Conditional Normal Distributions for Missing Visits

Why condition on observed visits?

In a clinical trial, a participant may have recorded outcomes at some visits and missing outcomes at others. A reference predictive distribution supplies a joint model for the visit outcomes; conditioning that model on the participant’s observed values gives a distribution for the missing values. This distribution accounts for both the model’s visit-to-visit covariance and the particular outcomes already observed. It is not generally correct to impute each missing visit from its marginal distribution alone.

The calculation in this tutorial is a property of a multivariate-normal model. It does not, by itself, establish which joint model or reference distribution is appropriate. The joint mean and covariance must be specified under the modeling assumptions being used. Given those quantities, conditioning is a matrix calculation.

Definition: Let $Y=(Y_o^\mathsf{T},Y_m^\mathsf{T})^\mathsf{T}$ be a jointly multivariate-normal vector of visit outcomes, partitioned into observed components $Y_o$ and missing components $Y_m$. Write its mean and covariance as $$ \begin{pmatrix} \mu_o\\ \mu_m \end{pmatrix}, \qquad \begin{pmatrix} \Sigma_{oo} & \Sigma_{om}\\ \Sigma_{mo} & \Sigma_{mm} \end{pmatrix}, $$ where $\Sigma_{mo}=\Sigma_{om}^\mathsf{T}$. If $\Sigma_{oo}$ is nonsingular, then $$ Y_m\mid Y_o=y_o \sim N\left( \mu_m+\Sigma_{mo}\Sigma_{oo}^{-1}(y_o-\mu_o),\ \Sigma_{mm}-\Sigma_{mo}\Sigma_{oo}^{-1}\Sigma_{om} \right). $$

The first parameter in this conditional normal distribution is its conditional mean vector; the second is its conditional covariance matrix. The observed residual $y_o-\mu_o$ shifts the mean from $\mu_m$. The covariance adjustment accounts for the information about the missing outcomes carried by the observed outcomes.

For the formula to apply as stated, the observed and missing components must be parts of a jointly multivariate-normal vector with the displayed mean and covariance, and $\Sigma_{oo}$ must be invertible. In the usual nondegenerate case, the full covariance matrix is positive definite, which also ensures that the observed covariance block is positive definite. If the model is singular, the ordinary inverse formula requires separate treatment. These distributional assumptions concern the outcome model; they do not say that missingness is ignorable or determine the choice of a reference distribution.

Why the conditional formula holds

A short derivation makes the two parts of the formula easier to interpret. Define the observed and missing residual vectors $U_o=Y_o-\mu_o$ and $U_m=Y_m-\mu_m$, and set $B=\Sigma_{mo}\Sigma_{oo}^{-1}$. Consider the residual left after predicting $U_m$ linearly from $U_o$:

$$ R=U_m-BU_o. $$

Its covariance with $U_o$ is zero:

\operatorname{Cov}(R,U_o) =\Sigma_{mo}-B\Sigma_{oo} =\Sigma_{mo}-\Sigma_{mo}\Sigma_{oo}^{-1}\Sigma_{oo} =0.

The pair $(R,U_o)$ is jointly normal because it is a linear transformation of a jointly normal vector. For jointly normal vectors, zero cross-covariance implies independence. Thus $R$ is independent of $U_o$; knowing the observed outcomes does not change the distribution of this residual. Its covariance is

\begin{aligned} \operatorname{Var}(R) &=\Sigma_{mm}-B\Sigma_{om}-\Sigma_{mo}B^\mathsf{T} +B\Sigma_{oo}B^\mathsf{T}\\ &=\Sigma_{mm}-\Sigma_{mo}\Sigma_{oo}^{-1}\Sigma_{om}. \end{aligned}

In the last equality, $\Sigma_{oo}$ and its inverse are symmetric, so the last three terms combine to subtract one copy of $\Sigma_{mo}\Sigma_{oo}^{-1}\Sigma_{om}$. Since $Y_m=\mu_m+BU_o+R$, conditioning on $Y_o=y_o$ fixes $U_o=y_o-\mu_o$ and leaves the independent residual unchanged. This gives both the stated conditional mean and conditional covariance.

Worked calculations

Worked Example: One Missing Visit Given Two Observed Visits

Suppose a hypothetical three-visit outcome vector, measured in points, has the joint model $Y\sim N(\mu,\Sigma)$, with

\mu= \begin{pmatrix}10\\8\\6\end{pmatrix}, \qquad \Sigma= \begin{pmatrix} 4&2&1\\ 2&9&3\\ 1&3&16 \end{pmatrix}.

