Delta Adjustments for Ordinal Outcomes

Why Adjust ordinal probabilities?

An ordinal outcome has categories with a meaningful order, such as mild, moderate, and severe. When some outcomes are missing, a fitted imputation model supplies a reference distribution for those categories conditional on the available information. A sensitivity analysis can ask how the conclusions would change if the unobserved outcomes were systematically distributed differently from that reference.

A delta adjustment expresses that departure through the probabilities assigned to the categories. The reference distribution is usually the distribution under a missing-at-random (MAR) assumption, conditional on the variables in the imputation model. The delta values are sensitivity assumptions, not quantities identified from the observed outcomes alone. The earlier material on anchoring delta adjustments to a MAR reference gives the general interpretation of that starting point.

Definition: Let an ordinal outcome $Y$ have ordered categories $1,\ldots,K$. For a record with observed information $x$, let $q_k(x)$ be the reference probability of category $k$, where $q_k(x)\geq 0$ and $\sum_{k=1}^K q_k(x)=1$. A common-score delta adjustment assigns category probabilities $$ p_k(x;\delta)=\frac{q_k(x)\exp(\delta s_k)}{\sum_{j=1}^K q_j(x)\exp(\delta s_j)}, $$ where $s_k$ is a prespecified score increasing with category order and $\delta$ is a sensitivity parameter. A category-specific adjustment replaces $\delta s_k$ with a category-specific shift $\delta_k$: $$ p_k(x;\boldsymbol{\delta})=\frac{q_k(x)\exp(\delta_k)}{\sum_{j=1}^K q_j(x)\exp(\delta_j)}. $$

The denominator is essential: it makes the adjusted probabilities sum to one. In the common-score form, a positive $\delta$ tilts probability toward categories with larger scores; a negative $\delta$ tilts it toward lower scores. The score values determine how the tilt is distributed across the categories. If equally spaced scores $s_k=k-1$ are used, each step up the category scale receives the same multiplicative change in its unnormalized weight.

The category-specific form allows more tailored assumptions. For example, an analyst may posit that moderate outcomes are more common among missing cases while severe outcomes are less common. Because shifts can differ by category, this form does not necessarily preserve a simple upward or downward ordering of the distribution.

How the adjustment changes category odds

For the common-score adjustment, the odds of category $k$ relative to category $l$ can be expressed directly in terms of the reference probabilities and the score difference. The identity is defined when $q_l(x)>0$:

\frac{p_k(x;\delta)}{p_l(x;\delta)} = \frac{q_k(x)}{q_l(x)} \exp\{\delta(s_k-s_l)\}, \qquad q_l(x)>0.

When $q_k(x)=0$, the adjusted probability $p_k(x;\delta)$ also equals zero, and the ratio on the left is zero if $q_l(x)>0$. If $q_l(x)=0$, the displayed ratio is undefined, so the identity is not asserted for that pair. The same zero-preservation property applies to category-specific shifts: multiplying a zero reference probability by a finite positive exponential weight leaves it zero.

This means that the adjustment redistributes probability only among categories with positive reference probability. It cannot introduce support for a category that the reference model assigns probability zero. That is an important limitation when probabilities are numerically zero because of sparse data or model fitting, rather than because the category is genuinely impossible.

Worked Example: A Common Shift Toward Higher Categories

Suppose a hypothetical three-category outcome has reference probabilities $q=(0.20,0.50,0.30)$, with scores $s=(0,1,2)$. Take $\delta=\log(2)$. The category weights are $\exp(\delta s_k)=(1,2,4)$, so the unnormalized probabilities are:

(0.20\times 1,\;0.50\times 2,\;0.30\times 4)=(0.20,\;1.00,\;1.20).

Their sum is $0.20+1.00+1.20=2.40$. Dividing each by 2.40 gives:

p=(0.20/2.40,\;1.00/2.40,\;1.20/2.40) =(0.0833,\;0.4167,\;0.5000).

