Why the Outcome Mechanism Matters
A simulation can generate carefully structured multi-omics measurements and still answer the wrong benchmark question if its outcome is disconnected from the signals those measurements contain. Outcome generation makes that connection explicit: it specifies which signals predict the outcome, the scale on which they act, and what randomness remains after conditioning on them. The resulting data are hypothetical by construction, so the mechanism should be described precisely enough that another analyst could reproduce it.
The focus here is the outcome-generating rule, not the design of signal patterns or a comparison of analysis methods. We treat the shared and block-specific signals as quantities already defined for each subject. A shared signal represents information common across modalities; a block-specific signal represents information associated with a particular modality or block. Their construction is separate from the outcome model.
Let \(i=1,\ldots,n\) index subjects. Write \(S_i\) for subject \(i\)'s shared signal and \(B_{im}\) for their block-specific signal in modality \(m\). These are not necessarily individual measured features; they may be summaries of a designed signal structure. A general linear predictor can be written as
The coefficients determine how the specified signals enter the outcome mechanism. Setting a coefficient to zero removes that term from this particular mechanism; it does not alter how the underlying omics data were generated. Coefficients and signals must use compatible scales, and their units should be stated when the outcome has physical units.
Continuous Outcomes
A common continuous-outcome mechanism is a linear model with additive noise:
Here \(Y_i\) is the simulated outcome, \(\eta_i\) is its conditional mean, and \(\sigma\) is the residual standard deviation in the outcome's units. The usual simple version assumes errors are mutually independent and independent of the generated predictors. Conditional on the signals, \(Y_i\sim N(\eta_i,\sigma^2)\). If the outcome must be positive or otherwise bounded, this Gaussian model may be inappropriate; transforming the outcome or choosing a distribution with a suitable support changes the mechanism and its interpretation.
Worked Example: A Continuous Outcome With Two Signal Sources
Suppose an outcome is measured in standardized laboratory units. Generate it from \(Y_i=2+0.8S_i-0.5B_{i1}+\varepsilon_i\), where independent errors follow \(N(0,0.6^2)\). For one hypothetical subject, let \(S_i=1.2\), \(B_{i1}=-0.4\), and the generated error be \(\varepsilon_i=0.3\).
Substitution gives \(Y_i=2+0.8(1.2)-0.5(-0.4)+0.3\). The terms are \(0.96\), \(0.20\), and \(0.30\), so \(Y_i=3.46\) standardized units. Before the error is drawn, the conditional mean is \(2+0.96+0.20=3.16\); the remaining difference, \(0.30\), is this subject's particular noise realization. The model assumes the errors have mean zero and standard deviation \(0.6\), not that every simulated value lies close to its conditional mean.
Binary Outcomes
For a binary outcome \(Y_i\in\{0,1\}\), a logistic model maps the linear predictor to a probability:
The model assumes each outcome is a Bernoulli draw conditional on its predictors; the simplest version also assumes conditional independence across subjects. Each coefficient is a change in log odds per one-unit increase in its predictor, holding the other terms fixed. It is not a change in probability of the same size: the probability change depends on the starting value of \(\eta_i\).
The probability formula follows by rearranging the logit definition. If \(\log\{p_i/(1-p_i)\}=\eta_i\), exponentiation gives \(p_i/(1-p_i)=\exp(\eta_i)\). Thus \(p_i=\exp(\eta_i)\{1-p_i\}\), so \(p_i\{1+\exp(\eta_i)\}=\exp(\eta_i)\), which yields the displayed expression.
Worked Example: Drawing a Binary Outcome
Let \(\eta_i=-1+0.9S_i-0.4B_{i1}\), and generate \(Y_i\) as a Bernoulli variable with probability \(p_i\). For a hypothetical subject with \(S_i=1\) and \(B_{i1}=0.5\), the linear predictor is \(-1+0.9(1)-0.4(0.5)=-0.3\). Therefore \(p_i=1/(1+\exp(0.3))\approx0.4256\).
