Continuation-Ratio Example: Educational Progression

Educational Attainment as a Sequence of Gates

Educational attainment can be represented by ordered categories, but a forward continuation-ratio model asks a more specific question than which category is most common: among people who have reached a given level, what is the probability of progressing to the next level? Those conditional probabilities describe successive gates. Their products give the probabilities of the final attainment categories.

Consider four categories for highest attainment: \(Y=0\), did not complete secondary school; \(Y=1\), completed secondary school but not a bachelor's degree; \(Y=2\), earned a bachelor's degree but not a graduate degree; and \(Y=3\), earned a graduate degree. For a covariate value \(x\), the first gate separates category 0 from categories 1, 2, and 3. The second concerns progression to at least category 2 among those at category 1 or above. The third concerns progression to category 3 among those at category 2 or above.

Definition: For \(r=1,2,3\), the forward continuation probability at gate \(r\) is \(q_r(x)=P(Y\ge r\mid Y\ge r-1, X=x)\), provided \(P(Y\ge r-1\mid X=x)>0\). It is the probability of reaching category \(r\) or higher, conditional on having reached category \(r-1\) or higher. A common regression specification is \(\operatorname{logit}\{q_r(x)\}=\alpha_r+\beta_r x\), where \(\operatorname{logit}(q)=\log\{q/(1-q)\}\).

The conditional event in the definition matters. For example, \(q_2(x)\) is not the probability of having a bachelor's degree or higher in the whole population at \(x\). It is that probability among those who completed secondary school. The overall probability of a bachelor's degree or higher will also depend on the chance of passing the first gate.

From Gates to Attainment Probabilities

To have category 0, a person must not pass gate 1. To have category 1, the person must pass gate 1 and then not pass gate 2. Category 2 requires passing the first two gates and not passing the third. Category 3 requires passing all three.

Theorem: For four ordered categories \(0,1,2,3\), the forward continuation probabilities determine the category probabilities as \(P(Y=0\mid X=x)=1-q_1(x)\), \(P(Y=1\mid X=x)=q_1(x)\{1-q_2(x)\}\), \(P(Y=2\mid X=x)=q_1(x)q_2(x)\{1-q_3(x)\}\), and \(P(Y=3\mid X=x)=q_1(x)q_2(x)q_3(x)\).

Proof. By definition, \(q_1(x)=P(Y\ge1\mid X=x)\), so \(P(Y=0\mid X=x)=1-q_1(x)\). For category 1, conditional on \(Y\ge1\), the probability of stopping below category 2 is \(1-q_2(x)\). The multiplication rule therefore gives \(P(Y=1\mid X=x)=P(Y\ge1\mid X=x)P(Y=1\mid Y\ge1,X=x)=q_1(x)\{1-q_2(x)\}\). Similarly, category 2 requires reaching category 2 and then failing to reach category 3, giving \(q_1(x)q_2(x)\{1-q_3(x)\}\). Reaching category 3 requires passing every gate, giving \(q_1(x)q_2(x)q_3(x)\).

These probabilities sum to one. Indeed, grouping the first two terms and the last two terms gives \[ \{1-q_1\}+q_1\{1-q_2\}+q_1q_2\{1-q_3\}+q_1q_2q_3 =1-q_1+q_1(1-q_2)+q_1q_2=1. \] Thus the construction gives a valid probability distribution. \(\square\)

Worked Example: Estimating Gates From One Attainment Table

In a sample of 100 people, the counts in categories \(0,1,2,3\) are \(40,30,20,10\), respectively. Estimate the three gate probabilities and recover the four category probabilities.

At gate 1, all 100 people are in the risk set, and 60 reached category 1 or higher. Thus \(\hat q_1=60/100=0.6\). At gate 2, the risk set consists of the 60 people in categories 1, 2, or 3; 30 of them reached category 2 or higher. Therefore \(\hat q_2=30/60=0.5\). At gate 3, the risk set has 30 people in categories 2 or 3, of whom 10 reached category 3, so \(\hat q_3=10/30=1/3\).