The probability of the highest category rises from 0.30 to 0.50, while the probability of the lowest falls from 0.20 to about 0.0833. The reference mean score is $0(0.20)+1(0.50)+2(0.30)=1.10$. The adjusted mean score is $0(0.0833)+1(0.4167)+2(0.5000)=1.4167$. Thus, under this hypothetical shift, the expected ordinal score rises by about 0.317 score units.

What a common shift guarantees

A common-score tilt has a useful mathematical property: as $\delta$ increases, the mean score of the adjusted distribution cannot decrease. To see why, write the normalizing denominator as $Z(\delta)=\sum_j q_j\exp(\delta s_j)$, and define the adjusted mean score $\mu_s(\delta)=\sum_k p_k(\delta)s_k$. Differentiating the normalized probability gives:

\frac{d p_k(\delta)}{d\delta} =p_k(\delta)\{s_k-\mu_s(\delta)\}.

Multiplying by $s_k$, summing across categories, and using $\sum_k p_k(\delta)=1$, gives a complete expression for the change in the mean score:

\begin{aligned} \frac{d\mu_s(\delta)}{d\delta} &=\sum_k s_k p_k(\delta)\{s_k-\mu_s(\delta)\}\\ &=\sum_k p_k(\delta)s_k^2-\left\{\sum_k p_k(\delta)s_k\right\}^2\\ &=\operatorname{Var}_{p(\delta)}(s)\geq 0. \end{aligned}

The derivative is strictly positive whenever the adjusted distribution assigns positive probability to at least two distinct scores. If only one score has positive probability, the distribution cannot move and the derivative is zero. This result concerns the mean of the specified scores; it does not say that every individual category probability changes in the same direction. Middle-category probabilities can rise or fall as probability shifts across the full distribution.

Worked Example: Category-Specific Shifts

Consider a hypothetical ordinal outcome with reference probabilities $q=(0.25,0.50,0.25)$ for low, middle, and high categories. Suppose the analyst uses category-specific shifts $\boldsymbol{\delta}=(0,\log 2,\log 0.5)$. The corresponding weights are $(1,2,0.5)$, giving unnormalized probabilities:

(0.25\times 1,\;0.50\times 2,\;0.25\times 0.5) =(0.25,\;1.00,\;0.125).

The normalizing sum is $0.25+1.00+0.125=1.375$. Therefore:

p=(0.25/1.375,\;1.00/1.375,\;0.125/1.375) =(0.1818,\;0.7273,\;0.0909).

The middle category increases from 0.50 to about 0.727, while the high category decreases from 0.25 to about 0.091. This is not a uniform upward shift; it encodes the specific assumption that middle outcomes become relatively more likely and high outcomes less likely. The shifts must be justified as a sensitivity scenario, not interpreted as estimates from the missing outcomes.

Apply the adjustment to missing outcomes

For each record with a missing ordinal outcome, the reference imputation model provides probabilities $q_k(x)$ using the information $x$ available for that record. A delta scenario transforms those probabilities into $p_k(x;\delta)$ or $p_k(x;\boldsymbol{\delta})$. The adjusted probabilities define the categorical distribution from which an imputed category is drawn. The same adjustment rule should be applied consistently within the scenario being examined.

The interpretation depends on the score and delta scale. With scores $0,1,\ldots,K-1$, a one-unit increase in $\delta$ multiplies the adjusted odds of category $k$ versus category $l$ by $\exp(s_k-s_l)$, when $q_l(x)>0$. For adjacent categories, where $s_k-s_l=1$, the multiplier is $\exp(1)$. With different scores, the same numerical delta has a different meaning. Scores and delta values therefore need to be reported together.

Worked Example: A Zero Reference Probability

Suppose a hypothetical three-category reference distribution is $q=(0.60,0,0.40)$, with scores $s=(0,1,2)$. Use the common shift $\delta=\log(1.5)$. The weights are $(1,1.5,2.25)$, so the unnormalized values are:

(0.60\times 1,\;0\times 1.5,\;0.40\times 2.25) =(0.60,\;0,\;0.90).

The normalizing sum is $0.60+0+0.90=1.50$, and the adjusted probabilities are:

p=(0.60/1.50,\;0/1.50,\;0.90/1.50)=(0.40,\;0,\;0.60).

