Why Shift a Distribution?
A mediator can take many values, and an intervention need not force every person to have the same mediator value. It may instead make some values more likely than others. For example, a program might increase the chance of higher treatment adherence without guaranteeing perfect adherence for each person. An incremental stochastic intervention describes this kind of change by reweighting the mediator’s existing conditional distribution.
This approach is useful when deterministic assignments are implausible or when the mediator is continuous. Rather than asking what would happen if everyone’s mediator were set to one chosen value, we ask what the expected outcome would be if mediator values were drawn from a specified, shifted distribution.
Defining an Incremental Tilt
Let \(A\) denote exposure, \(C\) baseline covariates, \(M\) a mediator, and \(Y\) the outcome. Write \(g(m\mid a,c)\) for the conditional distribution of \(M\) given \(A=a,C=c\). This notation can represent a probability mass function for a discrete mediator or a density for a continuous mediator. Let \(w_\delta(m,a,c)\) be a nonnegative weight, indexed by a shift parameter \(\delta\).
The weight changes the relative representation of mediator values, while normalization makes the result a valid probability distribution. A common choice is \(w_\delta(m,a,c)=\exp\{\delta s(m,a,c)\}\), where \(s\) is a specified score. The score determines which mediator values are favored: for example, \(s(m)=m\) favors larger values when \(\delta>0\). The weight family and its parameter scale must be stated; \(\delta\) has no universal interpretation independent of that choice.
To make \(\delta=0\) the no-shift intervention, require that \(w_0(m,a,c)\) is constant in \(m\), usually \(w_0=1\). Then \(g_0(m\mid a,c)=g(m\mid a,c)\). More generally, an arbitrary weight family need not have this property, so its value at zero cannot automatically be called “no shift.”
For a binary mediator \(M\in\{0,1\}\), a particularly interpretable tilt uses \(w_\lambda(m)=\lambda^m\), with \(\lambda>0\). It multiplies the conditional odds of \(M=1\) by \(\lambda\). When \(\lambda=1\), there is no shift; when \(\lambda>1\), the intervention favors \(M=1\); and when \(0<\lambda<1\), it favors \(M=0\). It does not set \(M=1\) or \(M=0\) deterministically.
From the Tilt to an Outcome Effect
Let \(\mu(a,m,c)=E[Y\mid A=a,M=m,C=c]\) be the conditional outcome mean. Under a stochastic intervention that assigns \(M\) from \(g_\delta(\cdot\mid a,c)\), define \(\psi(a;\delta)\) as the mean outcome under exposure level \(a\) and that mediator distribution. Its identifying functional is the mediational g-formula:
Here \(F_C\) is the baseline covariate distribution in the target population. For discrete variables, the corresponding integrals are sums. With \(w_0=1\), the incremental effect at exposure level \(a\) is \(\psi(a;\delta)-\psi(a;0)\). This is a contrast between two mediator distributions while holding the exposure level fixed; it is not automatically a natural indirect effect.
Interpreting this functional causally from observed data requires assumptions. Consistency requires observed outcomes and mediators to agree with the corresponding potential quantities under the received exposure and mediator values. Conditional exchangeability requires no unmeasured confounding of exposure with mediator or outcome, and no unmeasured mediator–outcome confounding after conditioning on \(A,C\). Positivity requires support for the exposure and mediator values used in the formula, within relevant covariate strata. Under these conditions, the observed conditional mean and mediator distribution identify the stated intervention mean. The intervention draws mediator values from the specified distribution; it does not require assigning a deterministic mediator value to each person.
Worked Calculations
Worked Example: Doubling the Odds of a Binary Mediator
Suppose, within a fixed exposure and covariate stratum, the observed probability of \(M=1\) is \(0.30\). Suppose the outcome mean is 10 points when \(M=0\) and 18 points when \(M=1\). Consider the tilt \(w_\lambda(m)=\lambda^m\) with \(\lambda=2\).
The original mediator probabilities are \(P(M=0)=0.70\) and \(P(M=1)=0.30\). Weighting gives unnormalized probabilities \(0.70\) and \(2(0.30)=0.60\), so the normalizing constant is \(0.70+0.60=1.30\). Therefore,
The original odds were \(0.30/0.70=3/7\); the shifted odds are \((6/13)/(7/13)=6/7\), exactly twice as large. The original outcome mean is \(0.70(10)+0.30(18)=12.40\) points. Under the tilt, it is
The intervention contrast is \(13.69-12.40\approx1.292\) points. The change is not \(8\) points, the outcome difference between mediator categories, because the intervention changes the probability of the higher-outcome category rather than assigning everyone to it.
