Forward and Backward Continuation-Ratio Models

Two Directions for Sequential Questions

For an ordered response, a sequential model can ask whether an observation stops at a category or proceeds to a later one. The direction of that sequence matters. A forward continuation-ratio model begins at the lowest category and moves upward; a backward model begins at the highest category and moves downward. Although both use conditional logits, their risk sets—and therefore their conditional questions—are different.

Let $Y$ take one of the ordered values $1,\ldots,K$, and write $p_j=P(Y=j)$. The category order is part of the model’s interpretation: category 1 is the starting end for a forward sequence, and category $K$ is the starting end for a backward sequence. The definitions below concern a fixed distribution, or the distribution at a fixed covariate value.

Definition: At forward step $j$, the risk set is the event $Y\ge j$. Within that set, the observation either stops at $Y=j$ or continues to $Y>j$. The forward continuation-ratio logit is $$ \phi_j=\log\left\{\frac{P(Y=j\mid Y\ge j)}{P(Y>j\mid Y\ge j)}\right\}, \qquad j=1,\ldots,K-1. $$ At backward step $j$, the risk set is $Y\le j$. Within it, the observation either stops at $Y=j$ or continues downward to $Y<j$. The backward continuation-ratio logit is $$ \psi_j=\log\left\{\frac{P(Y=j\mid Y\le j)}{P(Y<j\mid Y\le j)}\right\}, \qquad j=2,\ldots,K. $$

The edge categories do not have a logit in the direction in which there is no alternative category: the forward sequence has no continuation decision after $K$, and the backward sequence has none below 1. A displayed logit is finite when both alternatives have positive conditional probability. If either alternative has probability zero, its odds may be zero or infinite, and the corresponding finite logit is not defined.

Deriving the Risk-Set Formulas

Define the forward tail probability $S_j=\sum_{r=j}^{K}p_r$ and the backward cumulative probability $T_j=\sum_{r=1}^{j}p_r$. The tail or cumulative probability in the conditioning event cancels from the conditional odds.

Result: Whenever the probabilities in the relevant numerator and denominator are positive, the forward and backward logits can be written directly in terms of the category probabilities: $$ \phi_j=\log\left(\frac{p_j}{S_{j+1}}\right) =\log\left(\frac{p_j}{\sum_{r=j+1}^{K}p_r}\right), \qquad j=1,\ldots,K-1, $$ and $$ \psi_j=\log\left(\frac{p_j}{T_{j-1}}\right) =\log\left(\frac{p_j}{\sum_{r=1}^{j-1}p_r}\right), \qquad j=2,\ldots,K. $$

Derivation. For the forward step, the risk-set probability is $P(Y\ge j)=S_j$. Thus $P(Y=j\mid Y\ge j)=p_j/S_j$, while $P(Y>j\mid Y\ge j)=S_{j+1}/S_j$. Dividing these conditional probabilities cancels $S_j$, leaving $p_j/S_{j+1}$. Taking the logarithm gives the formula for $\phi_j$.

For the backward step, $P(Y\le j)=T_j$. The two conditional probabilities are $p_j/T_j$ and $T_{j-1}/T_j$. Their ratio is $p_j/T_{j-1}$, giving the formula for $\psi_j$. This cancellation is why the denominators in the two formulas contain the categories beyond the current category in the direction of travel, rather than the whole risk set.

Worked Examples

Worked Example: Forward and Backward Steps in Four Categories

Suppose an ordered outcome has probabilities $p_1=0.10$, $p_2=0.20$, $p_3=0.30$, and $p_4=0.40$. These sum to 1. We can calculate each forward decision from the lower end and each backward decision from the upper end.

Forward step 1. Everyone is at risk. The conditional probabilities of stopping at 1 and continuing above 1 are $0.10/1.00=0.10$ and $0.90/1.00=0.90$. Therefore $\phi_1=\log(0.10/0.90)=\log(1/9)\approx-2.1972$.

Forward step 2. The risk set consists of categories 2, 3, and 4, with total probability $0.90$. The conditional probabilities are $0.20/0.90$ and $0.70/0.90$. Hence $\phi_2=\log(0.20/0.70)=\log(2/7)\approx-1.2528$.

Forward step 3. The risk set consists of categories 3 and 4, with total probability $0.70$. The conditional probabilities are $0.30/0.70$ and $0.40/0.70$. Thus $\phi_3=\log(0.30/0.40)=\log(0.75)\approx-0.2877$.

Backward step 4. Everyone is at risk. The conditional probabilities of stopping at 4 and continuing below 4 are $0.40$ and $0.60$. Thus $\psi_4=\log(0.40/0.60)=\log(2/3)\approx-0.4055$.

Backward step 3. Categories 1, 2, and 3 are at risk, with total probability $0.60$. The conditional probabilities are $0.30/0.60$ and $0.30/0.60$, so $\psi_3=\log(1)=0$.

Backward step 2. Categories 1 and 2 are at risk, with total probability $0.30$. The conditional probabilities are $0.20/0.30$ and $0.10/0.30$. Therefore $\psi_2=\log(0.20/0.10)=\log(2)\approx0.6931$.

These values are not two estimates of the same set of odds. Each direction forms a different sequence of conditional comparisons.

