Why Work With Observation-Level Scores?
A binary-response model assigns a conditional probability to each observed outcome. Maximum-likelihood estimation chooses parameters that make the observed outcomes collectively more likely under the model. The key calculation is therefore the contribution each observation makes to the log-likelihood, and how that contribution changes as the parameters change. Those derivatives are the observation-level score contributions.
Keeping the contributions separate is useful for understanding the total score, checking a gradient, and seeing which observations push an estimate in which direction. Logit and probit use different probability functions, but both begin with the same Bernoulli likelihood structure.
Model and Bernoulli Log-Likelihood
For observation \(i\), let \(Y_i\) be a binary outcome taking values 0 or 1, and let \(\boldsymbol{x}_i\) be a \(k\)-component vector of observed covariates, including an intercept if the model specifies one. Write \(\boldsymbol{\beta}\) for the \(k\)-component parameter vector and \(\eta_i=\boldsymbol{x}_i^\mathsf{T}\boldsymbol{\beta}\) for the index. Conditional on the covariates, assume \(Y_i\) is Bernoulli with success probability \(p_i\), determined by the chosen link. For the likelihood across observations, assume the outcomes are conditionally independent given their covariates and the parameters.
If \(y_i=1\), the contribution is \(\log p_i\); if \(y_i=0\), it is \(\log(1-p_i)\). Conditional independence means the sample likelihood is the product of the individual contributions. Taking logarithms turns that product into a sum:
Here \(L\) denotes the likelihood and \(\ell\) the log-likelihood. In both models, \(p_i=F(\eta_i)\), where \(F\) is the specified response function. For logit, \(F(\eta)=1/(1+\exp(-\eta))\); for probit, \(F(\eta)=\Phi(\eta)\), the standard normal cumulative distribution function. In the probit score, \(\phi\) denotes the standard normal density.
Deriving the General Score Contribution
The observation-level score is the vector of derivatives of \(\ell_i\) with respect to \(\boldsymbol{\beta}\). First differentiate with respect to \(p_i\), then use the chain rule through \(p_i=F(\eta_i)\) and \(\eta_i=\boldsymbol{x}_i^\mathsf{T}\boldsymbol{\beta}\). The derivative of \(\eta_i\) with respect to \(\boldsymbol{\beta}\) is \(\boldsymbol{x}_i\).
Let \(f(\eta)=F'(\eta)\). Since \(\partial p_i/\partial\boldsymbol{\beta}=f(\eta_i)\boldsymbol{x}_i\), the chain rule gives the general result:
This expression applies when the modeled probability is differentiable and strictly between 0 and 1. Its scalar factor depends on the observed outcome and the link; the covariate vector determines how that scalar becomes a contribution to each parameter’s score.
Logit and Probit Score Contributions
For logit, \(f(\eta_i)=p_i(1-p_i)\). Those factors cancel in the general score, giving a particularly simple contribution. For probit, \(f(\eta_i)=\phi(\eta_i)\), and the probability terms remain in the denominator.
In logit, an observation with \(y_i=1\) contributes a positive multiple of \(\boldsymbol{x}_i\) when its fitted probability is below 1; an observation with \(y_i=0\) contributes a negative multiple when its fitted probability is above 0. This sign description concerns the scalar multiplier: individual components also depend on the signs of the covariates. In probit, the same residual \(y_i-p_i\) appears, but the density and probability factors rescale the contribution.
Worked Example: A Logit Score for One Observation
Suppose a hypothetical observation has \(y_i=1\), \(\boldsymbol{x}_i=(1,2)^\mathsf{T}\), and current parameters \(\boldsymbol{\beta}=(0,0.2)^\mathsf{T}\). Then \(\eta_i=0+2(0.2)=0.4\), and the logit probability is \(p_i=1/(1+\exp(-0.4))\approx0.598688\).
The log-likelihood contribution is \(\ell_i=\log(0.598688)\approx-0.5130\). The logit score is \(\boldsymbol{s}_i=\boldsymbol{x}_i(y_i-p_i)\), so substitution gives \((1,2)^\mathsf{T}(1-0.598688)\).
Thus \(\boldsymbol{s}_i\approx(0.401312,0.802624)^\mathsf{T}\). The first component is the contribution to the intercept score and the second is the contribution to the slope score. Both are positive because this observation has \(y_i=1\) and its fitted probability is below 1.
