Multiplicative Delta Adjustments for Log-Scale Outcomes

From a Log-Scale Shift to an Original-Scale Change

A delta adjustment is specified on a chosen scale, so its practical interpretation depends on how that scale relates to the outcome being analyzed. For a strictly positive outcome, an additive change on the log scale becomes a multiplicative change on the original scale. This gives the delta a direct interpretation as a ratio, but that interpretation depends on using a consistent logarithm and on applying the shift to positive values.

Here, $Y^{\mathrm{ref}}$ denotes a reference conditional outcome distribution for a missing outcome, given covariates $X=x$ and $R=0$, where $R=0$ indicates that the outcome is missing. The adjusted outcome is denoted $Y^\delta$. The delta $\delta$ is a fixed, dimensionless shift on the natural-log scale. A positive delta shifts missing outcomes upward relative to the reference distribution; a negative delta shifts them downward.

Definition: A multiplicative delta adjustment for a strictly positive outcome specifies $$ \log(Y^\delta)\mid X=x,R=0 \ \overset{d}{=}\ \log(Y^{\mathrm{ref}})\mid X=x,R=0+\delta, $$ where $\overset{d}{=}$ means equality in distribution after the stated shift. Equivalently, the adjusted outcome distribution is the reference outcome distribution multiplied by $e^\delta$. The same fixed $\delta$ is used across the conditional group considered here.

The displayed model uses the natural logarithm. If a different log base were used, the corresponding shift would have to be defined on that base and converted before interpreting it as a ratio. Because the outcome is logged, it must be strictly positive. The shift changes the conditional distribution by scaling its values; it does not add a fixed number of outcome units.

Deriving the Effect on Means and Medians

Exponentiating the log-scale relationship gives the original-scale relationship. In distributional terms, the adjusted outcome has the same distribution as the reference outcome multiplied by the positive constant $e^\delta$. This result is not restricted to lognormal outcomes.

Y^\delta\mid X=x,R=0 \ \overset{d}{=}\ e^\delta Y^{\mathrm{ref}}\mid X=x,R=0.

If the reference conditional mean exists and is finite, expectation scales by a constant. Thus the adjusted conditional mean is exactly $e^\delta$ times the reference conditional mean. For medians, multiplication by a positive constant multiplies the median by that constant: the ordering of outcome values is preserved, and the point containing half the distribution scales along with them. The same argument applies to any conditional quantile when it is defined.

E(Y^\delta\mid X=x,R=0) = e^\delta E(Y^{\mathrm{ref}}\mid X=x,R=0), \qquad \operatorname{Median}(Y^\delta\mid X=x,R=0) = e^\delta\operatorname{Median}(Y^{\mathrm{ref}}\mid X=x,R=0).

These equalities require a finite reference mean for the mean statement and a well-defined conditional median for the median statement. They do not require the conditional outcome to be lognormal. If the reference mean is infinite, multiplying it by a finite positive constant does not produce a finite adjusted mean.

The scale factor $e^\delta$ is a ratio. The exact percentage change in a positive conditional mean or median is $100(e^\delta-1)\%$, not $100\delta\%$. For small values of $\delta$, $e^\delta$ is approximately $1+\delta$, so $100\delta\%$ can be a useful rough approximation—but it is not the exact interpretation.

Key result: Under a constant natural-log delta, the adjusted conditional mean and median are each multiplied by $e^\delta$. A delta of $\log(1.20)$ represents a 20% increase; a delta of $\log(0.80)$ represents a 20% decrease.

Worked Examples

Worked Example: A Lognormal Reference Distribution

Suppose, hypothetically, a positive laboratory outcome $Y^{\mathrm{ref}}$, measured in mg/L, has conditional log distribution $N(4.2,0.36)$ for missing outcomes in a particular covariate group. Apply a log-scale delta of $\delta=0.20$. Find the reference and adjusted conditional means and medians.