Substitution gives \(\hat P(Y=0)=1-0.6=0.4\), \(\hat P(Y=1)=0.6(1-0.5)=0.3\), \(\hat P(Y=2)=0.6(0.5)(1-1/3)=0.2\), and \(\hat P(Y=3)=0.6(0.5)(1/3)=0.1\). These reproduce the observed proportions \(40/100,30/100,20/100,10/100\).

Estimating Gate Probabilities With Covariates

A regression model lets the conditional progression probabilities vary with covariates. With one covariate \(x\), a separate logistic regression at each gate is \[ \operatorname{logit}\{q_r(x)\}=\alpha_r+\beta_r x,\qquad r=1,2,3. \] The coefficients are gate-specific: a covariate may have a different association with each conditional progression probability. At gate \(r\), the observations at risk are those with \(Y\ge r-1\); within that risk set, the successes are those with \(Y\ge r\).

For grouped counts \(n_0,n_1,n_2,n_3\) at a fixed covariate value, the gate 1 risk set has size \(n_0+n_1+n_2+n_3\), with \(n_1+n_2+n_3\) successes. Gate 2 has risk-set size \(n_1+n_2+n_3\), with \(n_2+n_3\) successes. Gate 3 has risk-set size \(n_2+n_3\), with \(n_3\) successes. This describes the binomial contributions used to estimate the gate probabilities.

Worked Example: Comparing Progression by a Binary Covariate

Suppose \(x=0\) and \(x=1\) identify two groups, each with 100 people. Their category counts, ordered \(0,1,2,3\), are \((50,30,15,5)\) for \(x=0\) and \((30,30,25,15)\) for \(x=1\). Estimate the gate probabilities in each group, then calculate the coefficients of separate logistic regressions with \(x\) coded 0 or 1.

For \(x=0\), the gate estimates are \(\hat q_1=50/100=0.5\), \(\hat q_2=(15+5)/(30+15+5)=20/50=0.4\), and \(\hat q_3=5/(15+5)=5/20=0.25\). For \(x=1\), they are \(\hat q_1=70/100=0.7\), \(\hat q_2=(25+15)/(30+25+15)=40/70=4/7\), and \(\hat q_3=15/(25+15)=15/40=3/8\).

With a binary predictor, the fitted intercept is the logit at \(x=0\), and the slope is the difference between the logits at \(x=1\) and \(x=0\). At gate 1, \(\hat\alpha_1=\operatorname{logit}(0.5)=0\) and \(\hat\beta_1=\operatorname{logit}(0.7)-0=\log(7/3)\). At gate 2, \(\hat\alpha_2=\operatorname{logit}(0.4)=\log(2/3)\), while \(\hat\beta_2=\log(4/3)-\log(2/3)=\log 2\). At gate 3, \(\hat\alpha_3=\operatorname{logit}(0.25)=\log(1/3)\) and \(\hat\beta_3=\log(3/5)-\log(1/3)=\log(9/5)\).

For example, at \(x=1\), the fitted category probabilities are \(0.3\), \(0.7(3/7)=0.3\), \(0.7(4/7)(5/8)=0.25\), and \(0.7(4/7)(3/8)=0.15\). They match the group’s observed proportions \(30/100,30/100,25/100,15/100\).

Why the Sequential Likelihood Works

The connection to binomial regression can be made precise. Suppose the observations are independent, the four categories are mutually exclusive and exhaustive, and each observation has a fixed covariate value \(x\). Write \(n_j\) for the count in category \(j\), and let \(N=n_0+n_1+n_2+n_3\). Under the model, the grouped category counts have a multinomial likelihood.

Result: For one covariate group, the multinomial likelihood equals the product of the three sequential binomial likelihoods, including their combinatorial coefficients: \[ \frac{N!}{n_0!n_1!n_2!n_3!}\prod_{j=0}^3 P(Y=j\mid X=x)^{n_j} = {N\choose n_1+n_2+n_3} {n_1+n_2+n_3\choose n_2+n_3} {n_2+n_3\choose n_3} \prod_{r=1}^3q_r(x)^{S_r}\{1-q_r(x)\}^{R_r-S_r}, \] where \((R_1,S_1)=(N,n_1+n_2+n_3)\), \((R_2,S_2)=(n_1+n_2+n_3,n_2+n_3)\), and \((R_3,S_3)=(n_2+n_3,n_3)\).

