Why a Predicted Probability Needs an Interval
A fitted binary-response model can report a probability such as 0.62 for an individual with specified covariates. That number is an estimate, not a certainty: the model coefficients were estimated from a finite sample. A confidence interval describes the resulting sampling uncertainty about the conditional probability at those covariate values. It does not say that the individual outcome itself is uncertain by exactly the interval’s width.
The interval is usually constructed on the model’s index scale, where coefficient uncertainty is approximately normal, and then transformed back to the probability scale. This is especially useful because it respects the probability range and naturally allows uncertainty to be asymmetric around the estimated probability.
The Model and the Target
Let \(Y_i\) be a binary outcome and \(x_i\) a vector of observed covariates, including an intercept if the model has one. In a logit or probit model, the conditional mean of \(Y_i\), which is also the conditional probability that \(Y_i=1\), is linked to a linear index:
For logit, \(G(t)=\Lambda(t)=1/(1+\exp(-t))\); for probit, \(G(t)=\Phi(t)\), the standard normal cumulative distribution function. Both links are increasing and map any real index to a probability strictly between zero and one.
Suppose the fitted coefficient vector is \(\hat{\beta}\), with estimated covariance matrix \(\widehat V\). The estimated index at \(x\) is \(\hat{\eta}(x)=x^\top\hat{\beta}\), and the estimated probability is \(\hat p(x)=G(\hat{\eta}(x))\). For the usual large-sample interval, assume the binary-response model is correctly specified, observations provide independent information, the coefficients are identified, and the estimator is in a regular large-sample setting with nonsingular information. The covariance estimate must be appropriate for the sampling design and dependence structure. The covariate vector \(x\) is treated as fixed for this pointwise calculation.
Constructing an Interval by Transforming the Index
Because the index is linear in the estimated coefficients, its estimated variance follows directly from the coefficient covariance matrix. A large-sample confidence interval for the index is then transformed using the model’s increasing link function.
The calculation has four steps: evaluate the index at the covariate values of interest, find its standard error, form the index-scale interval, and apply \(G\) to both endpoints. For a 95% interval, \(z_{0.975}\) is approximately 1.96. For logit, apply \(\Lambda\); for probit, apply \(\Phi\). Do not transform the standard error itself or apply the link to the coefficient limits separately.
Why the Transformation Works
The justification uses the large-sample normal approximation for the coefficient estimator and the fact that the link is increasing. At fixed \(x\), the index is a linear combination of the coefficients, so its estimated variance is \(x^\top\widehat Vx\). Under the regularity conditions above, the standardized index error is approximately standard normal:
Consequently, the index interval from \(\hat{\eta}(x)-z_{1-\alpha/2}\widehat{SE}\) to \(\hat{\eta}(x)+z_{1-\alpha/2}\widehat{SE}\) covers \(\eta(x)\) with approximate probability \(1-\alpha\) over repeated samples. Since \(G\) is increasing, the event that the index lies between these endpoints is exactly the event that \(G(\eta(x))\) lies between the transformed endpoints. Thus the transformed interval inherits the index interval’s approximate coverage for \(p(x)\). The approximation comes from large-sample inference; transforming does not make it an exact finite-sample interval.
Worked Example: A Logit Probability at Specified Covariates
Suppose a hypothetical logit model estimates the probability of an event using an intercept and one covariate, with \(\hat{\beta}=(-0.8,\,0.4)^\top\). At the covariate value \(x=(1,\,2)^\top\), suppose the estimated coefficient covariance matrix is \(\widehat V=\begin{pmatrix}0.04&0\\0&0.01\end{pmatrix}\). Construct a pointwise 95% interval. Assume the model is correctly specified for this illustration, observations are independent, and the estimates satisfy the large-sample conditions for the normal approximation.
Index and estimate. The estimated index is \(\hat{\eta}=x^\top\hat{\beta}=1(-0.8)+2(0.4)=0\). The fitted event probability is \(\hat p=\Lambda(0)=1/(1+\exp(0))=0.5000\).
Index standard error. Substituting into the covariance formula gives \(x^\top\widehat Vx=(1,2)\begin{pmatrix}0.04&0\\0&0.01\end{pmatrix}(1,2)^\top =0.04+4(0.01)=0.08\). Therefore, \(\widehat{SE}(\hat{\eta})=\sqrt{0.08}=0.2828\).
Index interval and transformation. The 95% index interval is \(0\pm1.96(0.2828)=0\pm0.5543\), or \([-0.5543,\,0.5543]\). Applying the logit link to each endpoint gives \([\Lambda(-0.5543),\,\Lambda(0.5543)]=[0.3648,\,0.6352]\).
Under the stated model and large-sample assumptions, this procedure gives an approximate 95% confidence interval for the event probability at \(x=(1,2)^\top\). It is not an interval for an individual’s eventual binary response.
Worked Example: A Probit Probability at a Different Covariate Profile
In a hypothetical probit model, suppose the estimated index at a specified profile is \(\hat{\eta}=0.30\), with estimated index standard error \(0.20\). Assume correct probit specification, independent observations, fixed covariates for the target profile, and a valid large-sample covariance estimate. Find a pointwise 95% interval for the conditional probability.
