631 lines
44 KiB
TeX
631 lines
44 KiB
TeX
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\documentclass[a4paper, 10pt]{article}
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\usepackage[utf8]{inputenc}
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\usepackage[T1]{fontenc}
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\usepackage{fullpage}
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\usepackage{amsmath, amssymb, amstext, amsthm}
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\usepackage{bm} % \boldsymbol and italic corrections, ...
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\usepackage[pdftex]{hyperref}
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\usepackage{makeidx} % Index (Symbols, Names, ...)
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\usepackage[
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% backend=bibtex,
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style=authoryear-comp
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]{biblatex}
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% Document meta into
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\title{Derivation of Gradient Descent Algorithm for K-PIR}
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\author{Daniel Kapla}
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\date{November 24, 2021}
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% Set PDF title, author and creator.
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\AtBeginDocument{
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\hypersetup{
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pdftitle = {Derivation of Gradient Descent Algorithm for K-PIR},
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pdfauthor = {Daniel Kapla},
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pdfcreator = {\pdftexbanner}
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}
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}
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\makeindex
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% Bibliography resource(s)
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\addbibresource{main.bib}
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% Setup environments
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% Theorem, Lemma
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\theoremstyle{plain}
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\newtheorem{theorem}{Theorem}
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\newtheorem{lemma}{Lemma}
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\newtheorem{example}{Example}
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% Definition
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\theoremstyle{definition}
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\newtheorem{defn}{Definition}
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% Remark
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\theoremstyle{remark}
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\newtheorem{remark}{Remark}
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% Define math macros
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\newcommand{\mat}[1]{\boldsymbol{#1}}
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\newcommand{\ten}[1]{\mathcal{#1}}
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\renewcommand{\vec}{\operatorname{vec}}
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\DeclareMathOperator{\kron}{\otimes} % Kronecker Product
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\DeclareMathOperator{\hada}{\odot} % Hadamard Product
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\newcommand{\ttm}[1][n]{\times_{#1}} % n-mode product (Tensor Times Matrix)
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\DeclareMathOperator{\df}{\operatorname{df}}
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\DeclareMathOperator{\tr}{\operatorname{tr}}
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\DeclareMathOperator{\var}{Var}
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\DeclareMathOperator{\cov}{Cov}
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\DeclareMathOperator{\E}{\operatorname{\mathbb{E}}}
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% \DeclareMathOperator{\independent}{{\bot\!\!\!\bot}}
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\DeclareMathOperator*{\argmin}{{arg\,min}}
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\DeclareMathOperator*{\argmax}{{arg\,max}}
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\newcommand{\D}{\textnormal{D}}
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\renewcommand{\d}{\textnormal{d}}
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\renewcommand{\t}[1]{{{#1}'}}
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\newcommand{\pinv}[1]{{{#1}^{\dagger}}} % `Moore-Penrose pseudoinverse`
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\newcommand{\todo}[1]{{\color{red}TODO: #1}}
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\begin{document}
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\maketitle
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%% Introduction %%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\section{Introduction}
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We assume the model
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\begin{displaymath}
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\mat{X} = \mat{\mu} + \mat{\beta}\mat{f}_y \t{\mat{\alpha}} + \mat{\epsilon}
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\end{displaymath}
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where the dimensions of all the components are listed in Table~\ref{tab:dimensions}.
