Tensor-Normal Marginal and Conditional Laws

Why Marginal and Conditional Laws Matter

Tensor data are often only partly available: an analysis may retain selected levels of each mode, or observe complete slices for some levels while predicting the rest. For tensor-normal models, these operations can preserve the model’s separable covariance structure, but not every pattern of missing or observed entries does. The key is to identify which covariance-factor blocks are involved and exactly what has been conditioned on.

We use the tensor-normal convention that a random $K$-way tensor $X$, with dimensions $d_1\times\cdots\times d_K$, has mean tensor $M$ and covariance factors $\Sigma_1,\ldots,\Sigma_K$ when $\operatorname{vec}(X)$ is multivariate normal with covariance $\Sigma_K\otimes\cdots\otimes\Sigma_1$. The vectorization order has mode 1 varying fastest. Assume each $\Sigma_k$ is positive definite. Fixing a scale convention for the factors is useful for parameter interpretation, but the marginal and conditional calculations below use the specified factors directly.

Definition: Write $X\sim TN_{d_1,\ldots,d_K}(M;\Sigma_1,\ldots,\Sigma_K)$ if $\operatorname{vec}(X)\sim N(\operatorname{vec}(M),\Sigma_K\otimes\cdots\otimes\Sigma_1)$. For a matrix $A$ acting along mode 1, the mode product is defined by $(X\times_1 A)_{b,i_2,\ldots,i_K}=\sum_{a=1}^{d_1}A_{b,a}X_{a,i_2,\ldots,i_K}$.

Marginal Laws for Cartesian Subarrays

Choose index sets $S_k\subseteq\{1,\ldots,d_k\}$ for each mode, and retain the entries whose indices lie in the Cartesian product $S_1\times\cdots\times S_K$. Denote this subarray by $X_{S_1,\ldots,S_K}$, and let $\Sigma_k[S_k,S_k]$ be the principal submatrix of $\Sigma_k$ indexed by $S_k$. The mean is restricted to the same entries, and each covariance factor is restricted to its corresponding index set.

Theorem: If $X\sim TN_{d_1,\ldots,d_K}(M;\Sigma_1,\ldots,\Sigma_K)$, then $X_{S_1,\ldots,S_K}\sim TN_{|S_1|,\ldots,|S_K|}(M_{S_1,\ldots,S_K};\Sigma_1[S_1,S_1],\ldots,\Sigma_K[S_K,S_K])$.

Proof. The retained vector is a subvector of $\operatorname{vec}(X)$, so it is multivariate normal. For two retained entries indexed by $(i_1,\ldots,i_K)$ and $(j_1,\ldots,j_K)$, their covariance is $\prod_{k=1}^K(\Sigma_k)_{i_k,j_k}$, by the Kronecker covariance definition. Restricting each index to $S_k$ therefore gives exactly the covariance whose factors are the principal blocks $\Sigma_k[S_k,S_k]$. Its mean is the corresponding restriction of $M$. This establishes the stated tensor-normal law. $\square$

Worked Example: Marginalizing to Selected Levels

Let $X\sim TN_{3,2}(M;\Sigma_1,\Sigma_2)$, where $\Sigma_1=\begin{pmatrix}2&1&0\\1&3&1\\0&1&4\end{pmatrix}$ and $\Sigma_2=\begin{pmatrix}5&2\\2&6\end{pmatrix}$. Retain mode-1 indices $S_1=\{1,3\}$ and both mode-2 indices. The retained mean is $M_{S_1,\{1,2\}}$, and the mode-1 factor becomes $\Sigma_1[S_1,S_1]=\begin{pmatrix}2&0\\0&4\end{pmatrix}$. The mode-2 factor remains $\Sigma_2$. Thus $X_{S_1,\{1,2\}}\sim TN_{2,2}(M_{S_1,\{1,2\}};\begin{pmatrix}2&0\\0&4\end{pmatrix},\begin{pmatrix}5&2\\2&6\end{pmatrix})$. In particular, the covariance between the retained entries $X_{1,1}$ and $X_{3,2}$ is $0\times 2=0$, using the corresponding entries of the two restricted factors. The calculation uses principal blocks, not inverses or Schur complements: no conditioning has taken place.

