The SMC-FCS Conditional Density

Why the Conditional Density Has Two Parts

When a covariate $X$ is incomplete, its imputation distribution should reflect both how $X$ is distributed among individuals with the same covariates and what the observed outcome $Y$ tells us about $X$. The SMC-FCS target conditional density combines those two sources of information. The covariate model supplies a distribution for $X$; the substantive model supplies a likelihood contribution that changes the relative weight assigned to each possible value of $X$.

This tutorial considers one incomplete covariate $X$, an observed outcome $Y$, and a vector $Z$ of observed covariates. The substantive model specifies $f(y\mid x,z;\psi)$, while the covariate model specifies $f(x\mid z;\beta)$. The parameters $\psi$ and $\beta$ are treated as given when writing the conditional density; in an imputation procedure, they are estimated or drawn as part of the iterative process.

Definition: Under the factorization $f(y,x\mid z;\psi,\beta)=f(y\mid x,z;\psi)f(x\mid z;\beta)$, the SMC-FCS target conditional density for $X$, given observed $Y=y$ and $Z=z$, is $$ f(x\mid y,z;\psi,\beta)= \frac{f(y\mid x,z;\psi)f(x\mid z;\beta)} {\displaystyle\int_{\mathcal X} f(y\mid u,z;\psi)f(u\mid z;\beta)\,du}. $$ Here $\mathcal X$ is the support of $X$. For discrete $X$, replace the integral in the denominator by a sum over its possible values.

The numerator has a direct interpretation. The term $f(x\mid z;\beta)$ is the covariate-model contribution for a candidate value $x$. The term $f(y\mid x,z;\psi)$ is the substantive-model likelihood for the observed outcome if $X$ took that value. The denominator, which does not depend on $x$, makes the weighted contributions integrate or sum to one. It is also the model-implied density $f(y\mid z;\psi,\beta)$.

Deriving and Using the Target

The result follows from the product rule and Bayes’ rule. Under the stated factorization, the joint density of $Y$ and $X$ conditional on $Z=z$ is the product of the outcome density and the covariate density. Dividing this joint density by its marginal over $X$ gives the conditional density of $X$ given $Y$ and $Z$.

\begin{aligned} f(x\mid y,z;\psi,\beta) &=\frac{f(y,x\mid z;\psi,\beta)}{f(y\mid z;\psi,\beta)}\\ &=\frac{f(y\mid x,z;\psi)f(x\mid z;\beta)} {\displaystyle\int_{\mathcal X}f(y\mid u,z;\psi)f(u\mid z;\beta)\,du}, \end{aligned}

For a discrete covariate, the same argument uses a sum in place of the integral. These expressions describe the conditional distribution implied by the two models; they are not simply the covariate model reused unchanged. If the substantive likelihood is constant in $x$ for the observed $y,z$, its contribution cancels during normalization and the target equals $f(x\mid z;\beta)$. Otherwise, the outcome changes the conditional distribution.

The factorization is a modeling assumption: it asserts that the joint distribution of $Y$ and $X$, conditional on $Z$, is represented by the specified outcome and covariate models. The substantive model must describe the analysis relationship of interest, and the covariate model must be an adequate model for $X\mid Z$. For an incomplete $X$, using this conditional distribution as the imputation target also relies on an ignorable missingness mechanism: missingness in $X$ is missing at random given the observed variables used in the analysis, and the missingness parameters are distinct from the data-model parameters. If missingness still depends on the unseen value of $X$ after conditioning on observed information, this density alone does not account for that dependence.

Worked Example: A Binary Covariate and Binary Outcome

Suppose, hypothetically, $X$ indicates whether a participant has a particular exposure, and $Y$ is a binary outcome. At a fixed value of $Z=z$, the covariate model gives $P(X=1\mid z)=0.30$, so $P(X=0\mid z)=0.70$. The substantive model gives $P(Y=1\mid X=1,z)=0.80$ and $P(Y=1\mid X=0,z)=0.20$. For an individual with observed $Y=1$, calculate the target probability that $X=1$.

The two unnormalized weights are the covariate probabilities multiplied by the outcome likelihoods:

\begin{aligned} w_1&=P(Y=1\mid X=1,z)P(X=1\mid z)=0.80(0.30)=0.24,\\ w_0&=P(Y=1\mid X=0,z)P(X=0\mid z)=0.20(0.70)=0.14. \end{aligned}

The normalizing constant is $0.24+0.14=0.38$. Therefore,

P(X=1\mid Y=1,z)=\frac{0.24}{0.38}=0.6316.

Thus, under these hypothetical models and at this value of $z$, the conditional probability of exposure for an individual with $Y=1$ is about 0.632. It is greater than the covariate-model probability of 0.30 because the substantive model assigns a higher likelihood to $Y=1$ when $X=1$.

Worked Example: A Normal Covariate Under a Normal Substantive Model

Suppose, hypothetically, that $X\mid z\sim N(2,1)$ and $Y\mid X=x,z\sim N(1+x,1)$. The second argument of $N$ is the variance. For an observed $y=5$, derive the target conditional distribution of $X\mid y,z$. Treat the model parameters as known for this calculation.

Ignoring factors that do not depend on $x$, the covariate density contributes $\exp\{-(x-2)^2/2\}$, and the substantive likelihood contributes $\exp\{-(5-(1+x))^2/2\}$. Multiplying them adds their quadratic terms in the exponent:

-\frac{1}{2}\left[(x-2)^2+(4-x)^2\right] =-\frac{1}{2}\left[2x^2-12x+20\right] =-\frac{1}{2}\left[2(x-3)^2+2\right].

