Log-Odds Delta Adjustments for Binary Outcomes

Why Shift Binary Outcomes on the Log-Odds Scale?

For a binary outcome, a missing value represents an unobserved event or non-event. A reference model supplies a probability for that outcome; a sensitivity analysis asks how conclusions would change if the missing outcomes had a different distribution. For binary outcomes, adding a delta to the log odds is one way to specify that departure. It lets the analyst express a simple, constant change on the log-odds scale while ensuring the adjusted probability remains between zero and one.

The adjustment is a sensitivity assumption, not a quantity identified by the observed outcomes alone. The reference risk and the delta play different roles: the reference risk describes the starting distribution, while the delta describes how far the missing-outcome distribution is shifted from it.

Definition: Let $Y$ be a binary outcome, with $Y=1$ denoting the event of interest. Let $p_0$ be the reference probability of that event for a missing outcome, and let $\delta$ be a specified additive shift on the log-odds scale. The log-odds delta adjustment is $$ \operatorname{logit}(p_\delta)=\operatorname{logit}(p_0)+\delta, \qquad \operatorname{logit}(p)=\log\left(\frac{p}{1-p}\right), $$ where $0<p_0<1$. The adjusted probability is $p_\delta$, the probability assigned to the event for the missing outcome under this sensitivity assumption.

A positive $\delta$ raises the event odds; a negative $\delta$ lowers them. The direction of the clinical interpretation depends on what $Y=1$ means: raising the odds of an adverse event is not the same as raising the odds of a beneficial event. The delta is dimensionless. Its exponent, $e^\delta$, is the multiplier applied to the reference odds, not an additive change in probability.

Converting Adjusted Odds to a Probability

The reference odds are $p_0/(1-p_0)$. Adding $\delta$ to the log odds multiplies those odds by $e^\delta$. Thus the adjusted odds are $e^\delta p_0/(1-p_0)$. Converting those odds back to a probability gives a formula that can be used directly.

\frac{p_\delta}{1-p_\delta} =e^\delta\frac{p_0}{1-p_0} \quad\Longrightarrow\quad p_\delta =\frac{p_0e^\delta}{1-p_0+p_0e^\delta}.

To justify the conversion, call the adjusted odds $o_\delta$. By definition, $o_\delta=p_\delta/(1-p_\delta)$, so $o_\delta(1-p_\delta)=p_\delta$. Rearranging gives $o_\delta=p_\delta(1+o_\delta)$, hence $p_\delta=o_\delta/(1+o_\delta)$. Substituting $o_\delta=e^\delta p_0/(1-p_0)$ and multiplying numerator and denominator by $1-p_0$ yields the displayed probability formula.

For $\delta=0$, the formula returns $p_0$. For any finite $\delta$, if $0<p_0<1$, it gives $0<p_\delta<1$. Also, the transformation is increasing in $\delta$: differentiating the logistic form gives $dp_\delta/d\delta=p_\delta(1-p_\delta)>0$. Thus a positive shift increases the risk and a negative shift decreases it, without allowing a valid probability to exceed one or fall below zero.

Worked Example: A Negative Shift in Event Odds

Suppose the reference event risk for each of 80 missing outcomes is $p_0=0.20$, and the sensitivity assumption is $\delta=-0.70$. Calculate the adjusted risk and the expected number of events among these missing outcomes.

The reference odds are $0.20/(1-0.20)=0.25$. The odds multiplier is $e^{-0.70}$, so the adjusted odds are $0.25e^{-0.70}$. Since $e^{-0.70}\approx0.4965853$, the arithmetic is

0.25e^{-0.70}\approx0.1241463, \qquad p_\delta=\frac{0.1241463}{1+0.1241463} \approx0.1104361.

The adjustment therefore assigns an event probability of about 11.04% to each of the 80 missing outcomes, rather than the reference probability of 20.00%. The expected event count is $80(0.1104361)\approx8.8349$ events. This is an expected count, not a claim that a fractional number of events will be observed.

For context, suppose 24 events were observed among 120 people with observed outcomes, and the total group contained 200 people. Combining observed events with the adjusted expected missing events gives $24+8.8349=32.8349$ expected events, or $32.8349/200\approx0.1642$, about 16.42%. The 24 observed events are not shifted; only the missing-outcome distribution is adjusted.