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and its vectorized form
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\begin{displaymath}
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\vec\mat{X} = \vec\mat{\mu} + (\mat{\alpha}\kron\mat{\beta})\vec\mat{f}_y + \vec\mat{\epsilon}
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\end{displaymath}
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\begin{table}[!htp]
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\centering
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\begin{minipage}{0.8\textwidth}
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\centering
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\begin{tabular}{l l}
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$\mat X, \mat\mu, \mat R, \mat\epsilon$ & $p\times q$ \\
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$\mat{f}_y$ & $k\times r$ \\
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$\mat\alpha$ & $q\times r$ \\
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$\mat\beta$ & $p\times k$ \\
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$\mat\Delta$ & $p q\times p q$ \\
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$\mat\Delta_1$ & $q\times q$ \\
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$\mat\Delta_2$ & $p\times p$ \\
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$\mat{r}$ & $p q\times 1$ \\
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\hline
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$\ten{X}, \ten{R}$ & $n\times p\times q$ \\
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$\ten{F}$ & $n\times k\times r$ \\
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\end{tabular}
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\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 $\widehat{\mat\Delta}_2$.}
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\end{minipage}
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\end{table}
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The log-likelihood $l$ given $n$ i.i.d. observations assuming that $\mat{X}_i\mid(Y = y_i)$ is normal distributed as
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\begin{displaymath}
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\vec\mat{X}_i \sim \mathcal{N}_{p q}(\vec\mat\mu + (\mat\alpha\kron\mat\beta)\vec\mat{f}_{y_i}, \Delta)
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\end{displaymath}
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Replacing all unknown by there estimates gives the (estimated) log-likelihood
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\begin{equation}\label{eq:log-likelihood-est}
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\hat{l}(\mat\alpha, \mat\beta) = -\frac{n q p}{2}\log 2\pi - \frac{n}{2}\log|\widehat{\mat\Delta}| - \frac{1}{2}\sum_{i = 1}^n \t{\mat{r}_i}\widehat{\mat\Delta}^{-1}\mat{r}_i
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\end{equation}
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where the residuals are
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\begin{displaymath}
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\mat{r}_i = \vec\mat{X}_i - \vec\overline{\mat{X}} - (\mat\alpha\kron\mat\beta)\vec{\mat f}_{y_i}\qquad (p q \times 1)
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\end{displaymath}
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and the MLE estimate assuming $\mat\alpha, \mat\beta$ known for the covariance matrix $\widehat{\mat\Delta}$ as solution to the score equations is
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\begin{equation}\label{eq:Delta}
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\widehat{\mat\Delta} = \frac{1}{n}\sum_{i = 1}^n \mat{r}_i \t{\mat{r}_i} \qquad(p q \times p q).
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\end{equation}
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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.
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%% Derivative %%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\section{Derivative of the Log-Likelihood}
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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
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\begin{displaymath}
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\mat{r}_i = \vec(\mat X_i - \mat\mu - \mat\beta\mat{f}_{y_i}\t{\mat\alpha}) \sim \mathcal{N}_{p q}(\mat 0, \mat\Delta).
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\end{displaymath}
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The MLE estimates of $\mat\mu, \mat\Delta$ are
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\begin{displaymath}
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\widehat{\mat\mu} = \overline{\mat X} = \frac{1}{n}\sum_{i = 1}^n \mat X_i {\color{gray}\qquad(p\times q)},
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\qquad \widehat{\mat\Delta} = \frac{1}{n}\sum_{i = 1}^n \mat r_i\t{\mat r_i} {\color{gray}\qquad(p q\times p q)}.
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\end{displaymath}
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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
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\begin{displaymath}
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\hat{l}(\mat\alpha, \mat\beta) = -\frac{n q p}{2}\log 2\pi - \frac{n}{2}\log|\widehat{\mat\Delta}| - \frac{1}{2}\sum_{i = 1}^n \t{\mat{r}_i}\widehat{\mat\Delta}^{-1}\mat{r}_i.
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\end{displaymath}
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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 $\hat{l}$.
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\begin{align}
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\d\hat{l}(\mat\alpha, \mat\beta)
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&= -\frac{n}{2}\log|\widehat{\mat{\Delta}}| - \frac{1}{2}\sum_{i = 1}^n \big(\t{(\d \mat{r}_i)}\widehat{\mat{\Delta}}^{-1} \mat{r}_i + \t{\mat{r}_i}(\d\widehat{\mat{\Delta}}^{-1}) \mat{r}_i + \t{\mat{r}_i}\widehat{\mat{\Delta}}^{-1} \d \mat{r}_i\big) \nonumber\\
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&= \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}
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- \sum_{i = 1}^n \t{\mat{r}_i}\widehat{\mat{\Delta}}^{-1} \d \mat{r}_i.