One equivalent simulation procedure draws \(U_i\sim\operatorname{Uniform}(0,1)\) independently and sets \(Y_i=1\) when \(U_i<p_i\), and \(Y_i=0\) otherwise. If this subject's hypothetical draw is \(U_i=0.31\), then \(0.31<0.4256\), so \(Y_i=1\). This single result does not mean that the subject's outcome was certain to be one; it is one draw from the stated conditional distribution.
Time-to-Event Outcomes and Censoring
A time-to-event outcome has an event time \(T_i\), which is positive, and may have a censoring time \(C_i\). One clear construction uses an exponential event-time distribution with a proportional-hazards signal effect:
Here \(\lambda_0\) and \(\lambda_i\) are rates per unit time, so \(T_i\) is measured in that time unit. The model assumes a constant event hazard over time for a subject conditional on the predictor, with the predictor multiplying the baseline hazard by \(\exp(\eta_i)\). To simulate \(T_i\), draw \(U_i\sim\operatorname{Uniform}(0,1)\) and use \(T_i=-\log(U_i)/\lambda_i\). Censoring is a separate mechanism. A common simple choice draws \(C_i\) independently of \(T_i\) and the predictors, conditional on the scenario. The observed time is \(\tilde T_i=\min(T_i,C_i)\), and the event indicator is \(\Delta_i=I(T_i\leq C_i)\).
Independent censoring is a substantive assumption, not an automatic feature of simulated data. If censoring depends on predictors, or on event risk after conditioning on them, that dependence must be specified. The event-time and censoring models should be recorded separately so that the observed follow-up process is interpretable.
Worked Example: Event Time With Independent Censoring
Suppose time is measured in years, \(\lambda_0=0.10\) per year, and the event rate is \(\lambda_i=0.10\exp(0.7S_i)\). For a hypothetical subject with \(S_i=0.8\), \(\lambda_i=0.10\exp(0.56)\approx0.1751\) per year. If the event-time uniform draw is \(U_i=0.40\), then \(T_i=-\log(0.40)/0.1751\approx0.9163/0.1751=5.23\) years.
Now suppose an independent censoring time is generated as exponential with rate \(0.05\) per year. With a separate draw \(V_i=0.50\), \(C_i=-\log(0.50)/0.05\approx13.86\) years. Thus \(\tilde T_i=\min(5.23,13.86)=5.23\) years and \(\Delta_i=1\): the event is observed before censoring. The separate independent draws implement the stated censoring assumption; merely reporting the minimum would not establish that assumption.
Longitudinal Outcomes
Repeated observations on the same subject are usually dependent. A random-intercept model is one transparent way to generate this dependence:
where \(k\) indexes measurement occasion. In this basic model, the random intercept \(b_i\) is shared across occasions for subject \(i\); residual errors are independent across occasions and independent of \(b_i\) and the predictors. Conditional on the signals, two different measurements from the same subject have covariance \(\tau^2\), while each measurement has variance \(\tau^2+\sigma^2\). The within-subject correlation is consequently \(\tau^2/(\tau^2+\sigma^2)\). This is an exchangeable correlation structure: every pair of occasions has the same correlation.
When observations nearer in time should be more correlated, residuals can instead follow a first-order autoregressive structure: \(\operatorname{Cov}(\varepsilon_{ik},\varepsilon_{il})=\sigma^2\rho^{|k-l|}\), where \(|\rho|<1\). If the random intercept and residual process are independent, the total covariance between two occasions is \(\tau^2+\sigma^2\rho^{|k-l|}\). This adds serial dependence without changing the subject-level signal contribution.
Worked Example: Generating Three Repeated Measurements
Let \(\eta_i=10+2S_i-B_{i1}\), with a random intercept \(b_i\sim N(0,1)\) and residual variance \(\sigma^2=0.25\). For a hypothetical subject, take \(S_i=0.5\), \(B_{i1}=1\), and a generated random intercept \(b_i=0.4\). Then \(\eta_i=10+2(0.5)-1=10\), so the subject-specific level before occasion errors is \(10.4\). Suppose the three generated residuals are \(-0.2\), \(0.1\), and \(0.3\). The measurements are \(10.4-0.2=10.2\), \(10.4+0.1=10.5\), and \(10.4+0.3=10.7\).