Worked Example: A Probit Score for One Observation
For a hypothetical observation, let \(y_i=0\), \(\boldsymbol{x}_i=(1,2)^\mathsf{T}\), and \(\eta_i=0.4\). Then \(p_i=\Phi(0.4)\approx0.655422\), \(1-p_i\approx0.344578\), and \(\phi(0.4)\approx0.368270\). The observed outcome’s log-likelihood contribution is \(\log(1-p_i)\approx-1.0655\).
For \(y_i=0\), the scalar multiplying \(\boldsymbol{x}_i\) in the probit score is \[ \frac{\phi(0.4)(0-0.655422)} {0.655422(0.344578)} =-\frac{\phi(0.4)}{1-\Phi(0.4)} \approx-1.06876. \]
Therefore \(\boldsymbol{s}_i\approx(-1.06876,-2.13751)^\mathsf{T}\). The negative signs reflect that the observed zero is below the fitted success probability. The two components differ by a factor of 2 because the second covariate is 2 while the intercept covariate is 1.
From Individual Scores to Maximum Likelihood
Because the sample log-likelihood is a sum, its score is also a sum of observation-level score contributions:
For an interior, finite maximum at which the log-likelihood is differentiable, a necessary first-order condition is \(\boldsymbol{S}(\widehat{\boldsymbol{\beta}})=\boldsymbol{0}\). Maximum-likelihood software typically searches for a parameter vector satisfying this condition, using the score together with curvature information and numerical optimization methods. A zero score alone does not establish that the point is a maximum: it could be a minimum or another stationary point, and numerical convergence should be checked against the likelihood and other diagnostics.
Worked Example: Summed Logit Score and a Newton Update
Consider three hypothetical binary outcomes \(y=(1,1,0)\) with an intercept-only logit model. At the current value \(\beta=0.6667\), all three fitted probabilities are \(p=1/(1+\exp(-0.6667))\approx0.660764\). The individual logit score contributions are \(y_i-p\), so they are approximately \(0.339236\), \(0.339236\), and \(-0.660764\).
Adding them gives the sample score \(S=2-3p\approx2-3(0.660764)=0.01771\). The score is slightly positive, so the likelihood is locally increasing as \(\beta\) increases at this current value.
For this intercept-only example, differentiating the score \(S(\beta)=\sum_i(y_i-p)\) gives \(H(\beta)=-3p(1-p)\). At the stated value, \(H\approx-3(0.660764)(0.339236)=-0.67247\). A Newton update for maximizing the log-likelihood is \(\beta_{\mathrm{new}}=\beta-S/H\), hence \[ \beta_{\mathrm{new}}\approx0.6667-\frac{0.01771}{-0.67247} \approx0.6930. \]
This is one numerical update, not a claim that the updated value is the final estimate. An estimation routine would continue iterating and check convergence. The negative curvature here is consistent with a local maximum in this one-parameter example, but the score equation and curvature must be considered in the fitted model as a whole.
A Useful Score Check and a Common Pitfall
Under the stated model, the score has conditional expectation zero at the true parameter. For a single observation, conditional on \(\boldsymbol{x}_i\), the probability that \(Y_i=1\) is \(p_i\), so \(E(Y_i-p_i\mid\boldsymbol{x}_i)=p_i-p_i=0\). The general score formula therefore has conditional expectation zero as well, provided its probability factors are defined. This is a useful theoretical check on the score expression; it does not mean each realized observation has a zero score.
A common mistake is to use the logit simplification \(\boldsymbol{x}_i(y_i-p_i)\) for probit because both models have a residual \(y_i-p_i\). That drops the probit rescaling factor \(\phi(\eta_i)/[p_i(1-p_i)]\). The correction is to differentiate the specific response function: its derivative cancels the Bernoulli denominator for logit, but not for probit. Another mistake is to treat a small total score as proof of a valid maximum. It is a first-order condition; likelihood behavior and convergence still require checking.
Check Your Understanding
Use the definitions and formulas above to answer each question.
- Write the Bernoulli log-likelihood contribution when \(y_i=0\), and explain why it reduces to that expression.
- For logit, what feature of the response function makes the score simplify to \(\boldsymbol{x}_i(y_i-p_i)\)?
- For probit, identify the scalar multiplying \(\boldsymbol{x}_i\) in the observation-level score.
- Why does conditional independence allow the sample score to be written as a sum of observation-level scores?
- Why does satisfying the score equation not, by itself, prove that an estimate is a maximum of the log-likelihood?