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\end{align}
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The next step is to compute $\d \mat{r}_i$ which depends on both $\mat\alpha$ and $\mat\beta$
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\begin{align*}
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\d\mat{r}_i(\mat\alpha, \mat\beta)
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&= -\d(\mat\alpha\kron \mat\beta)\vec\mat{f}_{y_i} \\
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&= -\vec\!\big( \mat{I}_{p q}\,\d(\mat\alpha\kron \mat\beta)\vec\mat{f}_{y_i} \big) \\
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&= -(\t{\vec(\mat{f}_{y_i})}\kron \mat{I}_{p q})\,\d\vec(\mat\alpha\kron \mat\beta) \\
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\intertext{using the identity \ref{eq:vecKron}, to obtain vectorized differentials, gives}
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\dots
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&= -(\t{\vec(\mat{f}_{y_i})}\kron \mat{I}_{p q})(\mat{I}_r\kron\mat{K}_{k,q}\kron\mat{I}_p) \,\d(\vec \mat\alpha\kron \vec \mat\beta) \\
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&= -(\t{\vec(\mat{f}_{y_i})}\kron \mat{I}_{p q})(\mat{I}_r\kron\mat{K}_{k,q}\kron\mat{I}_p) \big((\d\vec \mat\alpha)\kron \vec \mat\beta + \vec \mat\alpha\kron (\d\vec \mat\beta)\big) \\
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&= -(\t{\vec(\mat{f}_{y_i})}\kron \mat{I}_{p q})(\mat{I}_r\kron\mat{K}_{k,q}\kron\mat{I}_p) \big(\mat{I}_{r q}(\d\vec \mat\alpha)\kron (\vec \mat\beta)\mat{I}_1 + (\vec \mat\alpha)\mat{I}_1\kron \mat{I}_{k p}(\d\vec \mat\beta)\big) \\
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&= -(\t{\vec(\mat{f}_{y_i})}\kron \mat{I}_{p q})(\mat{I}_r\kron\mat{K}_{k,q}\kron\mat{I}_p) \big((\mat{I}_{r q}\kron\vec \mat\beta)\d\vec \mat\alpha + (\vec \mat\alpha\kron \mat{I}_{k p})\d\vec \mat\beta\big)
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\end{align*}
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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})
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\begin{align*}
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\nabla_{\mat\alpha}\hat{l}(\mat\alpha, \mat\beta) &=
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\sum_{i = 1}^n (\t{\vec(\mat{f}_{y_i})}\kron\t{\mat{r}_i}\widehat{\mat\Delta}^{-1}) (\mat{I}_r\kron\mat{K}_{k,q}\kron\mat{I}_p) (\mat{I}_{r q}\kron\vec \mat\beta),
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{\color{gray}\qquad(q\times r)} \\
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\nabla_{\mat\beta}\hat{l}(\mat\alpha, \mat\beta) &=
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\sum_{i = 1}^n (\t{\vec(\mat{f}_{y_i})}\kron\t{\mat{r}_i}\widehat{\mat\Delta}^{-1}) (\mat{I}_r\kron\mat{K}_{k,q}\kron\mat{I}_p) (\vec \mat\alpha\kron \mat{I}_{k p}).
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{\color{gray}\qquad(p\times k)}
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\end{align*}
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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
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\begin{equation}\label{eq:permTransResponse}
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\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)}
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\end{equation}
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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}
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\begin{align*}
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\nabla_{\mat\alpha}\hat{l}(\mat\alpha, \mat\beta) &=
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\vec_{q}^{-1}(\mat{G}\vec{\mat\beta}),
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{\color{gray}\qquad(q\times r)} \\
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\nabla_{\mat\beta}\hat{l}(\mat\alpha, \mat\beta) &=
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\vec_{p}^{-1}(\t{\mat{G}}\vec{\mat\alpha}).
