Identity Links And Direct Risk Differences

Modeling Probability Directly

For a binary outcome, the conditional mean is a probability: if $Y$ records whether an event occurred, then $E[Y\mid X=x]=P(Y=1\mid X=x)$. The identity link models that probability directly as a linear predictor. This makes coefficients easy to interpret on the risk scale, but creates a constraint that does not arise for an unrestricted linear predictor: every predicted probability must remain between zero and one.

Let $Y_i$ be a binary response and let $x_i$ be a column vector of predictors, including an intercept if required. Under independent observations, the identity-link Bernoulli regression model is

Y_i\mid x_i\sim\operatorname{Bernoulli}(p_i),\qquad p_i=E[Y_i\mid x_i]=x_i^\top\beta.

The linear predictor is $x_i^\top\beta$, and the identity link sets it equal to the conditional mean. Unlike links whose inverse maps every real number into $(0,1)$, the identity inverse does not automatically produce a probability. Its predictions are valid only at covariate values for which $0\le x_i^\top\beta\le1$. If probabilities are required to be strictly between zero and one, the inequalities must be strict.

Definition: An identity-link binary regression model specifies $P(Y=1\mid X=x)=x^\top\beta$ on a stated covariate domain. The model is probability-valid on that domain if $0\le x^\top\beta\le1$ for every $x$ in the domain.

At a boundary value, a Bernoulli distribution is still well-defined: $p=0$ means $Y=0$ with probability one, and $p=1$ means $Y=1$ with probability one. Some fitting procedures or interpretations require probabilities strictly inside the interval, so it is important to state whether boundary values are allowed.

Coefficients on the Risk-Difference Scale

A risk is the probability of the event in a specified group or at a specified covariate value. A risk difference is the difference between two such probabilities. With the identity link, a change in the linear predictor is exactly a change in risk; no derivative of a nonlinear inverse link is needed.

Theorem: Suppose $P(Y=1\mid X=x)=x^\top\beta$ is valid at two covariate vectors $x$ and $x'$. Then the conditional risk difference for changing the covariate vector from $x$ to $x'$ is $$ P(Y=1\mid X=x')-P(Y=1\mid X=x)=(x'-x)^\top\beta. $$ If one coordinate is a predictor $z$ that enters linearly with coefficient $\beta_z$, and the other design coordinates are held fixed, increasing $z$ by $\Delta z$ changes the conditional risk by $\Delta z\,\beta_z$.

Proof. By the model definition, the two conditional probabilities are $x'^\top\beta$ and $x^\top\beta$. Subtracting gives $x'^\top\beta-x^\top\beta=(x'-x)^\top\beta$. If only the coordinate for $z$ changes by $\Delta z$, the difference vector has that coordinate equal to $\Delta z$ and all other coordinates equal to zero. Its inner product with $\beta$ is therefore $\Delta z\,\beta_z$. $\square$

For a binary treatment indicator $T$ coded zero and one, if the model includes $T$ additively and all other predictors are held fixed, its coefficient is the treated-minus-untreated conditional risk difference. Multiplying by 100 expresses that difference in percentage points. A coefficient of $0.06$, for example, means a six-percentage-point difference, not a six-percent relative increase.

This is a model-based conditional association. It is a causal treatment effect only under additional assumptions about treatment assignment, confounding, consistency, and the target population. The identity link itself supplies no causal interpretation.

Worked Example: A Direct Difference Between Two Groups

Suppose $T=0$ denotes usual care and $T=1$ denotes a new treatment. The fitted model is $p(T)=0.18+0.07T$. Calculate both risks and their difference.

For usual care, $T=0$, so $p(0)=0.18+0.07(0)=0.18$, or 18%. For the new treatment, $T=1$, so $p(1)=0.18+0.07(1)=0.25$, or 25%. Both are valid probabilities.

The risk difference is $p(1)-p(0)=0.25-0.18=0.07$, which is seven percentage points. This agrees with the coefficient of $T$. It is not a 7% relative increase: the relative increase is $0.07/0.18\approx0.389$, or about 38.9%.

Adjusted Risk Differences

A direct risk difference can also be conditional on other predictors. The comparison must hold those predictors fixed: it contrasts the model’s predictions at two covariate vectors that differ in the exposure of interest. The following example makes the adjustment and the probability checks explicit.

Worked Example: Treatment Difference Adjusted for Age

Suppose $T$ is a binary treatment indicator and $A$ is age in years. Consider the model

$$ p(T,A)=0.10+0.06T+0.004(A-50). $$

The age range of interest is 40 to 70 years. At age 40, the predicted risks are $p(0,40)=0.10+0+0.004(-10)=0.06$ and $p(1,40)=0.12$. At age 70, they are $p(0,70)=0.10+0.004(20)=0.18$ and $p(1,70)=0.24$. The predictions at both ends are between zero and one; because the model is affine in age, its predictions between these endpoints are also between those endpoint predictions.

At any fixed age in this range, the treatment contrast is $p(1,A)-p(0,A)=0.06$. For instance, at age 65 the untreated prediction is $0.10+0.004(15)=0.16$ and the treated prediction is $0.22$, again a difference of 0.06, or six percentage points. The age coefficient gives a different contrast: increasing age by one year while holding treatment fixed changes predicted risk by $0.004$, or 0.4 percentage points.

The probability condition applies to every covariate combination at which predictions are to be used, not merely to an average prediction. A model may give valid predictions for the observed data and invalid ones for a proposed extrapolation. For a linear predictor, checking the covariate domain is therefore part of specifying and interpreting the model.

