\caption{\label{tab:dimensions}\small Summary listing of dimensions with the corresponding sample versions $\mat{X}_i, \mat{R}_i, \mat{r}_i, \mat{f}_{y_i}$ for $i =1, ..., n$ as well as estimates $\widehat{\mat{\alpha}}, \widehat{\mat{\beta}}, \widehat{\mat\Delta}, \widehat{\mat\Delta}_1$ and $\widehat{\mat\Delta}_2$.}
\end{minipage}
\end{table}
The log-likelihood $l$ given $n$ i.i.d. observations assuming that $\mat{X}_i\mid(Y = y_i)$ is normal distributed as
Note that the log-likelihood estimate $\hat{l}$ only depends on $\mat\alpha, \mat\beta$. Next, we compute the gradient for $\mat\alpha$ and $\mat\beta$ of $\hat{l}$ used to formulate a Gradient Descent base estimation algorithm for $\mat\alpha, \mat\beta$ as the previous algorithmic. The main reason is to enable an estimation for bigger dimensions of the $\mat\alpha, \mat\beta$ coefficients since the previous algorithm does \emph{not} solve the high run time problem for bigger dimensions.
Start with the general case of $\mat X_i|(Y_i = y_i)$ is multivariate normal distributed with the covariance $\mat\Delta$ being a $p q\times p q$ positive definite symmetric matrix \emph{without} an further assumptions. We have $i =1, ..., n$ observations following
Substitution of the MLE estimates into the log-likelihood $l(\mat\mu, \mat\Delta, \mat\alpha, \mat\beta)$ gives the estimated log-likelihood $\hat{l}(\mat\alpha, \mat\beta)$ as
We are interested in the gradients $\nabla_{\mat\alpha}\hat{l}(\mat\alpha, \mat\beta)$, $\nabla_{\mat\beta}\hat{l}(\mat\alpha, \mat\beta)$ of the estimated log-likelihood. Therefore, we consider the differential of $\hat{l}$.
&= \underbrace{-\frac{n}{2}\log|\widehat{\mat{\Delta}}| - \frac{1}{2}\sum_{i = 1}^n \t{\mat{r}_i}(\d\widehat{\mat{\Delta}}^{-1}) \mat{r}_i}_{=\,0\text{ due to }\widehat{\mat{\Delta}}\text{ beeing the MLE}}\label{eq:deriv1}
Now, substitution of $\d\mat{r}_i$ into \eqref{eq:deriv1} gives the gradients (not dimension standardized versions of $\D\hat{l}(\mat\alpha)$, $\D\hat{l}(\mat\beta)$) by identification of the derivatives from the differentials (see: \todo{appendix})
These quantities are very verbose as well as completely unusable for an implementation. By detailed analysis of the gradients we see that the main parts are only element permutations with a high sparsity. By defining the following compact matrix
\begin{equation}\label{eq:permTransResponse}
\mat G = \vec^{-1}_{q r}\bigg(\Big( \sum_{i = 1}^n \vec\mat{f}_{y_i}\otimes\widehat{\mat\Delta}^{-1}\mat{r}_i \Big)_{\pi(i)}\bigg)_{i = 1}^{p q k r}{\color{gray}\qquad(q r \times p k)}
\end{equation}
with $\pi$ being a permutation of $p q k r$ elements corresponding to permuting the axis of a 4D tensor of dimensions $p\times q\times k\times r$ by $(2, 4, 1, 3)$. As a generalization of transposition this leads to a rearrangement of the elements corresponding to the permuted 4D tensor with dimensions $q\times r\times p\times k$ which is then vectorized and reshaped into a matrix of dimensions $q r \times p k$. With $\mat G$ the gradients simplify to \todo{validate this mathematically}