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{\color{gray}\qquad(p\times k)}
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\end{align*}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%% Kronecker Covariance Structure %%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\section{Kronecker Covariance Structure}
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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
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\begin{displaymath}
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\mat{R}_i = \mat{X}_i - \mat\mu - \mat\beta\mat{f}_{y_i}\t{\mat\alpha} \sim \mathcal{MN}_{p\times q}(\mat 0, \mat\Delta_2, \mat\Delta_1).
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\end{displaymath}
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The density of the Matrix Normal (with mean zero) is equivalent to the vectorized quantities being multivariate normal distributed with Kronecker structured covariance
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\begin{align*}
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f(\mat R)
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&= \frac{1}{\sqrt{(2\pi)^{p q}|\mat\Delta|}}\exp\left(-\frac{1}{2}\t{\vec(\mat{R})} \mat\Delta^{-1}\vec(\mat{R})\right) \\
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&= \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{R}}\mat\Delta_2^{-1}\mat{R})\right)
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\end{align*}
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which leads for given data to the log-likelihood
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\begin{displaymath}
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l(\mat{\mu}, \mat\Delta_1, \mat\Delta_2) =
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-\frac{n p q}{2}\log 2\pi
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-\frac{n p}{2}\log|\mat{\Delta}_1|
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-\frac{n q}{2}\log|\mat{\Delta}_2|
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-\frac{1}{2}\sum_{i = 1}^n \tr(\mat\Delta_1^{-1}\t{\mat{R}_i}\mat\Delta_2^{-1}\mat{R}_i).
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\end{displaymath}
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\subsection{MLE covariance estimates}
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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
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\begin{align*}
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\d l(\mat\Delta_1, \mat\Delta_2) &=
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-\frac{n p}{2}\d\log|\mat{\Delta}_1|
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-\frac{n q}{2}\d\log|\mat{\Delta}_2|
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-\frac{1}{2}\sum_{i = 1}^n
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\tr( (\d\mat\Delta_1^{-1})\t{\mat{R}_i}\mat\Delta_2^{-1}\mat{R}_i
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+ \mat\Delta_1^{-1}\t{\mat{R}_i}(\d\mat\Delta_2^{-1})\mat{R}_i) \\
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&=
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-\frac{n p}{2}\tr\mat{\Delta}_1^{-1}\d\mat{\Delta}_1
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-\frac{n q}{2}\tr\mat{\Delta}_2^{-1}\d\mat{\Delta}_2 \\
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&\qquad\qquad
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+\frac{1}{2}\sum_{i = 1}^n
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\tr( \mat\Delta_1^{-1}(\d\mat\Delta_1)\mat\Delta_1^{-1}\t{\mat{R}_i}\mat\Delta_2^{-1}\mat{R}_i
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+ \mat\Delta_1^{-1}\t{\mat{R}_i}\mat\Delta_2^{-1}(\d\mat\Delta_2)\mat\Delta_2^{-1}\mat{R}_i) \\
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&= \frac{1}{2}\tr\!\Big(\Big(
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-n p \mat{I}_q + \mat\Delta_1^{-1}\sum_{i = 1}^n \t{\mat{R}_i}\mat\Delta_2^{-1}\mat{R}_i
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\Big)\mat{\Delta}_1^{-1}\d\mat{\Delta}_1\Big) \\
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&\qquad\qquad
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+ \frac{1}{2}\tr\!\Big(\Big(
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-n q \mat{I}_p + \mat\Delta_2^{-1}\sum_{i = 1}^n \mat{R}_i\mat\Delta_1^{-1}\t{\mat{R}_i}
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\Big)\mat{\Delta}_2^{-1}\d\mat{\Delta}_2\Big) \overset{!}{=} 0.
