Common-Slope Adjacent-Category Models

Why Share Slopes Across Adjacent Categories?

For an ordered response, adjacent-category logits compare each category with the one immediately below it. With three response levels, for example, the comparisons are level 2 versus level 1 and level 3 versus level 2. A covariate may affect both comparisons, but a model with a common slope requires its effect on the log odds to be the same in each comparison. The intercepts can still differ, so the odds themselves need not be equal.

This restriction is useful when a scientific question suggests a consistent shift across neighboring levels and a more economical model is desired. Its precise meaning is about how the odds change across covariate values—not about equal category probabilities or equal probability changes.

Definition of the Common-Slope Model

Let $Y$ take ordered values $1,\ldots,J$, where $J\geq 2$, and write $\pi_j(x)=P(Y=j\mid x)$. Assume these conditional probabilities are positive. Let $x$ be a covariate vector, and let $\alpha_1,\ldots,\alpha_{J-1}$ be category-specific intercepts. We use the orientation in which the adjacent-category odds compare the higher category with the lower one.

Definition: A common-slope adjacent-category model has the form $$ \log\left(\frac{\pi_{j+1}(x)}{\pi_j(x)}\right) = \alpha_j+x^\mathsf{T}\beta, \qquad j=1,\ldots,J-1. $$ The intercept $\alpha_j$ may vary with the adjacent pair; the slope vector $\beta$ is shared by all pairs.

The corresponding unrestricted linear adjacent-category regression would instead use a separate slope vector $\beta_j$ for each pair: $\log\{\pi_{j+1}(x)/\pi_j(x)\}=\alpha_j+x^\mathsf{T}\beta_j$. The common-slope model imposes the restrictions $\beta_1=\cdots=\beta_{J-1}=\beta$. Thus it is the slopes—not the intercepts—that are constrained to agree.

Worked Example: One Covariate and Three Response Levels

Suppose the response levels are low, medium, and high, and the model is $\log\{\pi_2(x)/\pi_1(x)\}=-0.2+0.4x$ and $\log\{\pi_3(x)/\pi_2(x)\}=0.7+0.4x$. The common slope is $0.4$, while the two intercepts differ.

At $x=0$, the adjacent odds are $\pi_2/\pi_1=\exp(-0.2)\approx0.8187$ and $\pi_3/\pi_2=\exp(0.7)\approx2.0138$. At $x=1$, they are $\exp(0.2)\approx1.2214$ and $\exp(1.1)\approx3.0042$.

For the first pair, the odds multiplier from $x=0$ to $x=1$ is $1.2214/0.8187\approx1.4918$. For the second pair it is $3.0042/2.0138\approx1.4918$, which equals $\exp(0.4)$. The adjacent odds differ from one pair to the next, but the one-unit covariate change multiplies both by the same amount.

What the Shared Slope Implies

Theorem: In the common-slope model, for any two covariate values $x$ and $x'$, the odds ratio comparing the adjacent odds at $x'$ with those at $x$ is the same for every adjacent pair: $$ \frac{\pi_{j+1}(x')/\pi_j(x')}{\pi_{j+1}(x)/\pi_j(x)} = \exp\left\{(x'-x)^\mathsf{T}\beta\right\}, \qquad j=1,\ldots,J-1. $$ Conversely, if these ratios are equal across all adjacent pairs for every covariate contrast in the model, then the adjacent logits have a common slope vector, with pair-specific intercepts.

Proof. Subtract the log adjacent odds at $x$ from those at $x'$. Under the model,

\begin{aligned} &\log\left(\frac{\pi_{j+1}(x')}{\pi_j(x')}\right) -\log\left(\frac{\pi_{j+1}(x)}{\pi_j(x)}\right)\\ &\qquad = \left(\alpha_j+(x')^\mathsf{T}\beta\right) -\left(\alpha_j+x^\mathsf{T}\beta\right) =(x'-x)^\mathsf{T}\beta. \end{aligned}

Exponentiating gives the stated ratio, which does not depend on $j$. For the converse, write the adjacent log odds as $g_j(x)$. If the change $g_j(x')-g_j(x)$ is the same across pairs for every covariate contrast, then each $g_j$ has the same covariate-dependent part. In a linear model, that means the coefficient vector of $x$ is identical across $j$; the remaining pair-specific constant is $\alpha_j$. $\square$

For a single predictor $x$, a one-unit increase therefore multiplies every neighboring-category odds by $\exp(\beta)$. For a vector of predictors, the multiplier for a change $\Delta x=x'-x$ is $\exp(\Delta x^\mathsf{T}\beta)$. This also clarifies why the claim is about a particular covariate contrast: changing several covariates at once combines their contributions through the inner product.

Worked Example: A Joint Change in Two Covariates

Suppose there are three ordered response levels and two predictors: a smoking indicator $s$ and age $a$ measured in decades. Let $\beta=(\log(1.5),\log(0.8))^\mathsf{T}$, with intercepts $\alpha_1=\log 2$ and $\alpha_2=0$.

Consider moving from $x=(s,a)=(0,0)$ to $x'=(1,1)$: smoking changes from no to yes, and age increases by ten years. The common multiplier is \[ \exp\{(1,1)\beta\} =\exp\{\log(1.5)+\log(0.8)\} =1.5(0.8)=1.2. \]

At $x=(0,0)$, the odds for level 2 versus level 1 are $\exp(\log 2)=2$, and those for level 3 versus level 2 are $\exp(0)=1$. At $x'=(1,1)$, these odds are $2(1.2)=2.4$ and $1(1.2)=1.2$. Each neighboring odds comparison increases by 20%, even though the predictors changed together and the two original odds were different.