Now we assume the residuals covariance has the form $\mat\Delta=\mat\Delta_1\otimes\mat\Delta_2$ where $\mat\Delta_1$, $\mat\Delta_2$ are $q\times q$, $p\times p$ covariance matrices, respectively. This is analog to the case that $\mat{R}_i$'s are i.i.d. Matrix Normal distribution
The density of the Matrix Normal (with mean zero) is equivalent to the vectorized quantities being multivariate normal distributed with Kronecker structured covariance
Out first order of business is to derive the MLE estimated of the covariance matrices $\mat\Delta_1$, $\mat\Delta_2$ (the mean estimate $\widehat{\mat\mu}$ is trivial). Therefore, we look at the differentials with respect to changes in the covariance matrices as
We first take a closer look at the sum. After a bit of algebra using $\d\mat A^{-1}=-\mat A^{-1}(\d\mat A)\mat A^{-1}$ and the definitions of $\widehat{\mat\Delta}_1$, $\widehat{\mat\Delta}_2$ the sum can be rewritten
\paragraph{Comparison to the general case:} There are two main differences, first the general case has a closed form solution for the gradient due to the explicit nature of the MLE estimate of $\widehat{\mat\Delta}$ compared to the mutually dependent MLE estimates $\widehat{\mat\Delta}_1$, $\widehat{\mat\Delta}_2$. On the other hand the general case has dramatically bigger dimensions of the covariance matrix ($p q \times p q$) compared to the two Kronecker components with dimensions $q \times q$ and $p \times p$. This means that in the general case there is a huge performance penalty in the dimensions of $\widehat{\mat\Delta}$ while in the other case an extra estimation is required to determine $\widehat{\mat\Delta}_1$, $\widehat{\mat\Delta}_2$.
which means that $\E\widetilde{\mat\Delta}_1=\mat\Delta_1\tr(\mat\Delta_2)$ and in analogy $\E\widetilde{\mat\Delta}_2=\mat\Delta_2\tr(\mat\Delta_1)$. Now, we need to handle the scaling which can be estimated unbiased by
because with $\|\mat{R}_i\|_F^2=\tr\mat{R}_i\t{\mat{R}_i}=\tr\t{\mat{R}_i}\mat{R}_i$ the scale estimate $\tilde{s}=\tr(\widetilde{\mat\Delta}_1)=\tr(\widetilde{\mat\Delta}_2)$. Then $\E\tilde{s}=\tr(\E\widetilde{\mat\Delta}_1)=\tr{\mat\Delta}_1\tr{\mat\Delta}_2=\tr({\mat\Delta}_1\otimes{\mat\Delta}_2)$. Leading to the estimate of the covariance as
\todo{ prove they are consistent, especially $\widetilde{\mat\Delta}=\tilde{s}^{-1}(\widetilde{\mat\Delta}_1\otimes\widetilde{\mat\Delta}_2)$!}
The hoped for a benefit is that these covariance estimates are in a closed form which means there is no need for an additional iterative estimations step. Before we start with the derivation of the gradients define the following two quantities
Now, the matrix normal with the covariance matrix of the vectorized quantities of the form $\mat{\Delta}= s^{-1}(\mat{\Delta}_1\otimes\mat{\Delta}_2)$ has the form
The second form is due to the property of the determinant for scaling and the Kronecker product giving that $|\widetilde{\mat\Delta}| =(\tilde{s}^{-1})^{p q}|\widetilde{\mat{\Delta}}_1|^p |\widetilde{\mat{\Delta}}_2|^q$ as well as an analog Kronecker decomposition as in the MLE case.
The last one is tedious but straight forward. Its computation extensively uses the symmetry of $\widetilde{\mat{\Delta}}_1$, $\widetilde{\mat{\Delta}}_2$, the cyclic property of the trace and the relation $\d\mat{A}^{-1}=-\mat{A}^{-1}(\d\mat{A})\mat{A}^{-1}$.