Visits 1 and 2 are observed as $y_o=(12,5)^\mathsf{T}$; visit 3 is missing. First check that this is a valid nondegenerate covariance matrix. Its leading principal minors are $4$, $4(9)-2(2)=32$, and

\det(\Sigma) =4(9\cdot16-3\cdot3)-2(2\cdot16-3\cdot1)+(2\cdot3-9\cdot1) =540-58-3=479.

All three are positive, so $\Sigma$ is positive definite. The observed covariance is $\Sigma_{oo}=\begin{pmatrix}4&2\\2&9\end{pmatrix}$, with determinant $32>0$, and its inverse is

\Sigma_{oo}^{-1} =\frac{1}{32} \begin{pmatrix}9&-2\\-2&4\end{pmatrix}.

Here $\Sigma_{mo}=(1,3)$, $\Sigma_{om}=(1,3)^\mathsf{T}$, and the observed residual is $y_o-\mu_o=(2,-3)^\mathsf{T}$. The conditional mean is

\begin{aligned} E(Y_3\mid Y_1=12,Y_2=5) &=6+(1,3)\frac{1}{32} \begin{pmatrix}9&-2\\-2&4\end{pmatrix} \begin{pmatrix}2\\-3\end{pmatrix}\\ &=6+(1,3)\begin{pmatrix}24/32\\-16/32\end{pmatrix}\\ &=6-\frac{24}{32}=5.25. \end{aligned}

The conditional variance is

\begin{aligned} \operatorname{Var}(Y_3\mid Y_1,Y_2) &=16-(1,3)\frac{1}{32} \begin{pmatrix}9&-2\\-2&4\end{pmatrix} \begin{pmatrix}1\\3\end{pmatrix}\\ &=16-(1,3)\begin{pmatrix}3/32\\10/32\end{pmatrix}\\ &=16-\frac{33}{32} =\frac{479}{32}=14.96875. \end{aligned}

Therefore $Y_3\mid(Y_1=12,Y_2=5)\sim N(5.25,14.96875)$, where the variance is in squared points. The standard deviation is $\sqrt{14.96875}\approx3.869$ points. Conditioning shifts the mean below its marginal value of 6 because the observed residuals and the covariances imply a net negative adjustment. It also reduces uncertainty relative to the marginal variance of 16.

Worked Example: Two Missing Visits Given One Observed Visit

Use the same hypothetical three-visit model, but suppose only visit 2 is observed, at 11 points. Visits 1 and 3 are missing. The observed residual is $11-8=3$, and $\Sigma_{oo}=(9)$. In the order $(Y_1,Y_3)$, the missing mean and covariance are

\mu_m= \begin{pmatrix}10\\6\end{pmatrix}, \qquad \Sigma_{mm}= \begin{pmatrix}4&1\\1&16\end{pmatrix}, \qquad \Sigma_{mo}= \begin{pmatrix}2\\3\end{pmatrix}.

The conditional mean vector is

\begin{aligned} E\left[ \begin{pmatrix}Y_1\\Y_3\end{pmatrix} \middle|Y_2=11\right] &= \begin{pmatrix}10\\6\end{pmatrix} +\begin{pmatrix}2\\3\end{pmatrix}\frac{1}{9}(3)\\ &= \begin{pmatrix}10+2/3\\7\end{pmatrix} = \begin{pmatrix}10.667\\7\end{pmatrix}. \end{aligned}

The conditional covariance is

\begin{aligned} \operatorname{Var}\left[ \begin{pmatrix}Y_1\\Y_3\end{pmatrix} \middle|Y_2\right] &= \begin{pmatrix}4&1\\1&16\end{pmatrix} -\begin{pmatrix}2\\3\end{pmatrix}\frac{1}{9} \begin{pmatrix}2&3\end{pmatrix}\\ &= \begin{pmatrix}4&1\\1&16\end{pmatrix} -\begin{pmatrix}4/9&6/9\\6/9&9/9\end{pmatrix}\\ &= \begin{pmatrix}32/9&1/3\\1/3&15\end{pmatrix}. \end{aligned}

This covariance remains positive definite: its determinant is $(32/9)(15)-(1/3)^2=479/9>0$, and its first diagonal entry is positive. The off-diagonal conditional covariance is $1/3$ points squared, so the two missing outcomes are still associated after conditioning on visit 2. A conditional distribution for multiple missing visits must preserve their joint uncertainty; separate marginal draws would not do so.

Worked Example: A Two-Visit Calculation in Measurement Units

Suppose, hypothetically, that systolic blood pressure at two visits has joint mean $(50,40)^\mathsf{T}$ mmHg and covariance $\begin{pmatrix}25&10\\10&16\end{pmatrix}$ in squared mmHg. Visit 1 is observed at 60 mmHg and visit 2 is missing. The model is nondegenerate because the covariance determinant is $25(16)-10^2=300>0$, and the observed variance is 25, which is positive. Thus the conditioning formula applies.