Under the independent-residual version, each measurement has variance \(1+0.25=1.25\), and measurements at different occasions have covariance \(1\) from the shared random intercept. Their correlation is \(1/1.25=0.80\). The repeated values are not three independent subjects: they share both the same signal-based level and the same random intercept.
Design Choices, Diagnostics, and Alternatives
Before drawing outcomes, state the conditional distribution and the dependence assumptions. A useful specification names the signal terms, coefficient units, outcome scale, noise distribution, and, where relevant, event and censoring distributions or repeated-measure covariance. Then generate the outcomes in an order consistent with the data-generating story: construct complete predictors, generate outcomes conditional on them, and apply a separately specified observation or missingness process if the scenario requires one. If observation depends on the outcome or its unobserved value, it is not enough to apply missingness as though it were independent.
Check whether the simulated data exhibit the intended structure. For continuous outcomes, inspect conditional means and residual spread across signal levels, as well as the marginal range and outliers. For binary outcomes, check the range of generated probabilities and the fraction of outcomes equal to one; extreme probabilities may leave very few observations in one class. For time-to-event data, examine event fractions, observed follow-up times, and how censoring varies across predictors. For longitudinal data, compare empirical within-subject correlations at different lags with the intended covariance pattern. These checks diagnose implementation and scenario mismatches; they do not by themselves establish that a mechanism is realistic for every application.
Alternatives answer different design needs. A Gaussian linear outcome is convenient when an unbounded continuous scale and additive errors are defensible; a generalized linear model can respect outcomes such as counts or binary indicators. Logistic regression is natural for binary outcomes, while a probit link offers a different mapping from the linear predictor to probability. Exponential survival times provide a constant conditional hazard; other baseline hazard distributions are needed when hazard changes over time. A random intercept gives exchangeable repeated-measure dependence, whereas an autoregressive residual structure represents dependence that declines with lag. Choosing among these is a modeling decision, not a cosmetic change: it changes what the generated outcomes mean and which assumptions hold.
Other failure modes include using a Gaussian model for a binary response, generating censoring times but omitting the event indicator, and interpreting a logistic coefficient as a probability difference. A further pitfall is letting a signal affect outcomes unintentionally through duplicated or correlated predictor terms. Check the exact linear predictor used in code and document which signal sources enter it. The mechanism is a controlled abstraction: it can clarify what a benchmark tests, but it cannot represent every feature of a clinical process. In particular, a simple conditional model may omit nonlinear effects, interactions, time-varying predictors, or dependence between event and censoring times. Add such features only when they serve the benchmark question and can be specified unambiguously.
Check Your Understanding
Answer each question, then compare with the solutions below.
- For \(Y_i=1+2S_i+\varepsilon_i\), with \(\varepsilon_i\sim N(0,3^2)\), what is the conditional mean and residual standard deviation when \(S_i=0.5\)?
- A logistic model gives \(\eta_i=0\). What is \(P(Y_i=1\mid\text{predictors})\)?
- In the exponential event model, what does \(\lambda_i\) measure, and what additional quantity is needed to determine whether an event was observed?
- For the random-intercept model with \(\tau^2=2\) and \(\sigma^2=2\), what is the correlation between two different occasions under independent residual errors?
Solutions. (1) The conditional mean is \(1+2(0.5)=2\) outcome units; the residual standard deviation is \(3\) units. (2) The logistic probability is \(\exp(0)/(1+\exp(0))=1/2=0.5\). (3) \(\lambda_i\) is the conditional event rate, with units inverse to the time unit; one also needs a censoring time \(C_i\), since the event is observed only when \(T_i\leq C_i\). (4) The covariance across different occasions is \(\tau^2=2\), and each occasion has variance \(\tau^2+\sigma^2=4\), so the correlation is \(2/4=0.5\).