criterion: for the zero, the contribution is $\log(1-0.50)$; for the one, it is $\log(0.50)$; and for either fractional value, the two weighted logarithms sum to $\log(0.50)$. Thus $$ Q_n(0)=4\log(0.50)=-2.7726. $$ The score is $$ S_n(0)=(0-0.50)+(1-0.50)+(0.25-0.50)+(0.75-0.50)=0. $$ In this intercept-only model, the score equation sets the fitted mean equal to the sample mean. Here the maximizer is $\hat{\beta}=0$, since $\Lambda(0)=0.50$. Both boundary observations have finite criterion contributions, and neither is discarded or replaced.

Conditions for QMLE and Inference

The fractional logit criterion is useful when a logistic conditional mean is a reasonable specification for a response in $[0,1]$. Its statistical justification still depends on assumptions about the data and the mean model. A standard set of conditions includes:

Correct conditional mean: $E[Y_i\mid X_i]=\Lambda(X_i'\beta_0)$ for some finite parameter vector $\beta_0$.
Regressors and identification: the regressors are appropriately observed, and different parameter vectors do not produce the same conditional means over the relevant covariate support. The design must contain enough independent variation to identify the coefficients.
Sampling and regularity: observations are independent under the basic formulation, and moments and other regularity conditions needed for consistency and asymptotic approximation hold.

Under these conditions, maximizing the sample criterion estimates the parameter that describes the specified conditional mean. Crucially, the variance of $Y_i\mid X_i$ need not equal the Bernoulli variance $\mu_i(1-\mu_i)$. The Bernoulli expression supplies the criterion and score, not a required conditional variance model. If the variance differs, inference should use a variance estimator robust to that misspecification. The ordinary Bernoulli model-based variance formula generally is not justified just because the criterion has a Bernoulli form.

Worked Example: A Saturated Model With Fractional Groups

Suppose a hypothetical sample has three fractional outcomes at each of two regressor values:

Regressor $X_i$	Observed outcomes $Y_i$	Group mean
0	0, 0.20, 0.40	0.20
1	0.60, 0.80, 1	0.80

Fit a logit conditional mean with an intercept and $X_i$: $\mu_i=\Lambda(\beta_0+\beta_1X_i)$. This model has one fitted mean for each of the two regressor values. The group score equations require the sum of $Y_i-\mu_i$ to be zero in each group, so the fitted mean in each group equals that group’s sample mean. Therefore, $$ \hat{\beta}_0=\log\left(\frac{0.20}{1-0.20}\right)=-1.3863, \qquad \hat{\beta}_0+\hat{\beta}_1=\log\left(\frac{0.80}{1-0.80}\right)=1.3863, $$ and hence $\hat{\beta}_1=1.3863-(-1.3863)=2.7726$. These coefficients produce fitted means $0.20$ at $X_i=0$ and $0.80$ at $X_i=1$.

For the $X_i=0$ group, the criterion contribution is $$ (0+0.20+0.40)\log(0.20)+(3-0.60)\log(0.80) =0.60\log(0.20)+2.40\log(0.80)=-1.501. $$ For the $X_i=1$ group it is $$ (0.60+0.80+1)\log(0.80)+(3-2.40)\log(0.20) =2.40\log(0.80)+0.60\log(0.20)=-1.501. $$ The total criterion is $-1.501-1.501=-3.002$, to the displayed precision. The observed zero and one are admissible in this fit: their contributions are $\log(0.80)$ and $\log(0.80)$, respectively. The fitted means remain strictly inside the unit interval.

Boundary Outcomes: What They Do and Do Not Mean

A zero or one is a valid observed outcome in fractional logit. At a finite coefficient vector, a fitted logit mean is strictly between zero and one; that fact does not prevent the observed response from being at a boundary. The criterion remains finite at either boundary, and the score formula $X_i(Y_i-\mu_i)$ remains valid there.

The distinction between a boundary observation and a boundary conditional mean matters. A model with finite coefficients cannot assign a fitted mean exactly zero or one. If the true conditional mean is exactly at a boundary for some covariate values, or if the data pattern drives estimates toward infinite coefficients, the usual finite-parameter model and its regularity conditions may fail. Having a few observed zeros or ones alone does not establish either problem.

Common Mistake and Key Takeaway

Mistake: “The Bernoulli likelihood assumes the response is binary, so fractional observations must be rounded, expanded into trials, or excluded.”
Correction: Fractional logit maximizes the Bernoulli-form criterion as a quasi-likelihood. Its score and expected criterion target a correctly specified logit conditional mean for $Y_i\in[0,1]$, without requiring a Bernoulli distribution. Rounding changes the observed data, and treating a fraction as a count of trials requires additional information and assumptions.

The key point is that the criterion’s usefulness comes from its relationship to the conditional mean. It does not license a Bernoulli interpretation of fractional responses or justify Bernoulli model-based standard errors when the conditional variance is misspecified.

Check Your Understanding

Answer each question using the criterion and assumptions defined above.

For $Y_i=0$ and fitted mean $\mu_i=0.30$, what is the observation’s criterion contribution? Is it finite?
Write the fractional logit score contribution for an observation with regressor vector $X_i$, response $Y_i$, and fitted mean $\mu_i$.
Why does maximizing the expected criterion target $E[Y_i\mid X_i]$, even if $Y_i$ is not Bernoulli?
In an intercept-only model with observations $0,\ 0.20,\ 0.80,\ 1$, calculate the sample mean and the corresponding intercept estimate.
Does observing one or more outcomes equal to zero or one imply that the fitted conditional mean must also equal zero or one? Explain.