Worked Example: The Middle Category Has Two Different Comparisons

Suppose a three-category clinical outcome has probabilities $0.20$, $0.50$, and $0.30$. For category 2, the forward and backward logits have the same stopping category but different comparison groups.

In the forward sequence, the risk set is $Y\ge2$, consisting of categories 2 and 3. The probability of stopping at 2 conditional on this risk set is $0.50/(0.50+0.30)=5/8$; the probability of continuing to 3 is $3/8$. The forward odds are $5/3$, so $\phi_2=\log(5/3)\approx0.5108$.

In the backward sequence, the risk set is $Y\le2$, consisting of categories 1 and 2. The conditional probabilities of stopping at 2 and continuing below 2 are $0.50/(0.20+0.50)=5/7$ and $2/7$. The backward odds are $5/2$, so $\psi_2=\log(5/2)\approx0.9163$.

The forward question is, “Among outcomes in category 2 or higher, what are the odds of category 2 rather than category 3?” The backward question is, “Among outcomes in category 2 or lower, what are the odds of category 2 rather than category 1?” The different answers follow from the different conditioning events.

Reversing the Category Order

Reversing the labels of an ordered response changes which end is treated as the start of a forward sequence. This gives an exact relationship between backward logits for $Y$ and forward logits for the reversed response.

Result: Let $Z=K+1-Y$, so that $Z=1$ corresponds to $Y=K$, and the order of the categories is reversed. If $\widetilde{\phi}_\ell$ denotes the forward continuation-ratio logit for $Z$, then $$ \widetilde{\phi}_{K+1-j}=\psi_j, \qquad j=2,\ldots,K. $$

Proof. Fix $j\in\{2,\ldots,K\}$ and set $\ell=K+1-j$. The event $Z\ge\ell$ is equivalent to $Y\le j$: indeed, $K+1-Y\ge K+1-j$ exactly when $Y\le j$. Within this risk set, $Z=\ell$ is equivalent to $Y=j$, and $Z>\ell$ is equivalent to $Y<j$. Therefore

\widetilde{\phi}_{\ell} =\log\left\{\frac{P(Z=\ell\mid Z\ge\ell)} {P(Z>\ell\mid Z\ge\ell)}\right\} =\log\left\{\frac{P(Y=j\mid Y\le j)} {P(Y<j\mid Y\le j)}\right\} =\psi_j.

Thus, a backward sequence on the original ordering is a forward sequence on the reversed ordering, with the corresponding steps matched as above. This identity concerns the conditional questions and their logits; it does not say that an unchanged scientific interpretation survives relabeling. For instance, reversing “mild, moderate, severe” makes a forward sequence begin at “severe,” not “mild.”

Worked Example: Checking the Reversal with Four Categories

Let the categories be ordered $1,2,3,4$, with probabilities $(0.12,0.28,0.35,0.25)$. Consider the backward step at category $j=3$. Its risk set is $Y\le3$, with probability $0.12+0.28+0.35=0.75$. The probability of stopping at 3 is $0.35/0.75$, and the probability of continuing below 3 is $(0.12+0.28)/0.75=0.40/0.75$. The odds are $0.35/0.40=0.875$, so $\psi_3=\log(0.875)\approx-0.1335$.

Now reverse the order and define $Z=5-Y$. The probabilities for $Z=1,2,3,4$ are $(0.25,0.35,0.28,0.12)$. The matched forward step is $\ell=5-3=2$. Its risk set $Z\ge2$ has probability $0.35+0.28+0.12=0.75$. Stopping at $Z=2$ has probability $0.35/0.75$, while continuing to $Z>2$ has probability $(0.28+0.12)/0.75=0.40/0.75$. The odds are again $0.875$, and $\widetilde{\phi}_2=\psi_3$, as the result predicts.

Why the Direction Matters

A continuation-ratio model is meaningful when its successive risk sets reflect the question of interest. A forward sequence can represent progression from lower to higher categories, with each step comparing arrival at a category against proceeding beyond it. A backward sequence can represent assessment from the upper end, comparing a category against outcomes below it. The formula alone does not determine which account is scientifically appropriate; the direction and category order do.

A common mistake is to treat forward and backward logits at the same category as interchangeable because both use the phrase “stop at $j$.” They do not generally compare against the same alternative: the forward denominator sums probabilities above $j$, while the backward denominator sums probabilities below $j$. As the three-category calculation showed, the resulting odds can differ even though the stopping category is identical. Another mistake is to reverse category labels while continuing to describe the first forward step as if it still began at the original lowest category. After reversal, the first step begins at the original highest category.

Before interpreting a continuation-ratio coefficient, write down the risk set and the two alternatives at that step. This simple check identifies what is being conditioned on and prevents an order reversal from silently changing the scientific question.

Check Your Understanding

Answer each question by naming the risk set and the comparison being made.

For $K=5$, which categories are in the forward risk set at step $j=3$, and what are the two alternatives?
Write the backward logit at category $j=4$ using $p_1,\ldots,p_K$.
For a three-category response with probabilities $(0.25,0.25,0.50)$, calculate the forward logit at category 2 and identify its conditional comparison.
If $Z=K+1-Y$, which forward step for $Z$ corresponds to the backward step $j=2$ for $Y$?
Why can the forward and backward logits at the same interior category have different values?