Conditional Laws Given Complete Fibers

For a tensor-normal conditional law that remains separable, a useful conditioning pattern is to observe all entries along the other modes for selected levels of one mode. For example, split mode 1 into disjoint index sets $A$ and $B$, and observe the entire tensor $X_A$ formed by the mode-1 fibers indexed by $A$. The unknown part $X_B$ retains every level in modes 2 through $K$.

Use covariance-factor blocks $\Sigma_{1,AA}$, $\Sigma_{1,BA}$, and so on, with the first block index indicating the row indices. Define $G=\Sigma_{1,BA}\Sigma_{1,AA}^{-1}$ and $\Sigma_{1,B\mid A}=\Sigma_{1,BB}-\Sigma_{1,BA}\Sigma_{1,AA}^{-1}\Sigma_{1,AB}$. Positive definiteness of $\Sigma_1$ ensures that $\Sigma_{1,AA}$ is invertible and that the Schur complement $\Sigma_{1,B\mid A}$ is positive definite.

Theorem: For $X\sim TN_{d_1,\ldots,d_K}(M;\Sigma_1,\ldots,\Sigma_K)$, with $X_A=x_A$ observed across all other modes, $$ X_B\mid X_A=x_A \sim TN_{|B|,d_2,\ldots,d_K}\left( M_B+(x_A-M_A)\times_1 G;\ \Sigma_{1,B\mid A},\Sigma_2,\ldots,\Sigma_K \right). $$ Here $G$ has dimensions $|B|\times|A|$, and $M_A,M_B$ denote the corresponding mode-1 parts of $M$.

Proof. Put $q=\prod_{k=2}^K d_k$ and $\Sigma_{\mathrm{rest}}=\Sigma_K\otimes\cdots\otimes\Sigma_2$. Under the stated vectorization convention, the covariance blocks of $\operatorname{vec}(X_B)$ and $\operatorname{vec}(X_A)$ are $\Sigma_{\mathrm{rest}}\otimes\Sigma_{1,BB}$, $\Sigma_{\mathrm{rest}}\otimes\Sigma_{1,BA}$, and $\Sigma_{\mathrm{rest}}\otimes\Sigma_{1,AA}$, respectively. The conditional law of jointly normal vectors has regression matrix equal to the cross-covariance times the inverse conditioning covariance. Here that matrix is \[ (\Sigma_{\mathrm{rest}}\otimes\Sigma_{1,BA}) (\Sigma_{\mathrm{rest}}\otimes\Sigma_{1,AA})^{-1} =I_q\otimes G. \] Acting on the centered vector $\operatorname{vec}(x_A-M_A)$, $I_q\otimes G$ applies $G$ to the mode-1 fibers. Thus the conditional mean is $M_B+(x_A-M_A)\times_1G$. The conditional covariance is \[ \begin{aligned} &\Sigma_{\mathrm{rest}}\otimes\Sigma_{1,BB} -(\Sigma_{\mathrm{rest}}\otimes\Sigma_{1,BA}) (\Sigma_{\mathrm{rest}}\otimes\Sigma_{1,AA})^{-1} (\Sigma_{\mathrm{rest}}\otimes\Sigma_{1,AB})\\ &\qquad=\Sigma_{\mathrm{rest}}\otimes \left(\Sigma_{1,BB}-\Sigma_{1,BA}\Sigma_{1,AA}^{-1}\Sigma_{1,AB}\right). \end{aligned} \] This is precisely the covariance of the stated tensor-normal distribution. $\square$

Worked Example: Conditioning on One Complete Matrix Row

Let $X\sim TN_{2,2}(M;\Sigma_1,\Sigma_2)$, where $M=\begin{pmatrix}1&2\\0&1\end{pmatrix}$, $\Sigma_1=\begin{pmatrix}4&2\\2&3\end{pmatrix}$, and $\Sigma_2=\begin{pmatrix}1&0.2\\0.2&2\end{pmatrix}$. Observe the first row as $x_A=(3,0)$, so $A=\{1\}$, $B=\{2\}$. The mode-1 regression coefficient and conditional variance are $G=2/4=0.5$ and $\Sigma_{1,B\mid A}=3-(2)(1/4)(2)=2$.

The observed row minus its mean is $(3,0)-(1,2)=(2,-2)$. Applying $G$ gives $(1,-1)$, and adding the second mean row $(0,1)$ yields conditional mean $(1,0)$. Consequently, $X_B\mid X_A=(3,0)$ has tensor-normal dimensions $1\times2$, mean $(1,0)$, and covariance factors $2$ and $\begin{pmatrix}1&0.2\\0.2&2\end{pmatrix}$. For example, the conditional variance of its first entry is $2(1)=2$, and the covariance between its two entries is $2(0.2)=0.4$.