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\end{align*}
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Setting $\d l$ to zero yields the MLE estimates as
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\begin{displaymath}
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\widehat{\mat{\mu}} = \overline{\mat X}{\color{gray}\quad(p\times q)}, \qquad
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\widehat{\mat\Delta}_1 = \frac{1}{n p}\sum_{i = 1}^n \t{\mat{R}_i}\widehat{\mat\Delta}_2^{-1}\mat{R}_i{\color{gray}\quad(q\times q)}, \qquad
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\widehat{\mat\Delta}_2 = \frac{1}{n q}\sum_{i = 1}^n \mat{R}_i\widehat{\mat\Delta}_1^{-1}\t{\mat{R}_i}{\color{gray}\quad(p\times p)}.
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\end{displaymath}
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Next, analog to above, we take the estimated log-likelihood and derive gradients with respect to $\mat{\alpha}$, $\mat{\beta}$.
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The estimated log-likelihood derives by replacing the unknown covariance matrices by there MLE estimates leading to
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\begin{displaymath}
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\hat{l}(\mat\alpha, \mat\beta) =
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-\frac{n p q}{2}\log 2\pi
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-\frac{n p}{2}\log|\widehat{\mat{\Delta}}_1|
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-\frac{n q}{2}\log|\widehat{\mat{\Delta}}_2|
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-\frac{1}{2}\sum_{i = 1}^n \tr(\widehat{\mat{\Delta}}_1^{-1}\t{\mat{R}_i}\widehat{\mat{\Delta}}_2^{-1}\mat{R}_i)
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\end{displaymath}
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and its differential
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\begin{displaymath}
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\d\hat{l}(\mat\alpha, \mat\beta) =
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-\frac{n p}{2}\d\log|\widehat{\mat{\Delta}}_1|
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-\frac{n q}{2}\d\log|\widehat{\mat{\Delta}}_2|
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-\frac{1}{2}\sum_{i = 1}^n \d\tr(\widehat{\mat{\Delta}}_1^{-1}\t{\mat{R}_i}\widehat{\mat{\Delta}}_2^{-1}\mat{R}_i).
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\end{displaymath}
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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
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\begin{displaymath}
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\frac{1}{2}\sum_{i = 1}^n \d\tr(\widehat{\mat{\Delta}}_1^{-1}\t{\mat{R}_i}\widehat{\mat{\Delta}}_2^{-1}\mat{R}_i)
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= \sum_{i = 1}^n \tr(\widehat{\mat{\Delta}}_1^{-1}\t{\mat{R}_i}\widehat{\mat{\Delta}}_2^{-1}\d\mat{R}_i)
|
||
|
- \frac{np}{2}\d\log|\widehat{\mat\Delta}_1|
|
||
|
- \frac{nq}{2}\d\log|\widehat{\mat\Delta}_2|.