The middle category remains impossible under this adjustment because its reference probability is zero. The odds of the high category relative to the low category change from $0.40/0.60=0.667$ to $0.60/0.40=1.50$, a factor of $1.50/0.667=2.25$. This matches $\exp\{\delta(2-0)\}=1.5^2=2.25$. No odds ratio involving the zero-probability middle category as its denominator is defined.

Choosing and checking sensitivity scenarios

A sound analysis states the reference model, category scores, adjustment form, and delta values. It then compares the resulting category distributions and the substantive analysis across a set of plausible scenarios. The range of delta values is a matter of subject-matter judgment or external information; a numerical range cannot be selected from the observed outcomes alone when the missing outcomes are unobserved.

Useful checks include confirming that every adjusted distribution sums to one, that all probabilities are nonnegative, and that the distribution moves in the direction implied by the shifts. For common-score shifts, check whether the mean score changes monotonically with $\delta$, as the derivative result requires. Examine adjusted probabilities by relevant covariate patterns: a scenario that seems moderate on average may produce near-certain categories for some records. If the reference model produces zero probabilities in categories that are plausible, the tilt cannot repair that lack of support.

A common pitfall is to multiply each category probability by its weight and use the products directly as probabilities. For example, the values $(0.20,1.00,1.20)$ in the first example do not form a probability distribution because they sum to 2.40. The correction is to divide every weighted probability by their common sum. Another error is to apply the pairwise odds identity when the denominator category has reference probability zero; that ratio is undefined, not an infinite or otherwise interpretable odds comparison.

Approach	What it changes	Key consideration
Common-score probability tilt	Tilts the full category distribution according to ordered scores	A single delta has an interpretable direction, but its meaning depends on the chosen scores
Category-specific probability tilt	Applies a separate log-weight to each category	Flexible, but needs more assumptions and may not imply a single ordinal direction
Probability-scale adjustment for a binary outcome	Changes the probability of one of two categories	There is no multi-category distribution to tilt; the binary adjustment does not directly specify ordinal category shifts
Cumulative-odds model shift	Changes cumulative odds across ordinal cut-points	Uses a different parameterization and assumptions; it is not the same operation as tilting category probabilities

Probability tilting is convenient because it guarantees a normalized distribution for finite shifts and preserves the reference model’s zero probabilities. Its limitations follow from the same construction: it cannot add categories to the support, its common-shift interpretation depends on scores, and category-specific shifts can be difficult to justify. Also, sensitivity analysis does not remove uncertainty about the reference model itself. If that model poorly represents the observed data, delta scenarios built on it may not provide a reliable description of missing outcomes.

Key takeaway: An ordinal delta adjustment multiplies each reference category probability by a positive exponential weight and then renormalizes. A common score shift tilts probability in an ordered direction, while category-specific shifts encode more tailored redistributions. Both operate only over categories with positive reference probability.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions.

For reference probabilities $(0.30,0.40,0.30)$, scores $(0,1,2)$, and $\delta=0$, what are the adjusted probabilities?
Why must the weighted values be divided by their sum in a probability tilt?
If $q_l=0$, can the common-shift pairwise odds identity be used for category $k$ versus category $l$? Explain.
With a common-score tilt and positive probability on at least two distinct scores, what happens to the adjusted mean score as $\delta$ increases?
Can a finite category-specific shift give positive adjusted probability to a category whose reference probability is zero?

Solutions. (1) Since $\exp(0s_k)=1$ for every category, the weighted values are unchanged and already sum to one: the adjusted probabilities are $(0.30,0.40,0.30)$. (2) Multiplication changes the total probability mass; dividing by the sum makes the probabilities nonnegative and ensures they sum to one. (3) No. The denominator probability in the ratio is zero, so both the odds ratio and the displayed identity are undefined for that pair. (4) The mean score increases strictly: its derivative with respect to $\delta$ is the variance of the scores under the adjusted distribution, which is positive when at least two distinct scores have positive probability. (5) No. A zero reference probability multiplied by a finite positive exponential weight remains zero, and normalization leaves it zero.