Multiplicative Structural Mean Models On The Risk-Ratio Scale

Why Model a Risk Ratio?

For a binary outcome, the mean is a risk: if $Y(a)$ denotes the outcome that would occur under treatment level $a$, then $E[Y(a)\mid\cdot]$ is a conditional probability. A multiplicative structural mean model describes how that risk changes between treatment levels on a ratio scale. This can be useful when a proportional change in risk is the scientific target, rather than a fixed difference in risk.

The model is structural because it relates potential-outcome means, including means for outcomes that may not be observed. An instrumental variable helps estimate its parameter when treatment is endogenous: treatment may be associated with potential outcomes even after conditioning on measured covariates. The instrument supplies a moment restriction, not an automatic guarantee that the model or its assumptions are correct.

Model and Interpretation

Definition: Let $A\in\{0,1\}$ be treatment, $Y\in\{0,1\}$ the observed outcome, $Y(a)$ the potential outcome under treatment level $a$, $Z$ an instrument, and $L$ measured covariates. Consistency means $Y=Y(A)$. A constant-parameter multiplicative structural mean model specifies, for relevant strata of $A,Z,L$, $$ E[Y(a)\mid A,Z,L]=E[Y(0)\mid A,Z,L]\exp(\psi a),\qquad a\in\{0,1\}. $$ Where $E[Y(0)\mid A,Z,L]>0$, this is equivalent to $E[Y(1)\mid A,Z,L]/E[Y(0)\mid A,Z,L]=\exp(\psi)$. Thus $\exp(\psi)$ is the conditional risk ratio and $\psi$ is its logarithm.

The displayed mean equation is a complete version of the model. In strata where the baseline mean is zero, it requires $E[Y(1)\mid A,Z,L]=0$ as well. The ratio definition alone does not impose that requirement: a ratio with a zero denominator is undefined, so it places no restriction on the treated mean in those strata. The distinction matters for the estimating equation derived below.

The conditional risk ratio compares mean potential outcomes within the specified $A,Z,L$ strata. It is not an individual-level ratio, and it is not automatically the population risk ratio for everyone. For example, $\exp(\psi)=0.70$ means the modeled risk under treatment is 0.70 times the modeled risk without treatment in a stratum; it does not mean a 30-percentage-point reduction. The absolute difference depends on the baseline risk.

Worked Example: Interpreting the Risk-Ratio Parameter

Suppose a hypothetical stratum has $E[Y(0)\mid A,Z,L]=0.20$ and $E[Y(1)\mid A,Z,L]=0.10$. The model’s risk ratio is $$ \frac{0.10}{0.20}=0.50,\qquad \psi=\log(0.50)\approx -0.6931. $$

The model says treatment halves the risk in this stratum. The risk difference is $0.10-0.20=-0.10$, or a reduction of 10 percentage points. Those are different effect measures: the ratio is 0.50, while the absolute difference is minus 0.10. A risk ratio of 0.50 would correspond to a different risk difference if the baseline risk were different.

Assumptions for the Instrument-Based Equation

Write $A(z)$ for the treatment that would be received under instrument level $z$. The standard assumptions relevant to the moment equation are as follows.

Consistency and no interference: each person’s observed outcome equals the potential outcome under their received treatment, and one person’s treatment does not change another person’s outcome.
Exclusion: the instrument affects the outcome only through treatment, so potential outcomes can be written $Y(a)$, without a separate $z$ argument.
Conditional instrument independence: given $L$, $Z$ is independent of the potential outcomes and treatment assignments, $\{Y(0),Y(1),A(z):z\}$. In particular, $E[Y(0)\mid Z,L]=E[Y(0)\mid L]$.
Instrument positivity: each instrument level being compared has positive probability at covariate values in the target population. For a binary instrument, $0<P(Z=1\mid L)<1$ there.
Instrument relevance: the instrument changes the distribution of treatment for at least some covariate strata. Relevance is needed for information about the treatment effect, but does not establish independence or exclusion.
Structural mean model: the full mean equation above holds, including its zero-baseline-risk condition. A single $\psi$ additionally assumes a common log risk ratio across the strata represented by the model.