The reference distribution is lognormal. For $Z\sim N(\mu,\sigma^2)$ and $Y=e^Z$, the median of $Y$ is $e^\mu$, while its mean is $e^{\mu+\sigma^2/2}$. Thus the reference median is $e^{4.2}\approx66.69$ mg/L, and its mean is $e^{4.2+0.36/2}=e^{4.38}\approx79.84$ mg/L.

The multiplicative factor is $e^{0.20}\approx1.2214$. Therefore the adjustment corresponds to an exact increase of $100(1.2214-1)=22.14\%$, not 20%.

The adjusted log outcome has distribution $N(4.4,0.36)$, because adding 0.20 changes the log mean but not the log variance. Its median is $e^{4.4}\approx81.45$ mg/L. Its mean is $e^{4.4+0.36/2}=e^{4.58}\approx97.51$ mg/L.

The scaling result gives the same values, up to rounding: $66.6863\ldots\times e^{0.20}\approx81.45$ mg/L for the median, and $79.838\ldots\times e^{0.20}\approx97.51$ mg/L for the mean. The difference between rounded and unrounded operands is only a rounding effect; the reported results come from the unrounded quantities.

Worked Example: A Clinically Chosen Ratio

In a hypothetical analysis, the reference conditional mean and median of a positive biomarker are 120 mg/L and 100 mg/L, respectively. Investigators choose an adjusted distribution with values 20% lower than the reference. Find the required delta and the adjusted mean and median.

A 20% reduction corresponds to a factor of $1-0.20=0.80$. The natural-log delta is therefore $\delta=\log(0.80)\approx-0.2231$. Exponentiating gives $e^\delta=0.80$, confirming the intended factor.

The adjusted mean is $0.80(120)=96.0$ mg/L, and the adjusted median is $0.80(100)=80.0$ mg/L. Both are 20% below their respective reference values. The log-scale delta is approximately $-0.2231$, not $-0.20$; using $-0.20$ would instead give a factor of $e^{-0.20}\approx0.8187$, or an approximately 18.13% reduction.

This calculation uses only the chosen ratio and the scaling result. It does not require the reference distribution to be lognormal, but it does require a positive outcome on which a log transformation is meaningful.

Worked Example: Scaling Values Without Assuming Lognormality

Consider three hypothetical positive reference outcomes, 40, 80, and 120 mg/L, in a small conditional group. Apply a common delta $\delta=\log(1.10)$, so $e^\delta=1.10$. Find the adjusted values, sample mean, and sample median.

Each reference value is multiplied by 1.10: $40(1.10)=44$, $80(1.10)=88$, and $120(1.10)=132$ mg/L. The reference sample mean is $(40+80+120)/3=80$ mg/L, so the adjusted sample mean is $(44+88+132)/3=264/3=88$ mg/L, also $1.10(80)=88$ mg/L.

The reference sample median is 80 mg/L, and the adjusted sample median is 88 mg/L, again a factor of 1.10. This example illustrates the scale transformation directly; it does not assert that a three-value sample has a lognormal distribution. The factor is an exact 10% increase for every value in this example.

Conditional and Marginal Interpretation

The formulas above are conditional on $X=x$ and $R=0$. To obtain a mean for a broader group, the conditional means must be averaged over the covariate distribution appropriate to that group. If the same constant delta applies throughout that group, the common factor can be taken outside the average. For example, if the reference conditional mean is $m(x)$ and the group uses covariate distribution $F$, then the reference marginal mean is the average of $m(x)$ over $F$; the adjusted marginal mean is $e^\delta$ times that average. This uses the same covariate distribution for the two quantities being compared.

Worked Example: Averaging Across Two Covariate Groups

Suppose, hypothetically, that among missing outcomes, 75% of a target group is in stratum A and 25% is in stratum B. The reference conditional means are 50 mg/L in A and 150 mg/L in B. Apply the same delta $\log(1.20)$ in both strata. Find the target-group reference and adjusted means.