Checking Probability Constraints

For a finite set of observed covariate vectors, validity is checked by evaluating $x_i^\top\beta$ at each vector. For prediction across a continuous range, one must check the whole stated range. A useful exact result applies when the predictors range independently over a rectangle.

Theorem: Let $p(x)=\beta_0+\sum_{j=1}^q\beta_jx_j$, and suppose the covariate domain is the rectangle $a_j\le x_j\le b_j$ for each $j$. The minimum and maximum of $p(x)$ on this domain are attained at corners. Consequently, the model is probability-valid throughout the rectangle if and only if its predictions at all $2^q$ corners lie in $[0,1]$.

Proof. Fix all coordinates except $x_1$. As a function of $x_1$, $p(x)$ is affine. Every $x_1\in[a_1,b_1]$ can be written as $x_1=\lambda a_1+(1-\lambda)b_1$ for some $0\le\lambda\le1$. The corresponding prediction is the same weighted average of the predictions at $a_1$ and $b_1$, with the other coordinates held fixed. Repeat this argument for each coordinate. It follows that at any point in the rectangle, $p(x)$ is a weighted average of its values at the corners, with nonnegative weights summing to one. It cannot be below the smallest corner value or above the largest. Thus, if all corner values lie in $[0,1]$, every value in the rectangle does too. Conversely, if the model is valid throughout the rectangle, it is valid at each corner because the corners belong to the domain. $\square$

Worked Example: Checking a Two-Predictor Rectangle

Let $p(x_1,x_2)=0.20+0.015x_1-0.01x_2$, with $0\le x_1\le20$ and $0\le x_2\le10$. Check the four corners. They give $p(0,0)=0.20$, $p(20,0)=0.50$, $p(0,10)=0.10$, and $p(20,10)=0.40$. The minimum is 0.10 and the maximum is 0.50, so the model is probability-valid throughout this rectangle.

For a one-unit increase in $x_1$, holding $x_2$ fixed, the conditional risk difference is 0.015, or 1.5 percentage points. For a one-unit increase in $x_2$, holding $x_1$ fixed, it is $-0.01$, or a decrease of one percentage point. These interpretations apply on the specified domain; the validity check does not justify extrapolation beyond it.

Extrapolation and Estimation

The constraints matter both when interpreting a fitted model and when estimating its coefficients. For independent Bernoulli observations, the likelihood at a coefficient vector $\beta$, provided $p_i=x_i^\top\beta$ is valid, is

L(\beta)=\prod_{i=1}^n (x_i^\top\beta)^{y_i} (1-x_i^\top\beta)^{1-y_i}.

A coefficient vector that produces a value outside $[0,1]$ does not specify Bernoulli probabilities for that observation, so it cannot be treated as an ordinary valid model fit. In practice, algorithms and inferential procedures may impose or encounter additional restrictions, especially near boundary probabilities. The essential modeling requirement remains: define the covariate domain, and ensure the predicted mean is a probability throughout the domain relevant to the analysis.

Worked Example: Valid Observed Range, Invalid Extrapolation

Suppose a model uses a nonnegative exposure $x$ and predicts $p(x)=0.08+0.025x$. The observed exposure range is from 0 to 30. At its endpoints, the predictions are $p(0)=0.08$ and $p(30)=0.08+0.75=0.83$. Since this prediction increases with $x$, all predictions over the observed range lie between 0.08 and 0.83 and are valid.

If the model is extrapolated to $x=40$, it predicts $p(40)=0.08+1.00=1.08$, which is not a probability. Solving $0.08+0.025x\le1$ gives $x\le0.92/0.025=36.8$, so the model ceases to be probability-valid above 36.8. One must not report 1.08 as a risk or silently replace it by 1; either changes the specified model. The linear model is unsuitable for that extrapolation unless a different valid model is adopted.

Why the Link Matters

The identity link is attractive when a direct risk difference is the scientific quantity of interest. Coefficients are on the probability scale, adjusted contrasts are transparent, and predictions at selected covariate values are easy to communicate. Those advantages are strongest when the covariate domain is restricted and the fitted probabilities stay comfortably within the probability range.

The central trade-off is that a linear predictor is not intrinsically bounded. A coefficient can have a clear risk-difference meaning while a model with that coefficient gives impossible predictions elsewhere. Interpretation and validity must therefore be considered together: state the contrast, specify the covariates held fixed, and check the predictions over the population or range where they will be used.

Confusing percentage points with relative percent change. A coefficient of 0.05 is a five-percentage-point difference, not necessarily a 5% relative increase.
Reading a coefficient without specifying the comparison. A conditional risk difference compares predictions with other covariates held fixed.
Assuming a linear predictor is automatically a probability. Check that it lies between zero and one across the relevant covariate domain.
Calling an association causal. A link function defines a conditional mean model; causal interpretation requires additional design and identification assumptions.

Key takeaway: With the identity link, changes in the linear predictor are direct conditional risk differences. That simplicity comes with a strict domain requirement: every prediction used as a probability must lie in $[0,1]$.

Check Your Understanding

Answer each question using the model and covariate range specified.

In $p(T)=0.12+0.04T$ for $T\in\{0,1\}$, calculate both risks and state the treatment risk difference in percentage points.
For $p(x)=0.30-0.02x$, what is the risk difference for a two-unit increase in $x$, provided both predictions are valid?
Why does a coefficient of 0.08 in an identity-link model not, by itself, establish an 8% causal benefit?
For $p(x)=0.15+0.03x$, determine the largest $x\ge0$ for which the prediction is at most one.
Explain why checking fitted probabilities only at the mean covariate vector does not establish validity throughout a covariate rectangle.