Derivation. Substitute the four category probabilities into the multinomial probability product. The exponent of \(q_1\) is \(n_1+n_2+n_3\), and the exponent of \(1-q_1\) is \(n_0\). The exponents of \(q_2\) and \(1-q_2\) are \(n_2+n_3\) and \(n_1\); those of \(q_3\) and \(1-q_3\) are \(n_3\) and \(n_2\). These are exactly the success and failure counts in the three risk sets. Finally, multiplying the three binomial coefficients gives \(N!/(n_0!n_1!n_2!n_3!)\), by cancellation of the intermediate factorials. This proves the stated equality. \(\square\)

With covariate patterns, the same factorization applies separately within each pattern. Consequently, fitting logistic regressions to the gate contributions gives the corresponding likelihood for this sequential parameterization. If the regression parameters are unrestricted and gate-specific, each gate’s contribution can be estimated separately; if predictors or parameters are constrained across gates, the joint likelihood must respect those constraints.

Worked Example: Converting Fitted Logits to Predictions

Suppose a fitted model at \(x=5\) gives gate logits \(-0.5\), \(0.45\), and \(-1.6\). Calculate the conditional gate probabilities and the unconditional attainment probabilities.

Applying the inverse-logit function, \(\hat q_1=1/(1+e^{0.5})\approx0.3775\), \(\hat q_2=1/(1+e^{-0.45})\approx0.6106\), and \(\hat q_3=1/(1+e^{1.6})\approx0.1680\). Therefore, \(\hat P(Y=0\mid x=5)=1-0.3775\approx0.6225\).

For category 1, \(\hat P(Y=1\mid x=5)=0.3775(1-0.6106)\approx0.1470\). For category 2, \(\hat P(Y=2\mid x=5)=0.3775(0.6106)(1-0.1680)\approx0.1918\). For category 3, \(\hat P(Y=3\mid x=5)=0.3775(0.6106)(0.1680)\approx0.0387\). The rounded probabilities sum to \(0.6225+0.1470+0.1918+0.0387=1.0000\), as required.

Interpretation and a Common Pitfall

A gate coefficient describes a conditional association. For example, \(\beta_2\) is the change in log-odds of reaching category 2 or higher rather than stopping at category 1, among people in the gate-2 risk set, per one-unit increase in \(x\). It is not directly the change in the probability of having a bachelor's degree or higher in the full population. That unconditional probability is \(q_1(x)q_2(x)\), so both the first and second gates matter.

Confusing a gate probability with a category probability. \(q_2(x)\) is conditional on reaching category 1 or higher; it is not \(P(Y=2\mid X=x)\). For four categories, the latter is \(q_1(x)q_2(x)\{1-q_3(x)\}\).
Forgetting the stopping factor. Category 2 in the four-category model requires passing gates 1 and 2, then failing gate 3. Its probability includes all three factors.
Misreading a coefficient as a causal effect. A fitted logistic coefficient describes the modeled conditional association. A causal interpretation requires additional design and assumptions.

The category formula depends on how many gates the model contains. With categories \(0,1,2\), category 2 is the final category and has probability \(q_1(x)q_2(x)\). With categories \(0,1,2,3\), category 2 is not final and has probability \(q_1(x)q_2(x)\{1-q_3(x)\}\). Always identify whether the category is terminal before deciding whether it needs a stopping factor.

Key takeaway: Estimate each forward continuation probability within its own risk set. To obtain an unconditional category probability, multiply the probabilities of passing every earlier gate and, for a nonterminal category, the probability of stopping at that gate.

Check Your Understanding

Use the four-category convention \(0,1,2,3\) unless a question states otherwise.

In one covariate group, the counts are \((20,30,30,20)\). What are the three empirical continuation probabilities?
If \(q_1=0.8\), \(q_2=0.5\), and \(q_3=0.4\), calculate all four category probabilities and verify that they sum to one.
Explain why the gate-3 risk set contains only observations in categories 2 and 3.
For a binary covariate, what are the intercept and slope of a gate-specific logistic regression in terms of the two group logits?
In a three-category model with categories \(0,1,2\), write the probability of the final category 2 and explain why it has no stopping factor.