Recovering Probabilities from Adjacent Odds

Why Adjacent Odds Determine Probabilities

For an ordered response with several categories, adjacent-category odds describe local comparisons: the odds of category 2 relative to category 1, category 3 relative to category 2, and so on. These comparisons do not look like probabilities, because they need not lie between zero and one and do not add to one. Yet together they contain enough information to recover the entire probability distribution. The key is to express every category relative to the first, then normalize the resulting weights.

We use the convention that each adjacent ratio compares the higher-indexed category with the lower-indexed one. This direction matters: reversing it replaces each ratio by its reciprocal. The reconstruction itself is simple, but keeping the direction, the cumulative products, and the final normalization straight is essential.

Definition: Let $Y$ take one of the ordered categories $1,\ldots,J$, with category probabilities $p_j=P(Y=j)$. For $j=1,\ldots,J-1$, the adjacent probability ratio is $r_j=p_{j+1}/p_j$. When $p_j>0$ for every category, $r_j$ is positive and finite, and the adjacent-category logit is $\eta_j=\log(r_j)=\log(p_{j+1}/p_j)$.

The Reconstruction Formula

Suppose the $J-1$ adjacent ratios $r_1,\ldots,r_{J-1}$ are known and positive. Set the first category's relative weight to one. Each subsequent weight is obtained by multiplying by the ratios along the chain:

w_1=1,\qquad w_k=\prod_{m=1}^{k-1}r_m\quad (k=2,\ldots,J).

The weights encode relative size, not probability: they usually do not sum to one. Divide each by their total $D$ to obtain the probabilities:

D=\sum_{k=1}^{J}w_k =1+r_1+r_1r_2+\cdots+\prod_{m=1}^{J-1}r_m, \qquad p_k=\frac{w_k}{D}.

Because every ratio is positive, every weight and $D$ are positive. Thus the formula produces positive probabilities, and their sum is one. The next result establishes that these are not merely probabilities consistent with the ratios: they are the only such probabilities.

Theorem: For any positive finite ratios $r_1,\ldots,r_{J-1}$, the formula $p_k=w_k/\sum_{\ell=1}^{J}w_\ell$, with $w_1=1$ and $w_k=\prod_{m=1}^{k-1}r_m$, gives the unique strictly positive probability distribution satisfying $p_{j+1}/p_j=r_j$ for all $j=1,\ldots,J-1$.

Proof. The weights satisfy $w_{j+1}=w_jr_j$. Normalization does not change a ratio, so

\frac{p_{j+1}}{p_j} =\frac{w_{j+1}/D}{w_j/D} =\frac{w_{j+1}}{w_j} =r_j.

Also, each $p_k$ is positive and $\sum_{k=1}^{J}p_k=D/D=1$, so the formula gives a valid distribution with the required ratios.

To prove uniqueness, suppose $q_1,\ldots,q_J$ is any other strictly positive distribution with $q_{j+1}/q_j=r_j$. Repeatedly applying these equations gives $q_k=q_1w_k$ for every $k$: this holds at $k=1$, and if it holds at $k$, then $q_{k+1}=r_kq_k=r_kq_1w_k=q_1w_{k+1}$. Since the $q_k$ sum to one, $1=q_1\sum_{k=1}^{J}w_k=q_1D$, so $q_1=1/D$ and $q_k=w_k/D=p_k$. Therefore the distribution is unique. $\square$

Worked Example: Reconstructing Four Probabilities From Ratios

Suppose an ordered response has four categories and the adjacent ratios are $r_1=2$, $r_2=1/2$, and $r_3=3$. Recover the category probabilities and check that they reproduce the supplied ratios.

Step 1: Build relative weights. Starting with $w_1=1$, multiply successively by each ratio:

w_1=1,\qquad w_2=1(2)=2,\qquad w_3=2(1/2)=1,\qquad w_4=1(3)=3.

Step 2: Normalize. The weight total is $D=1+2+1+3=7$. Therefore

(p_1,p_2,p_3,p_4) =\left(\frac17,\frac27,\frac17,\frac37\right).

Step 3: Check. The reconstructed adjacent ratios are $p_2/p_1=2$, $p_3/p_2=1/2$, and $p_4/p_3=3$, as required. The probabilities are positive and sum to $(1+2+1+3)/7=1$.

Working With Adjacent Logits

It is often more convenient to be given adjacent logits $\eta_j$ than ratios $r_j$. Since $r_j=\exp(\eta_j)$, products of ratios become sums of logits on the log scale. Define cumulative log weights $s_1=0$ and $s_k=\sum_{m=1}^{k-1}\eta_m$ for $k\geq 2$. Then $w_k=\exp(s_k)$, and the probabilities can be written as a normalized exponential:

p_k=\frac{\exp(s_k)}{\sum_{\ell=1}^{J}\exp(s_\ell)}, \qquad s_k=\sum_{m=1}^{k-1}\eta_m,\quad s_1=0.

This is a softmax form, but its inputs are not a freely chosen set of category scores: they are cumulative sums of the adjacent logits. In particular, the first score is fixed at zero as a reference, and each successive score adds one adjacent logit.

Worked Example: Recovering Probabilities From Three Adjacent Logits

For four categories, suppose $(\eta_1,\eta_2,\eta_3)=(\log 2,\log 3,-\log 2)$. Find the probabilities.

The corresponding ratios are $r_1=2$, $r_2=3$, and $r_3=1/2$, because exponentiating each logit gives its ratio. Equivalently, the cumulative log weights are $s_1=0$, $s_2=\log 2$, $s_3=\log 2+\log 3=\log 6$, and $s_4=\log 6-\log 2=\log 3$. Exponentiating gives weights $1,2,6,3$. Their total is $12$, so

(p_1,p_2,p_3,p_4) =\left(\frac1{12},\frac2{12},\frac6{12},\frac3{12}\right) =\left(\frac1{12},\frac16,\frac12,\frac14\right).