Putting it all together
\begin{align*}
\d\tilde{l}(\mat{\alpha}, \mat{\beta})
&= \frac{n p q}{2}\Big(-\frac{2}{n\tilde{s}}\Big)\sum_{i = 1}^n \tr(\t{\mat{f}_{y_i}}\t{\mat{\beta}}\mat{R}_i\d\mat{\alpha} + \mat{f}_{y_i}\t{\mat{\alpha}}\t{\mat{R}_i}\d\mat{\beta}) \\
Observe that the bracketed expressions before $\d\mat{\alpha}$ and $\d\mat{\beta}$ are transposed of each other. Lets denote the expression for $\d\mat{\alpha}$ as $\mat{G}_i$ which has the form
\begin{displaymath}
\mat{G}_i
= (\tr(\widetilde{\mat{\Delta}}_1^{-1}\mat{S}_1) - p q \tilde{s}^{-1})\mat{R}_i
The first example (which by it self is \emph{not} exemplary) is the estimation with parameters $n =200$, $p =11$, $q =5$, $k =14$ and $r =9$. The ``true'' matrices $\mat\alpha$, $\mat\beta$ generated by sampling there elements i.i.d. standard normal like the responses $y$. Then, for each observation, $\mat{f}_y$ is computed as $\sin(s_{i, j} y + o_{i j})$\todo{ properly describe} to fill the elements of $\mat{f}_y$. Then the $\mat{X}$'s are samples as
In this section we summarize a few useful matrix identities, for more details see for example \cite{MatrixAlgebra-AbadirMagnus2005}.
For two matrices $\mat A$ of dimensions $q\times r$ and $\mat B$ of dimensions $p\times k$ holds
\begin{equation}\label{eq:vecKron}
\vec(\mat A\kron\mat B) = (\mat{I}_r\kron\mat{K}_{k,q}\kron\mat{I}_p)(\vec\mat A\kron\vec\mat B).
\end{equation}
Let $\mat A$ be a $p\times p$ dimensional non-singular matrix. Furthermore, let $\mat a, \mat b$ be $p$ vectors such that $\t{\mat b}A^{-1}\mat a\neq-1$, then
\det(\mat A + \mat a\t{\mat b}) = \det(\mat A)(1 + \t{\mat b}{\mat A}^{-1}\mat a)
\end{displaymath}
which even holds in the case $\t{\mat b}A^{-1}\mat a =-1$. This is known as Sylvester's determinant theorem.
\section{Commutation Matrix and Permutation Identities}
\begin{center}
Note: In this section we use 0-indexing for the sake of simplicity!
\end{center}
In this section we summarize relations between the commutation matrix and corresponding permutation. We also list some extensions to ``simplify'' or represent some term. This is mostly intended for implementation purposes and understanding of terms occurring in the computations.
Let $\mat A$ be an arbitrary $p\times q$ matrix. The permutation matrix $\mat K_{p, q}$ satisfies
\begin{displaymath}
\mat{K}_{p, q}\vec{\mat{A}} = \vec{\t{\mat{A}}}\quad\Leftrightarrow\quad (\vec{\mat{A}})_{\pi_{p, q}(i)} = (\vec{\t{\mat{A}}})_{i}, \quad\text{for } i = 0, ..., p q - 1
\end{displaymath}
where $\pi_{p, q}$ is a permutation of the indices $i =0, ..., p q -1$ such that
\begin{displaymath}
\pi_{p, q}(i + j p) = j + i q, \quad\text{for }i = 0, ..., p - 1; j = 0, ..., q - 1.