These assumptions do not require treatment to be independent of potential outcomes. That is precisely why an instrument is useful. However, the assumptions do not by themselves ensure that the parameter is uniquely identified: the resulting population moment equation must have a unique relevant solution. A weak instrument, a poorly fitting structural model, or multiple roots can undermine identification or estimation.

How the Instrument Creates a Moment Condition

Define a transformed outcome $H_\psi=Y\exp(-\psi A)$. Under consistency and the full structural mean equation, its conditional expectation given $A,Z,L$ is the untreated potential-outcome mean in that stratum. If $A=0$, then $H_\psi=Y=Y(0)$. If $A=1$, then $$ E[H_\psi\mid A,Z,L] =\exp(-\psi)E[Y(1)\mid A,Z,L] =E[Y(0)\mid A,Z,L]. $$ The full equation makes this equality valid in zero-baseline strata too.

Averaging over treatment conditional on $Z,L$ gives $E[H_\psi\mid Z,L]=E[Y(0)\mid Z,L]$. Conditional instrument independence then makes this equal to $E[Y(0)\mid L]$, which does not vary with $Z$. Consequently, the transformed outcome has no conditional mean association with the instrument:

E\left[\{Z-E(Z\mid L)\}Y\exp(-\psi A)\right]=0.

The multiplication by the centered instrument removes any component of the transformed outcome mean that is common across instrument levels within $L$. It leaves a condition that can be used to estimate $\psi$. With a binary instrument, $E(Z\mid L)=P(Z=1\mid L)$. If there are no covariates and the instrument is randomized with probability $q$, the equation becomes $E[(Z-q)Y\exp(-\psi A)]=0$.

Important qualification: The derivation uses $E[Y(1)\mid A,Z,L]=0$ whenever $E[Y(0)\mid A,Z,L]=0$. A ratio-only definition on positive-baseline strata does not imply this. If zero-baseline strata are present, either impose the stated zero-mean condition to use this equation as written, or use a model and estimating strategy whose assumptions explicitly address those strata.

Worked Instrumental-Variable Calculation

Worked Example: Solving a Sample Moment Equation

Consider hypothetical data from 200 individuals, with no covariates and a binary instrument assigned equally often. The table reports treatment counts and outcome-positive counts; outcomes are binary.

Instrument	Number	Treated	Outcome-positive among treated	Outcome-positive among untreated
$Z=1$	100	60	18	12
$Z=0$	100	20	8	16

The treatment rates are $60/100=0.60$ for $Z=1$ and $20/100=0.20$ for $Z=0$, so the sample shows instrument relevance. The instrument groups are equal in size, so the sample moment is zero when the sum of transformed outcomes in the $Z=1$ group equals that in the $Z=0$ group. Let $r=\exp(-\psi)$. Untreated outcomes have $A=0$ and receive multiplier 1; treated outcomes have multiplier $r$. Thus the equation is $$ 12+18r=16+8r. $$ Solving gives $10r=4$, so $r=0.40$, and therefore $$ \widehat{\psi}=-\log(0.40)\approx 0.9163,\qquad \exp(\widehat{\psi})=2.50. $$

Under the model, this estimate corresponds to a conditional risk ratio of 2.50 for treatment versus no treatment. For a quick check, the transformed outcome totals are $12+18(0.40)=19.2$ and $16+8(0.40)=19.2$, so the sample moment is zero. The observed outcome risks themselves are $30/100=0.30$ and $24/100=0.24$ across instrument groups; those are instrument-group risks, not the model’s treatment risk ratio. This arithmetic illustrates the estimating equation, but does not establish that independence, exclusion, or the structural mean model holds.

Zero Risks, Diagnostics, and Failure Modes

Worked Example: A Zero Baseline-Risk Stratum

Suppose, hypothetically, that in one $A,Z,L$ stratum the mean untreated potential outcome is zero. Because $Y(0)$ is binary and nonnegative, this means $Y(0)=0$ with probability one in that stratum. The ratio $E[Y(1)\mid A,Z,L]/E[Y(0)\mid A,Z,L]$ is undefined; it cannot establish a risk ratio.