The fitted probability is \(\hat p=\Phi(0.30)\approx0.6179\). The index interval is \(0.30\pm1.96(0.20)=0.30\pm0.392\), which gives \([-0.092,\,0.692]\).
Transform both endpoints using the standard normal cumulative distribution function: \(\Phi(-0.092)\approx0.4633\) and \(\Phi(0.692)\approx0.7555\). Thus the transformed interval is approximately \([0.4633,\,0.7555]\). In context, it is an approximate 95% confidence interval for the conditional probability of the event at this covariate profile, not a range containing 95% of individual outcomes.
Why the Interval Can Be Asymmetric
A symmetric interval for the index generally becomes asymmetric when mapped to the probability scale because the link is curved. This asymmetry is not a defect: probability changes for a given change in the index depend on where the index lies. The transformation also ensures that the endpoints remain in the probability range.
Worked Example: Avoiding an Impossible Probability Endpoint
Consider a hypothetical logit model at a fixed covariate profile with \(\hat{\eta}=2.2\) and estimated index standard error \(0.6\). Assume the same correct-specification, independence, fixed-profile, and large-sample conditions used for the earlier examples.
The fitted probability is \(\Lambda(2.2)=1/(1+\exp(-2.2))\approx0.9002\). The 95% index interval is \(2.2\pm1.96(0.6)=2.2\pm1.176=[1.024,\,3.376]\). Transforming gives \([\Lambda(1.024),\,\Lambda(3.376)]\approx[0.7358,\,0.9669]\).
For comparison, a first-order delta-method calculation uses the logit derivative \(\Lambda(\eta)\{1-\Lambda(\eta)\}\). At the fitted index, the approximate probability standard error is \(0.9002(1-0.9002)(0.6)\approx0.0539\). A symmetric 95% interval based on that standard error is \(0.9002\pm1.96(0.0539)=[0.7946,\,1.006]\), which extends above 1. The transformed-index interval avoids that impossible endpoint and reflects the curvature of the probability scale. Both are approximations; their differences can matter, especially near zero or one.
Conditional-Probability Uncertainty Is Not Outcome Uncertainty
For a fixed \(x\), the model says \(Y\mid x\) is Bernoulli with event probability \(p(x)\). Even if \(p(x)\) were known exactly, a future outcome would still vary: it would equal 1 with probability \(p(x)\) and 0 with probability \(1-p(x)\). That Bernoulli variation is distinct from sampling uncertainty in estimating \(p(x)\).
A confidence interval such as \([0.46,0.76]\) describes uncertainty about the conditional probability parameter at the specified profile. It does not mean that a future person’s outcome has a 95% chance of lying between 0.46 and 0.76; outcomes are zero or one. Nor does it imply that 95% of future outcomes will fall within the interval. Predicting an individual outcome requires a different target, while the interval here concerns the model’s conditional mean probability.
The target also matters across covariate profiles. A pointwise interval at one \(x\) is not automatically an interval for an average probability over a population, and it does not provide simultaneous coverage for probabilities at every possible \(x\). Those questions require defining the corresponding aggregate or joint target and its uncertainty.
Common Mistakes and Practical Checks
- Reporting a coefficient interval as a probability interval. Coefficients and indices are on the link scale. First form the interval for \(x^\top\beta\), then transform its endpoints with the inverse link.
- Using the wrong standard error. The standard error for the index is \(\sqrt{x^\top\widehat Vx}\), not the standard error of one coefficient unless the target depends only on that coefficient.
- Calling it an individual prediction interval. The interval quantifies uncertainty in \(P(Y=1\mid x)\), not the randomness of a future zero-or-one outcome.
- Ignoring the target profile. State the covariate values to which the probability refers. Changing \(x\) changes both the index and, generally, its standard error.
- Overstating coverage. This is an approximate pointwise interval under the model and inference assumptions, not a guarantee of exact coverage or simultaneous coverage over many profiles.
Before reporting an interval, check that the coefficient covariance estimate is appropriate, the index variance is nonnegative, the correct link has been used, and the result is described in context as a confidence interval for a conditional event probability. In a fitted binary model, the final probability bounds should lie between zero and one.
Check Your Understanding
Answer each question using the distinction between the index scale, probability scale, and binary outcome scale.
- For a logit model at fixed \(x\), what quantity is \(\sqrt{x^\top\widehat Vx}\) the standard error of?
- A 95% index interval is \([-0.4,\,0.8]\). How would you obtain the corresponding logit probability interval, and why do you transform both endpoints?
- Why is an index-based interval transformed through the link generally asymmetric around the fitted probability?
- In a probit model, a pointwise interval for \(p(x)\) is \([0.3,\,0.6]\). Does this mean a future binary outcome has a 95% chance of lying in that numerical interval? Explain.
- What additional interpretation problem arises if a pointwise interval at one covariate profile is reported as an interval for the average probability in a population?