The conditional mean and variance for the missing second visit are

\begin{aligned} E(Y_2\mid Y_1=60) &=40+\frac{10}{25}(60-50) =44\text{ mmHg},\\ \operatorname{Var}(Y_2\mid Y_1) &=16-\frac{10^2}{25} =12\text{ mmHg}^2. \end{aligned}

Hence $Y_2\mid Y_1=60\sim N(44,12)$. The conditional standard deviation is $\sqrt{12}\approx3.464$ mmHg. The observed first-visit value is 10 mmHg above its modeled mean, and the positive covariance raises the conditional mean of the second visit by $0.4(10)=4$ mmHg.

Checks, interpretation, and alternatives

Several checks help catch errors before a conditional distribution is used for imputation:

Check the partition and dimensions. If there are $q$ missing and $r$ observed outcomes, then $\Sigma_{mo}$ is $q\times r$, $\Sigma_{oo}^{-1}$ is $r\times r$, and the conditional covariance is $q\times q$. The conditional mean must have $q$ entries.
Check the observed block. The inverse exists only when $\Sigma_{oo}$ is nonsingular. A nearly singular block can also make numerical results unstable, even if software returns an answer.
Check the conditional covariance. It must be symmetric and positive semidefinite; in a nondegenerate model it is positive definite. A negative conditional variance or a materially negative eigenvalue usually signals a coding, partition, or covariance-model problem.
Check limiting behavior and scale. The covariance adjustment is subtracted, so conditioning cannot increase the covariance in the positive-semidefinite sense. The mean adjustment depends on the observed residuals and cross-covariances; it need not be positive. Variances have squared outcome units, while means have outcome units.

A common mistake is to use only the marginal distribution of each missing visit—for example, imputing the third visit from $N(6,16)$ in the first example. That ignores the observed outcomes and their covariance with the missing visit. The correction is to use the full conditional formula, including both the residual-based mean adjustment and the covariance reduction. A second mistake is to omit the off-diagonal entries of a conditional covariance when several visits are missing; doing so changes the joint distribution even if the individual conditional means and variances look right.

An equivalent computational route is to factor the joint normal density and complete the square in the missing components. Another is to use the residual argument derived above. Both yield the same conditional normal distribution. Directly simulating a complete vector from the joint model and retaining draws consistent with fixed continuous observed values is generally not a practical substitute; conditioning analytically or using an equivalent conditional simulation method is more direct. A mean-only imputation retains the conditional mean but discards the conditional uncertainty and covariance, so it is not a substitute when a distribution of missing outcomes is needed.

The calculation has important limits. It is correct only relative to the specified joint normal model and its parameters; it does not validate normality, covariance structure, or the reference model. Observed data can inform aspects of the outcome distribution, but they cannot by themselves verify assumptions about outcomes after dropout. Diagnostics such as residual and covariance checks can reveal some incompatibilities, but cannot prove the unobserved-outcome assumptions. The conditional formula is therefore a building block for a model-based imputation, not a stand-alone justification for the assumptions behind that imputation.

Key takeaway: Given a valid joint normal model, the missing visits conditional on observed visits have mean $\mu_m+\Sigma_{mo}\Sigma_{oo}^{-1}(y_o-\mu_o)$ and covariance $\Sigma_{mm}-\Sigma_{mo}\Sigma_{oo}^{-1}\Sigma_{om}$. Preserve the full covariance structure, verify the matrix calculations, and interpret the result as conditional on the chosen model.

Check Your Understanding

Answer each question, then compare your work with the solutions below.

For a jointly normal vector with $\mu_o=5$, $\mu_m=8$, $\Sigma_{oo}=4$, and $\Sigma_{mo}=2$, what is the conditional mean of the missing outcome when the observed value is 7?
In the same setup, suppose $\Sigma_{mm}=9$. What is the conditional variance? State its units if both outcomes are measured in kilograms.
Why must a two-outcome missing vector retain an off-diagonal entry in its conditional covariance when that entry is nonzero?
What does a negative conditional variance indicate, and what should be checked?

Solutions. (1) The observed residual is $7-5=2$, so the conditional mean is $8+(2/4)(2)=9$, in the outcome’s units. (2) The conditional variance is $9-(2/4)(2)=8$ kilograms squared. (3) The off-diagonal entry describes the conditional covariance between the missing outcomes; omitting it gives a different joint distribution and changes their dependence. (4) A valid conditional covariance cannot have a negative variance. Check the covariance partition, matrix arithmetic, positive definiteness of the specified covariance, and numerical stability of the inversion.

Conditional Normal Distributions for Missing Visits

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