Probabilities Implied by the Restriction

The shared-slope specification also determines how the covariates enter the relative weights of all response categories. This follows by adding the adjacent log odds along the ordered categories.

Result: Define $A_1=0$ and $A_j=\sum_{k=1}^{j-1}\alpha_k$ for $j=2,\ldots,J$. Then the common-slope model implies $$ \pi_j(x) = \frac{\exp\{A_j+(j-1)x^\mathsf{T}\beta\}} {\displaystyle\sum_{r=1}^J\exp\{A_r+(r-1)x^\mathsf{T}\beta\}}, \qquad j=1,\ldots,J. $$

Derivation. For $j=1$, $\log\{\pi_1(x)/\pi_1(x)\}=0$. For $j\geq2$, telescope the adjacent log odds:

\begin{aligned} \log\left(\frac{\pi_j(x)}{\pi_1(x)}\right) &=\sum_{k=1}^{j-1}\log\left(\frac{\pi_{k+1}(x)}{\pi_k(x)}\right)\\ &=\sum_{k=1}^{j-1}(\alpha_k+x^\mathsf{T}\beta)\\ &=A_j+(j-1)x^\mathsf{T}\beta. \end{aligned}

Thus each probability is proportional to $\exp\{A_j+(j-1)x^\mathsf{T}\beta\}$. Dividing each such weight by the sum of all $J$ weights makes the probabilities sum to one and gives the displayed formula. $\square$

The factor $j-1$ is important: the shared adjacent slope accumulates as one moves through successive categories. A common slope does not say that every category probability has the same covariate effect. Probabilities must also adjust together to sum to one.

Worked Example: Recovering Probabilities from Common Adjacent Odds

Let $J=3$, with $\alpha_1=\log 2$, $\alpha_2=\log 3$, and scalar slope $\beta=\log 2$. At $x=0$, the adjacent odds are $\pi_2/\pi_1=2$ and $\pi_3/\pi_2=3$. Taking the first category's weight to be 1 gives weights $1,2,6$. Their sum is 9, so the probabilities are $(1/9,2/9,6/9)$.

At $x=1$, each adjacent odds value is multiplied by $\exp(\beta)=2$. The odds are now $\pi_2/\pi_1=4$ and $\pi_3/\pi_2=6$, giving relative weights $1,4,24$. Since these sum to 29, the probabilities are $(1/29,4/29,24/29)$.

The adjacent odds both double, as required. But the category probabilities do not double: for example, the probability of category 3 changes from $6/9$ to $24/29$, not to $12/9$. The latter would not even be a valid probability. The model constrains odds changes, while normalization governs probabilities.

Checking the Restriction and Avoiding a Common Mistake

A direct way to assess the restriction is to compare the covariate effects across adjacent logits. If separate slopes are estimated, a common-slope model requires them to agree. Equivalently, for a fixed covariate contrast, the ratios of adjacent odds across the two covariate values must match across pairs.

Worked Example: Detecting Unequal Adjacent Slopes

Suppose a three-category model has $\log(\pi_2/\pi_1)=0+0.7x$ and $\log(\pi_3/\pi_2)=\log 2-0.2x$. The first adjacent logit has slope $0.7$; the second has slope $-0.2$. For a one-unit increase in $x$, the corresponding odds multipliers are $\exp(0.7)\approx2.014$ and $\exp(-0.2)\approx0.819$.

These multipliers are unequal, so the model does not have a common slope. No single $\beta$ can represent both effects: the first adjacent odds increase while the second decrease. By contrast, changing the intercept $\log 2$ would alter the starting odds for the second pair without changing the slope restriction.

A common mistake is to say that a one-unit increase in $x$ multiplies each category probability by $\exp(\beta)$. The model says that this multiplier applies to each adjacent odds, $\pi_{j+1}/\pi_j$. The probabilities are normalized jointly, so their changes depend on all categories and need not have a common multiplier or a common difference.

Another useful distinction is between equal effects and equal odds. The intercepts $\alpha_j$ allow the odds for one adjacent pair to be much larger or smaller than for another. What is constant across pairs is the multiplicative effect of a specified covariate contrast. This makes the restriction substantive: it asserts a consistent covariate effect on the local odds throughout the ordered response.

Key takeaway: A common-slope adjacent-category model assigns a separate intercept to each neighboring pair but one shared covariate slope vector. Therefore, any fixed covariate contrast multiplies every adjacent-category odds by the same factor, $\exp\{(x'-x)^\mathsf{T}\beta\}$. This is a restriction on odds, not on category probabilities.

Check Your Understanding

Use the stated odds orientation: category $j+1$ versus category $j$.

In a common-slope model, what changes across adjacent pairs, and what remains shared?
If a scalar covariate has slope $\beta=-0.3$, what factor multiplies every adjacent odds for a one-unit increase?
Why does a common slope not mean that every category probability is multiplied by the same factor?
For two predictors with slope vector $\beta$, write the multiplier for a contrast $\Delta x$.
If two estimated adjacent slopes are $0.4$ and $0.4$, but their intercepts differ, does that satisfy the common-slope restriction? Explain.