\end{displaymath}
\begin{table}[!htp]
\centering
\begin{minipage}{0.8\textwidth}
\centering
\begin{tabular}{l c l}
$\mat{K}_{p, q}$&$\hat{=}$&$\pi_{p, q}(i + j p)= j + i q$\\
$\mat{I}_r\kron\mat{K}_{p, q}$&$\hat{=}$&$\tilde{\pi}_{p, q, r}(i + j p + k p q)= j + i q + k p q$\\
$\mat{K}_{p, q}\kron\mat{I}_r$&$\hat{=}$&$\hat{\pi}_{p, q, r}(i + j p + k p q)= r(j + i q)+ k$
\end{tabular}
\caption{\label{tab:commutation-permutation}Commutation matrix terms and corresponding permutations. Indices are all 0-indexed with the ranges; $i =0, ..., p -1$, $j =0, ..., q -1$ and $k =0, ..., r -1$.}
\end{minipage}
\end{table}
\section{Matrix and Tensor Operations}
The \emph{Kronecker product}\index{Operations!Kronecker@$\kron$ Kronecker product} is denoted as $\kron$ and the \emph{Hadamard product} uses the symbol $\circ$. We also need the \emph{Khatri-Rao product}\index{Operations!KhatriRao@$\hada$ Khatri-Rao product}
$\hada$ as well as the \emph{Transposed Khatri-Rao product}$\odot_t$ (or \emph{Face-Splitting product}). There is also the \emph{$n$-mode Tensor Matrix Product}\index{Operations!ttm@$\ttm[n]$$n$-mode tensor product} denoted by $\ttm[n]$ in conjunction with the \emph{$n$-mode Matricization} of a Tensor $\mat{T}$ written as $\mat{T}_{(n)}$, which is a matrix. See below for definitions and examples of these operations.\todo{ Definitions and Examples}
\todo{ resolve confusion between Khatri-Rao, Column-wise Kronecker / Khatri-Rao, Row-wise Kronecker / Khatri-Rao, Face-Splitting Product, .... Yes, its a mess.}
\paragraph{Kronecker Product $\kron$:}
\paragraph{Khatri-Rao Product $\hada$:}
\paragraph{Transposed Khatri-Rao Product $\odot_t$:} This is also known as the Face-Splitting Product and is the row-wise Kronecker product of two matrices. If relates to the Column-wise Kronecker Product through
\paragraph{$n$-mode unfolding:}\emph{Unfolding}, also known as \emph{flattening} or \emph{matricization}, is an reshaping of a tensor into a matrix with rearrangement of the elements such that mode $n$ corresponds to columns of the result matrix and all other modes are vectorized in the rows. Let $\ten{T}$ be a tensor of order $m$ with dimensions $t_1\times ... \times t_n\times ... \times t_m$ and elements indexed by $(i_1, ..., i_n, ..., i_m)$. The $n$-mode flattening, denoted $\ten{T}_{(n)}$, is defined as a $(t_n, \prod_{k\neq n}t_k)$ matrix with element indices $(i_n, j)$ such that $j =\sum_{k =1, k\neq n}^m i_k\prod_{l =1, l\neq n}^{k -1}t_l$.
\todo{ give an example!}
\paragraph{$n$-mode Tensor Product $\ttm[n]$:}
The \emph{$n$-mode tensor product}$\ttm[n]$ between a tensor $\mat{T}$ of order $m$ with dimensions $t_1\times t_2\times ... \times t_n\times ... \times t_m$ and a $p\times t_n$ matrix $\mat{M}$ is defined element-wise as
where $i_1, ..., i_{n-1}, i_{n+1}, ..., i_m$ run from $1$ to $t_1, ..., t_{n-1}, t_{n+1}, ..., t_m$, respectively. Furthermore, the $n$-th fiber index $j$ of the product ranges from $1$ to $p$. This gives a new tensor $\mat{T}\ttm[n]\mat{M}$ of order $m$ with dimensions $t_1\times t_2\times ... \times p\times ... \times t_m$.
\begin{example}[Matrix Multiplication Analogs]
Let $\mat{A}$, $\mat{B}$ be two matrices with dimensions $t_1\times t_2$ and $p\times q$, respectively. Then $\mat{A}$ is also a tensor of order $2$, now the $1$-mode and $2$-mode products are element wise given by
In other words, the $1$-mode product equals $\mat{A}\ttm[1]\mat{B}=\mat{B}\mat{A}$ and the $2$-mode is $\mat{A}\ttm[2]\mat{B}=\t{(\mat{B}\t{\mat{A}})}$ in the case of the tensor $\mat{A}$ being a matrix.
