Deriving Probabilities from Continuation Ratios

From Sequential Decisions to Category Probabilities

A forward continuation-ratio model describes an ordered response as a sequence of decisions. At category \(j\), a subject who has reached that stage either stops at \(j\) or continues to a later category. The conditional probability of stopping at each stage is not itself the probability of the corresponding category in the full population: it applies only to subjects who reached that stage. To obtain an unconditional category probability, we must account for the chance of reaching the stage first.

This conversion is useful for interpreting a fitted continuation-ratio model on the scale of the response categories. It also provides a direct check that the probabilities implied by a set of stage-specific conditional probabilities form a coherent distribution.

The stages are nested: reaching stage \(j+1\) requires reaching stage \(j\) and then continuing. Thus the probability of category \(j\) consists of the probabilities of continuing through stages \(1,\ldots,j-1\), followed by the probability of stopping at \(j\). For the final category, the subject must continue through every modeled stage.

The Probability Conversion

Proof. Define \(A_j=\{Y\geq j\}\). Then \(A_1\) is the entire outcome space, and \(A_{j+1}\) is a subset of \(A_j\). For each nonterminal stage \(h\), $$ P(A_{h+1}\mid A_h)=P(Y>h\mid Y\geq h)=1-q_h. $$ Applying the multiplication rule for conditional probabilities to the nested events gives, for \(j\geq 2\), $$ P(A_j)=\prod_{h=1}^{j-1}P(A_{h+1}\mid A_h) =\prod_{h=1}^{j-1}(1-q_h). $$ For \(j=1\), the same expression is the empty product and equals \(P(A_1)=1\). Since \(\{Y=j\}\) is contained in \(A_j\), the definition of \(q_j\) yields $$ p_j=P(Y=j)=P(A_j)P(Y=j\mid A_j) =\left\{\prod_{h=1}^{j-1}(1-q_h)\right\}q_j $$ for \(j<K\). The event \(Y=K\) is exactly \(A_K\), so the nested-event calculation gives $$ p_K=P(A_K)=\prod_{h=1}^{K-1}(1-q_h). $$ This proves both formulas. \(\square\)

The factors in the formula have distinct roles. The product over earlier stages is the probability of arriving at category \(j\); \(q_j\) is the probability of stopping there once it is reached. These are conditional stages of one process, not independent category events.

Why the Conversion Always Normalizes

The product formulas do more than assign probabilities to each category: they automatically make the probabilities add to one. The reason is that each stage divides the probability of reaching it into two parts—stopping there and continuing beyond it. The following derivation makes that accounting explicit.

Derivation. Set \(S_1=1\) and \(S_j=\prod_{h=1}^{j-1}(1-q_h)\) for \(j=2,\ldots,K\). Here \(S_j\) is the probability of reaching stage \(j\). For each \(j<K\), \(p_j=S_jq_j\), while \(S_{j+1}=S_j(1-q_j)\). Therefore $$ p_j=S_jq_j=S_j-S_j(1-q_j)=S_j-S_{j+1}. $$ Summing the first \(K-1\) category probabilities and adding the terminal probability gives $$ \sum_{j=1}^{K}p_j =\sum_{j=1}^{K-1}(S_j-S_{j+1})+S_K =S_1 =1. $$ The sum telescopes: every intermediate \(S_j\) appears once with a positive sign and once with a negative sign. Also, each \(S_j\), each \(q_j\), and each \(1-q_j\) is nonnegative, so every \(p_j\) is nonnegative. Hence the formulas define a probability distribution. \(\square\)

For probabilities strictly between zero and one, the conversion is reversible. This matters because a category distribution and its forward continuation probabilities are two ways to describe the same information, provided every stage has a positive chance of being reached.

Proof. Begin with \(q_1,\ldots,q_{K-1}\in(0,1)\), and define \(S_j\) as above. The product formulas give \(p_j=S_jq_j>0\) for \(j<K\), and \(p_K=S_K>0\). The normalization derivation shows that these probabilities sum to one. Moreover, for each \(j<K\), the telescoping identity \(p_r=S_r-S_{r+1}\) for \(r<K\), together with \(p_K=S_K\), gives $$ \sum_{r=j}^{K}p_r=S_j. $$ Consequently \(p_j/\sum_{r=j}^Kp_r=(S_jq_j)/S_j=q_j\), so applying the proposed inverse recovers every original continuation probability.

Conversely, start with \(p_1,\ldots,p_K>0\) summing to one, and define \(T_j=\sum_{r=j}^Kp_r\). Then \(T_j>0\), and \(q_j=p_j/T_j\) lies strictly between zero and one because \(p_j>0\) and \(T_j-p_j=\sum_{r=j+1}^Kp_r>0\). Also \(1-q_j=T_{j+1}/T_j\). Since \(T_1=1\), multiplying these ratios gives $$ \prod_{h=1}^{j-1}(1-q_h) =\prod_{h=1}^{j-1}\frac{T_{h+1}}{T_h} =T_j. $$ The product formula then returns \(T_jq_j=T_j(p_j/T_j)=p_j\) for \(j<K\); the terminal product is \(T_K=p_K\). Thus the two transformations are inverses. \(\square\)

Zero Probabilities and a Common Mistake

The inverse formula divides by the probability of reaching a stage. If \(\sum_{r=j}^Kp_r=0\), no observation reaches stage \(j\), so \(P(Y=j\mid Y\geq j)\) is undefined: it conditions on an event of probability zero. A boundary distribution can therefore have well-defined continuation probabilities at its reached stages and undefined ones after the process has become impossible to continue. The product expressions still define category probabilities for values \(q_j\in[0,1]\), but when a stage has zero reach probability, later \(q\)'s cannot be recovered uniquely from the category distribution.

Common mistake: “If \(q_1=0.20\) and \(q_2=0.30\), then category 2 has probability \(0.30\).” This treats the conditional stopping probability at stage 2 as if it applied to the whole population. The correction is to multiply by the probability of reaching stage 2: \(p_2=(1-q_1)q_2=(0.80)(0.30)=0.24\). In general, the chance of stopping at a stage must be weighted by the chance of arriving there.

This distinction is important when reporting model results. Conditional stage probabilities describe the sequential decisions, while the \(p_j\)'s describe the population distribution across the outcome categories. Neither scale is a substitute for the other; the appropriate interpretation depends on whether the question concerns a decision among those at risk or the overall probability of an outcome.