The checks are $p_2/p_1=2$, $p_3/p_2=3$, and $p_4/p_3=1/2$. The negative third logit correctly indicates that category 4 has lower probability than category 3; it does not mean that the category 4 probability is negative.

What the Local Ratios Imply Beyond Adjacent Categories

Once the adjacent ratios are known, every odds comparison between two nonadjacent categories follows by multiplication. For $a<b$, cancellation of intermediate probabilities gives

\frac{p_b}{p_a} =\frac{p_{a+1}}{p_a}\frac{p_{a+2}}{p_{a+1}}\cdots \frac{p_b}{p_{b-1}} =\prod_{m=a}^{b-1}r_m.

Thus, the odds of category $b$ relative to category $a$ are not an additional free quantity once the adjacent ratios have been specified. If someone supplies an extra nonadjacent odds claim, it must agree with this product for all claims to describe the same distribution.

Worked Example: Five Categories and a Nonadjacent Odds Check

Suppose the four adjacent ratios are $(1/2,3,1,2)$. Recover the probabilities, then find the odds of category 5 relative to category 1.

The successive weights are

w_1=1,\quad w_2=\frac12,\quad w_3=\frac12(3)=\frac32,\quad w_4=\frac32(1)=\frac32,\quad w_5=\frac32(2)=3.

Their sum is $1+1/2+3/2+3/2+3=15/2$. Dividing each weight by $15/2$ yields

(p_1,p_2,p_3,p_4,p_5) =\left(\frac2{15},\frac1{15},\frac15,\frac15,\frac25\right).

The endpoint odds are $p_5/p_1=(2/5)/(2/15)=3$. Multiplying the adjacent ratios gives the same result: $(1/2)(3)(1)(2)=3$. This check uses a comparison across four category steps, not an additional normalization.

How One Adjacent Logit Redistributes Probability

The reconstruction also shows precisely how a change in one adjacent logit affects the distribution. Increasing $\eta_m$ increases the odds of category $m+1$ relative to category $m$. In the cumulative log weights, it adds the same amount to the scores of categories $m+1,\ldots,J$, while leaving the scores of categories $1,\ldots,m$ unchanged. Normalization then redistributes probability between these two groups.

Result: With $p_k=\exp(s_k)/\sum_{\ell=1}^{J}\exp(s_\ell)$ and $s_k=\sum_{q=1}^{k-1}\eta_q$, define $T_m=\sum_{\ell=m+1}^{J}p_\ell$. For $m=1,\ldots,J-1$, $$ \frac{\partial p_k}{\partial\eta_m} =p_k\left(\mathbf{1}\{k>m\}-T_m\right), $$ where $\mathbf{1}\{k>m\}$ is one if $k>m$ and zero otherwise.

Derivation. Let $Z=\sum_{\ell=1}^{J}\exp(s_\ell)$, so $p_k=\exp(s_k)/Z$. The cumulative definition gives $\partial s_k/\partial\eta_m=\mathbf{1}\{k>m\}$. Differentiating the log of the normalizing constant gives

\frac{\partial\log Z}{\partial\eta_m} =\frac{1}{Z}\sum_{\ell=1}^{J}\exp(s_\ell)\mathbf{1}\{\ell>m\} =\sum_{\ell=m+1}^{J}p_\ell =T_m.

Since $\log p_k=s_k-\log Z$, it follows that $\partial\log p_k/\partial\eta_m=\mathbf{1}\{k>m\}-T_m$. Multiplying by $p_k$ yields the stated derivative. For $k\leq m$, the derivative is $-p_kT_m<0$; for $k>m$, it is $p_k(1-T_m)>0$. Thus a small increase in an adjacent logit shifts probability toward categories above that boundary and away from categories at or below it. $\square$

For instance, in the first worked example, $m=2$ separates categories 1 and 2 from categories 3 and 4. There $T_2=p_3+p_4=4/7$. The derivatives with respect to $\eta_2$ are $-4/49$ for $p_1$, $-8/49$ for $p_2$, $3/49$ for $p_3$, and $9/49$ for $p_4$. The changes sum to zero, as they must because the probabilities continue to sum to one.

Common Mistakes and Practical Checks

The most common error is to treat an odds ratio as a probability. A ratio such as $r_1=2$ says $p_2=2p_1$; it does not say $p_2=2$, and it does not by itself determine either probability. The ratio provides relative size. The total probability constraint supplies the common scale.

A second error is to forget that the ratios must be chained. For example, category 4's weight relative to category 1 is $r_1r_2r_3$, not just $r_3$. Starting each category's calculation from the same reference makes this easy to check: write the cumulative product for each weight, add the weights, and divide all of them by that one total.

Finite adjacent logits imply positive finite ratios, and therefore a reconstructed distribution with every category probability strictly positive. A distribution with a zero category probability is a boundary case and cannot be represented by finite adjacent logits across that zero. In ordinary calculations with positive ratios, a reliable final check is to verify both that the probabilities sum to one and that every reconstructed adjacent ratio matches its input.

Check Your Understanding

Use the reconstruction formula and show enough work to make your checks clear.

For three categories with adjacent ratios $r_1=4$ and $r_2=1/2$, find all three probabilities.
For four categories with adjacent logits $(\log 2,0,\log 3)$, find the relative weights and normalized probabilities.
If $p_4/p_2$ is required from adjacent ratios $r_2$ and $r_3$, write it as a product and explain why no other ratios enter.
Explain why a finite adjacent logit cannot connect two categories when one of their probabilities is zero.
For a fixed boundary $m$, describe which categories gain and which lose probability when $\eta_m$ increases slightly.