Worked Example: Tilting a Three-Level Mediator
Suppose a mediator has levels \(M=0,1,2\) with probabilities \(0.50,0.30,0.20\), respectively. Conditional outcome means at those levels are 5, 12, and 25 points. Use \(w_\lambda(m)=\lambda^m\) with \(\lambda=2\).
The weights are 1, 2, and 4. The weighted probabilities before normalization are \(0.50(1)=0.50\), \(0.30(2)=0.60\), and \(0.20(4)=0.80\). Their sum is \(1.90\), giving shifted probabilities \(0.50/1.90\), \(0.60/1.90\), and \(0.80/1.90\). These are approximately \(0.2632,0.3158,0.4211\), which sum to one (up to rounding).
The original mean is \(0.50(5)+0.30(12)+0.20(25)=2.50+3.60+5.00=11.10\) points. The shifted mean is
The contrast is \(15.63-11.10\approx4.532\) points. This tilt favors higher mediator levels but still assigns positive probability to all three levels. As in the binary case, it is a stochastic redistribution, not a deterministic assignment.
Worked Example: Exponential Tilting of a Continuous Mediator
Suppose \(M\mid A=a,C=c\sim N(2,1)\), so its mean is 2 units and its variance is 1 square unit. Use the exponential weight \(w_\delta(m)=e^{\delta m}\) with \(\delta=0.5\) per mediator unit. Suppose the conditional outcome mean is \(\mu(a,m,c)=30+4m\) points.
The normal density is proportional to \(\exp\{-(m-2)^2/2\}\). Multiplying by the tilt gives an exponent that can be rearranged by completing the square:
After normalization, the constant term does not affect the density. Thus the tilted distribution is \(N(2.5,1)\): the mediator mean increases by \(0.5\) units, while the variance remains 1 square unit. Under the original distribution, the outcome mean is \(30+4(2)=38\) points. Under the tilt, it is \(30+4(2.5)=40\) points. The intervention contrast is \(40-38=2\) points. No individual’s mediator is fixed at 2.5; the whole mediator distribution is shifted.
Why the Exponential Tilt Shifts the Normal Mean
The preceding calculation is an instance of a general normal identity. If \(M\sim N(\mu,\sigma^2)\) and the weight is \(e^{\delta M}\), the original density multiplied by the weight has exponent
The final two terms do not depend on \(m\), so they are absorbed into the normalizing constant. The tilted distribution is therefore \(N(\mu+\delta\sigma^2,\sigma^2)\), provided the normalizing constant is finite. If the outcome mean is linear in the mediator, \(\mu_Y(a,m,c)=\beta_0(a,c)+\beta_1(a,c)m\), the conditional outcome-mean change in this stratum is \(\beta_1(a,c)\delta\sigma^2\). Averaging that change over \(C\) gives the population contrast when these distributional and outcome models apply.
Practical Interpretation and a Common Pitfall
An incremental shift is defined by three choices: the baseline mediator distribution, the weight function, and the shift parameter. Report all three. For a binary mediator, “multiply the odds by two” is interpretable; for a continuous mediator, a value of \(\delta\) depends on the score scale and units. The resulting contrast is an average outcome difference in the outcome’s units, relative to the stated reference distribution and target population.
A practical calculation follows a short sequence:
Choose \(w_\delta\), its parameter value, the exposure level, and the baseline conditional mediator distribution.
Calculate \(Z_\delta(a,c)\) separately for each relevant \(a,c\), then form \(g_\delta\).
Integrate or sum \(\mu(a,m,c)\) against the tilted distribution and the target covariate distribution.
Subtract the mean under \(g_0=g\), and describe the difference in context and units.
A common mistake is to treat a weight as though it directly specifies a shifted probability. For example, if the original probability of a binary mediator is \(0.30\), multiplying it by two does not make the new probability \(0.60\). The weight first produces unnormalized probabilities; the probability of \(M=1\) after normalization is \(0.60/(0.70+0.60)\approx0.4615\). Another mistake is to say that the intervention “sets the mediator to a higher value.” It changes the distribution of mediator values, and the effect is calculated by averaging outcomes over that distribution.
Check Your Understanding
Use the definitions and calculations above to answer the following.
- For a binary mediator with \(P(M=1)=0.20\), what probability of \(M=1\) results from an odds-multiplying tilt with \(\lambda=3\)?
- Why must the weight be normalized separately within each covariate stratum when it depends on \(C\)?
- What condition on \(w_0\) makes \(\psi(a;0)\) the mean under the original mediator distribution?
- If \(M\sim N(\mu,\sigma^2)\) and the weight is \(e^{\delta M}\), what are the tilted mean and variance?
- Name two assumptions needed to interpret the mediational g-formula as a causal intervention mean from observed data.