Applying the Adjustment Across Reference Risks

The same delta can produce different changes in probability when applied to different reference risks. It multiplies odds by the same factor $e^\delta$, but odds and probabilities are not interchangeable scales. When the reference probability varies across people or strata, apply the conversion to each specified reference risk and then aggregate the resulting probabilities or expected counts. In general, applying the formula once to an average reference risk will not give the average of the adjusted risks.

Worked Example: Two Strata With Different Reference Risks

A hypothetical group has 50 missing outcomes in stratum A, with reference risk $p_{0,A}=0.10$, and 30 in stratum B, with reference risk $p_{0,B}=0.40$. Use a common log-odds shift of $\delta=0.50$ to find the adjusted risks and expected total event count. The same outcome definition and shift are assumed in both strata.

Here $e^{0.50}\approx1.6487213$. In stratum A, the reference odds are $0.10/0.90=1/9$. Therefore,

p_{\delta,A} =\frac{0.10e^{0.50}}{0.90+0.10e^{0.50}} =\frac{1.6487213}{9+1.6487213} \approx0.1548281.

The expected number of events in stratum A is $50(0.1548281)\approx7.7414$. In stratum B, the reference odds are $0.40/0.60=2/3$, so

p_{\delta,B} =\frac{0.40e^{0.50}}{0.60+0.40e^{0.50}} =\frac{1.0991475}{1+1.0991475} \approx0.5236161.

The expected number of events in stratum B is $30(0.5236161)\approx15.7085$. The total expected number among the 80 missing outcomes is $7.7414+15.7085\approx23.4499$, and the weighted average adjusted risk is $23.4499/80\approx0.2931$, or 29.31%. Both risks increase because the common delta is positive, but their probability increases differ because their starting risks differ.

Choosing and Interpreting the Delta

The delta should be interpreted on the log-odds scale and, when useful, translated into its odds multiplier. For example, $\delta=0.50$ multiplies event odds by $e^{0.50}\approx1.649$, a 64.9% increase in odds. It does not mean a 50-percentage-point increase in risk, nor a 50% increase in risk. The probability change depends on $p_0$.

The direction also depends on the event coding. If $Y=1$ is death, a positive delta raises the assumed probability of death among missing outcomes. If $Y=1$ is recovery, the same positive delta raises the assumed probability of recovery. State the event and the relevant missing-outcome group whenever reporting a delta so its direction is unambiguous.

A delta may be chosen to represent a clinically interpretable departure from the reference distribution, or varied over a range as a sensitivity analysis. The observed data do not, by themselves, select the correct delta for outcomes that are missing. Consequently, results should be reported across substantively plausible values rather than presenting one convenient value as though it were empirically established.

Worked Example: Finding a Delta for a Target Risk

Suppose a reference risk is $p_0=0.30$, and a sensitivity scenario is intended to assign a missing outcome risk of $p_\delta=0.20$. Find the log-odds delta and interpret its odds multiplier. This calculation is a way to express the chosen target risk on the log-odds scale; it does not establish that the target is true.

The reference odds are $0.30/0.70=3/7$, and the target odds are $0.20/0.80=1/4$. Since adjusted odds equal reference odds multiplied by $e^\delta$,

e^\delta=\frac{1/4}{3/7}=\frac{7}{12}\approx0.5833, \qquad \delta=\log(7/12)\approx-0.5390.

Thus the target scenario corresponds to a log-odds shift of about $-0.5390$, multiplying the reference event odds by approximately 0.5833, a reduction of about 41.67% in odds. The target probability is 20.00%; it is not obtained by subtracting 0.5390 from the reference probability.

Assumptions, Diagnostics, and Limitations

A log-odds delta adjustment is well defined and interpretable under several conditions. First, the outcome is binary and the event coded as $Y=1$ is defined consistently. Second, the reference probability $p_0$ is appropriate for the missing outcomes being adjusted; it must be supplied by the analysis model or reference distribution. Third, the analyst specifies the delta and the group to which it applies before interpreting the result. In the examples above, one common delta applies to every missing outcome within the stated group or stratum. Finally, a finite logit requires $0<p_0<1$.