|
||
|
\end{displaymath}
|
||
|
This means that most of the derivative cancels out and we get
|
||
|
\begin{align*}
|
||
|
\d\hat{l}(\mat\alpha, \mat\beta)
|
||
|
&= \sum_{i = 1}^n \tr(\widehat{\mat{\Delta}}_1^{-1}\t{\mat{R}_i}\widehat{\mat{\Delta}}_2^{-1}\d\mat{R}_i) \\
|
||
|
&= \sum_{i = 1}^n \tr(\widehat{\mat{\Delta}}_1^{-1}\t{\mat{R}_i}\widehat{\mat{\Delta}}_2^{-1}((\d\mat\beta)\mat{f}_{y_i}\t{\mat\alpha} + \mat\beta\mat{f}_{y_i}\t{(\d\mat\alpha}))) \\
|
||
|
&= \sum_{i = 1}^n \t{\vec(\widehat{\mat{\Delta}}_2^{-1}\mat{R}_i\widehat{\mat{\Delta}}_1^{-1}\mat\alpha\t{\mat{f}_{y_i}})}\d\vec\mat\beta
|
||
|
+ \sum_{i = 1}^n \t{\vec(\widehat{\mat{\Delta}}_1^{-1}\t{\mat{R}_i}\widehat{\mat{\Delta}}_2^{-1}\mat\beta\mat{f}_{y_i})}\d\vec\mat\alpha
|
||
|
\end{align*}
|
||
|
which means the gradients are
|
||
|
\begin{align*}
|
||
|
\nabla_{\mat\alpha}\hat{l}(\mat\alpha, \mat\beta)
|
||
|
&= \sum_{i = 1}^n \widehat{\mat{\Delta}}_1^{-1}\t{\mat{R}_i}\widehat{\mat{\Delta}}_2^{-1}\mat\beta\mat{f}_{y_i}
|
||
|
= (\ten{R}\ttm[3]\widehat{\mat{\Delta}}_1^{-1}\ttm[2]\widehat{\mat{\Delta}}_2^{-1})_{(3)}\t{(\ten{F}\ttm[2]\mat\beta)_{(3)}}\\
|
||
|
\nabla_{\mat\beta}\hat{l}(\mat\alpha, \mat\beta)
|
||
|
&= \sum_{i = 1}^n \widehat{\mat{\Delta}}_2^{-1}\mat{R}_i\widehat{\mat{\Delta}}_1^{-1}\mat\alpha\t{\mat{f}_{y_i}}
|
||
|
= (\ten{R}\ttm[3]\widehat{\mat{\Delta}}_1^{-1}\ttm[2]\widehat{\mat{\Delta}}_2^{-1})_{(2)}\t{(\ten{F}\ttm[3]\mat\alpha)_{(2)}}
|
||
|
\end{align*}
|
||
|
|
||
|
\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 $\widehat{\mat\Delta}_1$, $\widehat{\mat\Delta}_2$.
|
||
|
|
||
|
|
||
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||
|
%%% Alternative covariance estimates %%%
|
||
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||
|
\subsection{Alternative covariance estimates}
|
||
|
An alternative approach is \emph{not} to use the MLE estimates for $\mat\Delta_1$, $\mat\Delta_2$ but (up to scaling) unbiased estimates.
|
||
|
\begin{displaymath}
|
||
|
\widetilde{\mat\Delta}_1 = \frac{1}{n}\sum_{i = 1}^n \t{\mat{R}_i}\mat{R}_i {\color{gray}\quad(q\times q)},\qquad
|
||
|
\widetilde{\mat\Delta}_2 = \frac{1}{n}\sum_{i = 1}^n \mat{R}_i\t{\mat{R}_i} {\color{gray}\quad(p\times p)}.
|
||
|
\end{displaymath}
|
||
|
The unbiasednes comes directly from the following short computation;
|
||
|
\begin{displaymath}
|
||
|
(\E\widetilde{\mat\Delta}_1)_{j,k} = \frac{1}{n}\sum_{i = 1}^n \sum_{l = 1}^p \E \mat{R}_{i,l,j}\mat{R}_{i,l,k}
|
||
|
= \frac{1}{n}\sum_{i = 1}^n \sum_{l = 1}^p (\mat{\Delta}_{2})_{l,l}(\mat{\Delta}_{1})_{j,k}
|
||
|
= (\mat\Delta_1\tr(\mat\Delta_2))_{j,k}.