The full multiplicative mean equation adds a substantive restriction: $E[Y(1)\mid A,Z,L]=0$. Binary nonnegativity then implies $Y(1)=0$ with probability one there as well. If instead the hypothetical treated mean were 0.20, the ratio-only statement would still be silent, but the full equation would be violated. In that case the blipped outcome would not have the untreated mean in that stratum, and the moment-condition derivation would not apply as stated.

In practice, the potential means are not both directly observed within each stratum, so diagnostics cannot verify the structural model simply by comparing observed treatment groups. They can still reveal problems or warn that results are fragile:

Check whether the instrument has meaningful treatment variation after accounting for $L$. Very small conditional differences indicate weak identification and can produce unstable estimates.
Inspect overlap in instrument probabilities across covariate patterns. Sparse or absent instrument levels make the conditional comparison dependent on extrapolation.
Evaluate whether the fitted root is finite, unique in a scientifically plausible range, and stable to reasonable changes in covariate adjustment and instrument specification. A moment equation with no root or several roots does not provide a straightforward estimate.
Remember that $Y\exp(-\psi A)$ is a transformed outcome, not necessarily a binary outcome or a probability. Its values may exceed 1; that is not itself a contradiction.
Consider whether a common risk ratio is scientifically plausible. If effects differ substantially across covariate strata, a constant-$\psi$ model can obscure that variation, even if an overall moment equation can be solved.

A frequent mistake is to read $\exp(\psi)$ as a risk difference, or to treat an instrument association as if it were a treatment effect. The correction is to keep the scales and assumptions distinct: $\exp(\psi)$ is a modeled conditional risk ratio, and the moment equation identifies its parameter only under the instrument assumptions and the structural mean model. Another mistake is to divide by a zero baseline mean and claim the ratio is zero or infinite. It is undefined; any restriction in that stratum must be stated separately.

How This Model Differs From Alternatives

An additive structural mean model uses a risk-difference scale: its treatment blip is additive rather than multiplicative. The two models therefore make different statements about how treatment effects vary with baseline risk. The risk-ratio model is not automatically preferable; the choice should reflect the target scientific contrast and whether its restrictions are credible. Logistic structural mean models for binary outcomes use a different effect scale. Ordinary regression that conditions on treatment and covariates is not automatically an instrumental-variable analysis and, with endogenous treatment, does not gain instrument-based identification merely by including $Z$ as another predictor.

The multiplicative model’s strengths are a direct risk-ratio interpretation and a simple transformed-outcome moment equation. Its limitations include the common-effect restriction, reliance on valid and relevant instruments, and sensitivity to zero-risk strata under the full equation. It also does not, by itself, provide every population marginal risk or risk difference; those require additional standardization or other estimands and methods.

Key takeaway: The multiplicative structural mean model sets a conditional potential-outcome risk ratio to $\exp(\psi)$. Under consistency, the full mean equation, and valid instrument assumptions, $Y\exp(-\psi A)$ has a conditional mean independent of the instrument given $L$. The zero-baseline-risk condition is essential to that derivation.

Check Your Understanding

Answer each question, then compare with the short solutions.

If $\psi=\log(0.8)$, what risk ratio does the model specify, and does it describe a 20-percentage-point decrease?
In a stratum with untreated mean 0.15, what treated mean follows from $\psi=\log(2)$?
Why does a ratio definition alone fail to imply the full mean equation when the untreated mean is zero?
In a no-covariate sample with equal-sized instrument groups, transformed outcome totals are 25 for $Z=1$ and 20 for $Z=0$ when $r=\exp(-\psi)=0.5$. Does this value solve the sample moment equation?
Name two distinct instrumental-variable assumptions that relevance does not establish.

Solutions. (1) The risk ratio is $\exp(\psi)=0.8$. It describes a 20% relative reduction, not a fixed 20-percentage-point decrease; the absolute difference depends on baseline risk. (2) The treated mean is $0.15\exp(\log 2)=0.30$. (3) The denominator is zero, so the ratio is undefined and places no restriction on the treated mean. The full equation separately requires the treated mean to be zero in that stratum. (4) No. Equal group sizes require equal transformed totals for the centered-instrument moment to be zero, and $25\ne20$. (5) Relevance does not establish conditional instrument independence or exclusion.