Check the calculation by verifying that $\delta=0$ reproduces the reference risk, that a positive shift increases the risk, and that a negative shift decreases it. Also verify the odds ratio $p_\delta/(1-p_\delta)\div[p_0/(1-p_0)]$ equals $e^\delta$. If expected counts are reported, confirm that each stratum's expected count is its size multiplied by its adjusted risk, and that totals use the correct denominators.

A reference probability of exactly zero or one is a boundary case: its logit is undefined, and multiplying its zero or infinite odds by any finite factor does not move it into the interior. The adjustment cannot repair a reference model that assigns boundary probabilities; the model or probability specification needs attention. Numerical calculations can also become unstable for very large absolute deltas, so software should use a stable logistic calculation rather than directly evaluating extreme exponentials.

This adjustment changes the assumed distribution of missing outcomes; it does not correct a misspecified reference model, capture uncertainty in the reference-risk estimates, or make the delta identifiable from observed outcomes. A single common delta may also be an inadequate description if substantively different missing-outcome groups require different assumptions. These are limitations to disclose, not reasons to treat the sensitivity parameter as an estimated treatment effect.

Approach	What it specifies	Key distinction
Reference distribution with $\delta=0$	Uses the specified reference risk without a log-odds departure.	It is the zero-shift case of this adjustment.
Log-odds delta	Adds $\delta$ to the reference log odds, multiplying odds by $e^\delta$.	Adjusted probabilities remain within zero and one for interior reference risks.
Risk-difference delta	Specifies a departure directly in probability units.	It uses a different scale, so the same numerical delta has a different interpretation.

A common mistake is to add a log-odds delta directly to the reference probability. For instance, with $p_0=0.20$ and $\delta=-0.70$, subtracting 0.70 from 0.20 would give an invalid negative probability. The correction is to add the delta to $\operatorname{logit}(p_0)$, or equivalently multiply the reference odds by $e^\delta$, and then convert the adjusted odds back to a probability. Another error is to call $e^\delta$ a risk ratio: it is an odds multiplier, and generally differs from a risk ratio.

Key takeaway: A log-odds delta adjustment assigns $p_\delta=p_0e^\delta/(1-p_0+p_0e^\delta)$ to a binary missing outcome with reference risk $p_0$. It changes odds multiplicatively, changes probabilities nonlinearly, and represents a specified sensitivity assumption rather than information learned from the missing outcomes.

Check Your Understanding

Answer each question, then compare with the solutions below.

If $p_0=0.25$ and $\delta=\log(2)$, what are the adjusted odds and adjusted risk?
If $p_0=0.60$ and $\delta=-\log(3)$, calculate $p_\delta$.
What delta changes a reference risk of 0.40 to a target risk of 0.25?
Explain why a positive delta does not mean a fixed increase in probability.

Solutions. (1) The reference odds are $0.25/0.75=1/3$; multiplying by 2 gives adjusted odds $2/3$, so $p_\delta=(2/3)/(1+2/3)=0.40$. (2) The reference odds are $0.60/0.40=1.5$; $e^{-\log(3)}=1/3$, so adjusted odds are 0.50 and $p_\delta=0.50/1.50=1/3$. (3) The reference odds are $0.40/0.60=2/3$, and target odds are $0.25/0.75=1/3$; thus $e^\delta=(1/3)/(2/3)=1/2$, giving $\delta=\log(1/2)\approx-0.6931$. (4) A positive delta multiplies odds by $e^\delta>1$; the resulting probability change depends on the starting risk because the log-odds-to-probability conversion is nonlinear.

Approach	What it specifies	Key distinction
Reference distribution with \(\delta=0\)	Uses the specified reference risk without a log-odds departure.	It is the zero-shift case of this adjustment.
Log-odds delta	Adds \(\delta\) to the reference log odds, multiplying odds by \(e^\delta\).	Adjusted probabilities remain within zero and one for interior reference risks.
Risk-difference delta	Specifies a departure directly in probability units.	It uses a different scale, so the same numerical delta has a different interpretation.