The final term that does not contain $x$ is absorbed into the normalizing constant. The remaining kernel is $\exp\{-(x-3)^2\}$, the kernel of a normal density with mean 3 and variance $1/2$, since a normal kernel with variance $v$ has exponent $-(x-m)^2/(2v)$. Hence $X\mid Y=5,z\sim N(3,0.5)$. The outcome shifts the conditional mean from 2 to 3, while the conditional variance is smaller than the prior variance of 1 because both model contributions provide information about $X$.

Worked Example: Normalizing a Three-Level Covariate

Consider a hypothetical three-level covariate $X\in\{A,B,C\}$ at a fixed $z$. The covariate model gives probabilities $0.20$, $0.50$, and $0.30$, respectively. For an individual with observed binary outcome $Y=1$, suppose the substantive-model likelihoods are $0.10$, $0.40$, and $0.80$ for $A$, $B$, and $C$. Find the target probabilities.

The unnormalized weights are $0.20(0.10)=0.020$ for $A$, $0.50(0.40)=0.200$ for $B$, and $0.30(0.80)=0.240$ for $C$. Their sum is $0.020+0.200+0.240=0.460$. Dividing each weight by this sum gives

\begin{aligned} P(X=A\mid Y=1,z)&=\frac{0.020}{0.460}=0.04348,\\ P(X=B\mid Y=1,z)&=\frac{0.200}{0.460}=0.4348,\\ P(X=C\mid Y=1,z)&=\frac{0.240}{0.460}=0.5217. \end{aligned}

The probabilities sum to one, up to rounding. Although $B$ had the largest covariate-model probability before using the outcome, $C$ has the largest target probability because its likelihood for the observed outcome is higher.

Assumptions, Diagnostics, and Alternatives

In practice, the formula is a target conditional distribution within an iterative imputation procedure. Its reliability depends on both factors in the numerator, not just on the act of normalization. The outcome model must represent the substantive analysis, and the covariate model must provide a credible distribution for the incomplete covariate conditional on the included predictors. A numerically valid density can still be a poor imputation model if either component is misspecified.

Several checks help reveal problems:

Confirm that the covariate model assigns probability or density over the scientifically possible support of $X$. Values outside a valid domain cannot be repaired by weighting with the outcome likelihood.
Check that the weighted contributions can be normalized. If the denominator is zero or numerically negligible, the observed outcome is incompatible with the candidate support or the specified models, or the computation is unstable.
Inspect whether the target changes plausibly with $Y$ and $Z$. A likelihood contribution that is accidentally omitted, evaluated at the wrong covariate value, or coded on the wrong scale can make the target silently reduce to the covariate model or distort it.
Assess the fitted substantive and covariate models and the stability of the imputation procedure. Strong sensitivity to modeling choices or poor mixing can indicate that the target is not being represented reliably.

The central comparison is with imputing from $f(x\mid z;\beta)$ alone. That distribution represents the covariate model but does not, in general, condition on the observed outcome. The SMC-FCS target explicitly incorporates the substantive likelihood. A different approach is to specify a joint model for $(Y,X)$ directly; the factorization used here is one way to construct that joint model. Passive imputation addresses deterministic derived variables, but it does not by itself supply this outcome-weighted conditional density for an incomplete covariate.

Common pitfall: “The covariate model already estimates $X$, so I can draw the missing value from $f(x\mid z)$ and then use the substantive model only for analysis.” This omits the observed outcome from the conditional distribution used to impute $X$. The correction is to weight each candidate value by $f(y\mid x,z;\psi)$, then normalize, so that the imputation target is proportional to $f(y\mid x,z;\psi)f(x\mid z;\beta)$.

The target has limitations. It is conditional on the specified model forms and parameter values, so parameter uncertainty must be represented in multiple imputation rather than treating fitted parameters as known without qualification. It also does not automatically address nonignorable missingness, a misspecified substantive model, or a misspecified covariate model. Finally, the displayed formula covers one incomplete covariate with the other conditioning variables observed; additional incomplete covariates require a more elaborate conditional updating scheme.

Check Your Understanding

Answer each question using the target-density formula. Brief solutions follow the questions.

For binary $X$, suppose $P(X=1\mid z)=0.40$, $P(Y=1\mid X=1,z)=0.60$, and $P(Y=1\mid X=0,z)=0.20$. Find $P(X=1\mid Y=1,z)$.
In one sentence, explain why the target density must be normalized.
If $f(y\mid x,z;\psi)$ is constant over $x$ for the observed $y,z$, what is the target distribution of $X$?
Name one assumption about missingness needed to use this conditional density as an imputation target, and describe what fails if that assumption is not plausible.

Solutions. (1) The weights are $0.40(0.60)=0.24$ for $X=1$ and $0.60(0.20)=0.12$ for $X=0$, so $P(X=1\mid Y=1,z)=0.24/(0.24+0.12)=0.6667$. (2) The product of the covariate density and likelihood is an unnormalized kernel; dividing by its integral or sum makes its total probability one. (3) The likelihood factor cancels from numerator and denominator, leaving $f(x\mid z;\beta)$. (4) Missingness in $X$ must be ignorable, such as missing at random given the observed variables used in the model, with distinct missingness and data-model parameters. If missingness still depends on the unobserved value of $X$ after conditioning on observed information, the missingness mechanism may alter the distribution of $X$ among missing cases, and this target alone does not account for that alteration.

The SMC-FCS Conditional Density

What You'll Learn