Conditioning Along More Than One Mode

The same result can be applied successively when the observations are complete slabs at each step. After conditioning on mode-1 levels $A$, the remaining tensor has mode-1 factor $\Sigma_{1,B\mid A}$ and its other mode factors are unchanged. One may then split a remaining mode, say mode 2, and condition on all entries in selected mode-2 fibers of that remaining tensor. At each step, use the relevant current covariance factor’s blocks, update the mean by the corresponding mode product, and replace that factor by its Schur complement. This describes nested slab conditioning; it does not imply that conditioning on an arbitrary collection of cells has a separable conditional law.

Worked Example: A Single Observed Cell Is a Different Conditioning Pattern

Consider a $2\times2$ matrix-normal tensor with zero mean and both covariance factors equal to $\Sigma=\begin{pmatrix}1&0.5\\0.5&1\end{pmatrix}$. Each entry has variance 1. Under the mode-1-fastest vectorization, the covariance between entries $(i,j)$ and $(i',j')$ is $\Sigma_{i,i'}\Sigma_{j,j'}$. Observe only $X_{1,1}$; the remaining three entries are $X_{2,1},X_{1,2},X_{2,2}$.

Before conditioning, their covariance matrix is $\begin{pmatrix}1&0.25&0.5\\0.25&1&0.5\\0.5&0.5&1\end{pmatrix}$. Their covariances with $X_{1,1}$ are $(0.5,0.5,0.25)^{\mathsf T}$, and $\operatorname{Var}(X_{1,1})=1$. The multivariate-normal conditional covariance therefore equals the original covariance minus the outer product of that vector with itself: $\begin{pmatrix}0.75&0&0.375\\0&0.75&0.375\\0.375&0.375&0.9375\end{pmatrix}$. The conditional mean is also changed: for observed value $X_{1,1}=x$, it is $(0.5x,0.5x,0.25x)^{\mathsf T}$. These are valid Gaussian conditional calculations, but the three remaining cells form an L-shaped set, not a Cartesian tensor subarray or a complete slab split. The slab theorem cannot be applied to claim an ordinary separable tensor-normal law for this pattern.

Common Mistake and Practical Interpretation

A frequent error is to confuse the regression matrix’s dimensions or to replace its mode product with a transpose. In the conditional theorem, $G=\Sigma_{1,BA}\Sigma_{1,AA}^{-1}$ maps $|A|$ observed mode-1 coordinates to $|B|$ unobserved coordinates. Thus $G$ is $|B|\times|A|$, and the update is $(x_A-M_A)\times_1G$. Using $G^{\mathsf T}$ instead has dimensions $|A|\times|B|$ and does not perform the required mapping. The block order in $\Sigma_{1,BA}$ matters for the same reason.

These laws are useful for prediction, handling incomplete tensor observations, and checking what structure a model implies after restricting or conditioning the data. A Cartesian marginal retains separability through principal covariance blocks. Conditioning on complete fibers retains it through a Schur complement in the split mode. Conditioning on an irregular set of entries generally requires a general multivariate-normal conditional calculation, and separability should not be assumed without verification.

Key takeaway: First identify the observation pattern. Use principal covariance blocks for a Cartesian marginal; use a cross-block regression matrix and Schur complement for a complete-slab conditional. The two operations are not interchangeable.

Check Your Understanding

Answer each question by identifying the relevant covariance blocks and the pattern of retained or observed entries.

For a Cartesian subarray, why are the marginal mode factors principal submatrices rather than Schur complements?
In the slab-conditioning theorem, what are the dimensions of $G$, and what does its mode product do to each observed fiber?
If $\Sigma_{1,AA}=\begin{pmatrix}4&0\\0&2\end{pmatrix}$, $\Sigma_{1,BA}=(2\ \ 1)$, and $\Sigma_{1,BB}=(5)$, calculate $G$ and $\Sigma_{1,B\mid A}$.
Which covariance factors change when complete fibers are conditioned on along mode 1, and which remain unchanged?
Why does observing a single cell of a matrix generally not meet the complete-slab condition in the conditional theorem?