\end{example}
\begin{example}[Order Three Analogs]
Let $\mat{A}$ be a tensor of the form $t_1\times t_2\times t_3$ and $\mat{B}$ a matrix of dimensions $p\times q$, then the $n$-mode products have the following look
\begin{align*}
(\mat{A}\ttm[1]\mat{B})_{i,j,k}&= \sum_{l = 1}^{t_1}\mat{A}_{l,j,k}\mat{B}_{i,l}&\text{for }t_1 = q, \\
(\mat{A}\ttm[2]\mat{B})_{i,j,k}&= \sum_{l = 1}^{t_2}\mat{A}_{i,l,k}\mat{B}_{j,l}\equiv (\mat{B}\mat{A}_{i,:,:})_{j,k}&\text{for }t_2 = q, \\
or in other words, the $i$-th slice of the tensor product $\ten{F}\ttm[2]\mat{\beta}\ttm[3]\mat{\alpha}$ contains $\mat{\beta}\mat{f}_{y_i}\t{\mat{\alpha}}$ for $i =1, ..., n$.
In this section we give a short summary of alternative but equivalent operations.
Using the notation $\widehat{=}$ to indicate that two expressions are identical in the sense that they contain the same element in the same order but may have different dimensions. Meaning, when vectorizing ether side of $\widehat{=}$, they are equal ($\mat{A}\widehat{=}\mat{B}\ :\Leftrightarrow\ \vec{\mat{A}}=\vec{\mat{B}}$).
Therefore, we use $\mat{A}, \mat{B}, \mat{X}, \mat{F}, \mat{R}, ...$ for matrices. 3-Tensors are written as $\ten{A}, \ten{B}, \ten{T}, \ten{X}, \ten{F}, \ten{R}, ...$.
if and only if $\vec\mat{X}\sim\mathcal{N}_{p q}(\vec\mat\mu, \mat\Delta_1\otimes\mat\Delta_2)$. Note the order of the covariance matrices $\mat\Delta_1, \mat\Delta_2$. Its density is given by
\begin{displaymath}
f(\mat{X}) = \frac{1}{(2\pi)^{p q / 2}|\mat\Delta_1|^{p / 2}|\mat\Delta_2|^{q / 2}}\exp\left(-\frac{1}{2}\tr(\mat\Delta_1^{-1}\t{(\mat X - \mat\mu)}\mat\Delta_2^{-1}(\mat X - \mat\mu))\right).
This section contains short summaries of the main references with each sub-section concerning one paper.
\subsection{}
\subsection{Generalized Tensor Decomposition With Features on Multiple Modes}
The \cite{TensorDecomp-HuLeeWang2022} paper proposes a multi-linear conditional mean model for a constraint rank tensor decomposition. Let the responses $\ten{Y}\in\mathbb{R}^{d_1\times ... \times\d_K}$ be an order $K$ tensor. Associated with each mode $k\in[K]$ they assume feature matrices $\mat{X}_k\in\mathbb{R}^{d_k\times p_k}$. Now, they assume that conditional on the feature matrices $\mat{X}_k$ the entries of the tensor $\ten{Y}$ are independent realizations. The rank constraint is specified through $\mat{r}=(r_1, ..., r_K)$, then the model is given by
The order $K$ tensor $\ten{C}\in\mathbb{R}^{r_1\times...\times r_K}$ is an unknown full-rank core tensor and the matrices $\mat{M}_k\in\mathbb{R}^{p_k\times r_k}$ are unknown factor matrices. The function $f$ is applied element wise and serves as the link function based on the assumed distribution family of the tensor entries. Finally, the operation $\times$ denotes the tensor-by-matrix product using a short hand
with $\ttm[k]$ denoting the $k$-mode tensor matrix product.
The algorithm for estimation of $\ten{C}$ and $\mat{M}_1, ..., \mat{M}_K$ assumes the individual conditional entries of $\ten{Y}$ to be independent and to follow a generalized linear model with link function $f$. The proposed algorithm is an iterative algorithm for minimizing the negative log-likelihood
where $b = f'$ it the derivative of the canonical link function $f$ in the generalized linear model the conditioned entries of $\ten{Y}$ follow. The algorithm utilizes the higher-order SVD (HOSVD) to enforce the rank-constraint.
The main benefit is that this approach generalizes well to a multitude of different structured data sets.
\todo{ how does this relate to the $\mat{X}=\mat{\mu}+\mat{\beta}\mat{f}_y\t{\mat{\alpha}}+\mat{\epsilon}$ model.}