The reference marginal mean is $0.75(50)+0.25(150)=37.5+37.5=75.0$ mg/L. In each stratum the multiplier is $e^{\log(1.20)}=1.20$, so the adjusted conditional means are 60 mg/L in A and 180 mg/L in B. Averaging with the same weights gives $0.75(60)+0.25(180)=45+45=90.0$ mg/L.

Equivalently, the adjusted marginal mean is $1.20(75.0)=90.0$ mg/L. It is 20% larger because the same factor was applied in both strata and the same weights were used. If different deltas or different averaging weights were used, a single common multiplier could not be taken outside the average in this way.

Assumptions, Diagnostics, and Alternatives

The adjustment has a clear interpretation when its assumptions match the outcome and the intended sensitivity question:

The outcome is strictly positive for the observations to which the log model applies. A zero or negative value cannot be logged without changing the outcome definition or adopting a different model.
The shift is additive on the specified log scale and is constant across the conditional group being discussed. Its original-scale multiplier is therefore the same for every outcome in that group.
The reference distribution is meaningful for the missing outcomes under consideration. The delta specifies a departure from that reference; it does not establish that the reference distribution is correct.
For a mean interpretation, the reference conditional mean is finite. For a median interpretation, the reference conditional median is defined.
The log base and direction of the shift are stated consistently. Here the log is natural and a positive delta raises the outcome.

Useful checks include confirming that all adjusted values remain positive, verifying that the factor is exactly $e^\delta$, and checking that a positive delta increases both the mean and median while a negative delta decreases them. When using a lognormal calculation, check the variance convention carefully: if the log variance is $\sigma^2$, the mean is $e^{\mu+\sigma^2/2}$, not $e^\mu$. The latter is the median in the lognormal case.

A common mistake is to interpret $\delta=0.20$ as exactly a 20% increase. The correct factor is $e^{0.20}\approx1.2214$, giving a 22.14% increase. Another mistake is to multiply only a lognormal median by the factor but leave its mean unchanged. Under this distributional adjustment, both are multiplied by $e^\delta$; the lognormal formula simply gives a way to calculate their starting values.

A multiplicative adjustment is not interchangeable with an additive delta in clinical units. An additive shift of 5 mg/L adds 5 mg/L regardless of the starting outcome, whereas a log-scale delta multiplies each value by the same ratio. Earlier material on standardized deltas expresses a shift relative to a standard deviation; it has a different interpretation and should not be confused with the ratio $e^\delta$. The log-scale approach is unsuitable without further modeling choices when zero values are substantively possible, and a single common ratio may be inadequate if the intended departure differs across conditional groups.

Key takeaway: For a positive outcome, a constant additive shift $\delta$ on the natural-log scale gives the exact original-scale multiplier $e^\delta$. Conditional means and medians are multiplied by that factor, subject to the existence of the mean and a defined median. Interpret the factor—not $\delta$ alone—as the percentage change.

Check Your Understanding

Try each question, then check the solutions below.

A log-scale delta is $0.10$. What is the exact multiplier and approximate percentage increase?
A reference conditional median is 25 mg/L and the chosen multiplier is 0.70. What are the adjusted median and natural-log delta?
For a lognormal reference outcome with log mean 3 and log variance 0.25, give its median and mean. If $\delta=\log(1.10)$, give the adjusted median and mean.
Why can a log-scale delta not be applied directly to an outcome that can equal zero?

Solutions. (1) The multiplier is $e^{0.10}\approx1.1052$, so the exact percentage increase is $100(e^{0.10}-1)\%\approx10.52\%$. (2) The adjusted median is $0.70(25)=17.5$ mg/L, and $\delta=\log(0.70)\approx-0.3567$. (3) The reference median is $e^3\approx20.09$ outcome units; the reference mean is $e^{3+0.25/2}=e^{3.125}\approx22.76$ units. The multiplier is 1.10, so the adjusted median is approximately 22.10 units and the adjusted mean is approximately 25.04 units. (4) The logarithm of zero is undefined, so the stated model cannot be applied to zero without redefining the outcome or choosing another approach.