|
||
|
\end{displaymath}
|
||
|
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
|
||
|
\begin{displaymath}
|
||
|
\tilde{s} = \frac{1}{n}\sum_{i = 1}^n \|\mat{R}_i\|_F^2
|
||
|
\end{displaymath}
|
||
|
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
|
||
|
\begin{displaymath}
|
||
|
\widetilde{\mat\Delta} = \tilde{s}^{-1}(\widetilde{\mat{\Delta}}_1\otimes\widetilde{\mat{\Delta}}_2)
|
||
|
\end{displaymath}
|
||
|
|
||
|
\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
|
||
|
\begin{displaymath}
|
||
|
\mat{S}_1 = \frac{1}{n}\sum_{i = 1}^n \t{\mat{R}_i}\widetilde{\mat{\Delta}}_2^{-1}\mat{R}_i\quad{\color{gray}(q\times q)}, \qquad
|
||
|
\mat{S}_2 = \frac{1}{n}\sum_{i = 1}^n \mat{R}_i\widetilde{\mat{\Delta}}_1^{-1}\t{\mat{R}_i}\quad{\color{gray}(p\times p)}.
|
||
|
\end{displaymath}
|
||
|
|
||
|
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
|
||
|
\begin{align*}
|
||
|
f(\mat R)
|
||
|
&= \frac{1}{\sqrt{(2\pi)^{p q}|\mat\Delta|}}\exp\left(-\frac{1}{2}\t{\vec(\mat{R})} \mat\Delta^{-1}\vec(\mat{R})\right) \\
|
||
|
&= \frac{s^{p q / 2}}{(2\pi)^{p q / 2}|\mat\Delta_1|^{p / 2}|\mat\Delta_2|^{q / 2}}\exp\left(-\frac{s}{2}\tr(\mat\Delta_1^{-1}\t{\mat{R}}\mat\Delta_2^{-1}\mat{R})\right)
|
||
|
\end{align*}
|
||
|
|
||
|
The approximated log-likelihood is then
|
||
|
\begin{align*}
|
||
|
\tilde{l}(\mat\alpha, \mat\beta)
|
||
|
&=
|
||
|
-\frac{n p q}{2}\log{2\pi}
|
||
|
-\frac{n}{2}\log|\widetilde{\mat{\Delta}}|
|
||
|
-\frac{1}{2}\sum_{i = 1}^n \t{\mat{r}_i}\widetilde{\mat{\Delta}}^{-1}\mat{r}_i \\
|
||
|
&=
|
||
|
-\frac{n p q}{2}\log{2\pi}
|
||
|
+\frac{n p q}{2}\log\tilde{s}
|
||
|
-\frac{n p}{2}\log|\widetilde{\mat{\Delta}}_1|
|
||
|
-\frac{n q}{2}\log|\widetilde{\mat{\Delta}}_2|
|
||
|
-\frac{\tilde{s}}{2}\sum_{i = 1}^n \tr(\widetilde{\mat{\Delta}}_1^{-1}\t{\mat{R}_i}\widetilde{\mat{\Delta}}_2^{-1}\mat{R}_i).
|
||
|
\end{align*}
|
||
|
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.
|
||
|
|
||
|
Note that with the following holds
|
||
|
\begin{displaymath}
|
||
|
\sum_{i = 1}^n \tr(\widetilde{\mat{\Delta}}_1^{-1}\t{\mat{R}_i}\widetilde{\mat{\Delta}}_2^{-1}\mat{R}_i)
|
||
|
= n \tr(\widetilde{\mat{\Delta}}_1^{-1}\mat{S}_1)
|
||
|
= n \tr(\widetilde{\mat{\Delta}}_2^{-1}\mat{S}_2)
|
||
|
= n \tr(\mat{S}_1\widetilde{\mat{\Delta}}_1^{-1})
|
||
|
= n \tr(\mat{S}_2\widetilde{\mat{\Delta}}_2^{-1}).
|
||
|
\end{displaymath}
|
||
|
|
||
|
The derivation of the Gradient of the approximated log-likelihood $\tilde{l}$ is tedious but straight forward. We tackle the summands separately;
|
||
|
\begin{align*}
|
||
|
\d\log\tilde{s} &= \tilde{s}^{-1}\d\tilde{s} = \frac{2}{n\tilde{s}}\sum_{i = 1}^n \tr(\t{\mat{R}_i}\d\mat{R}_i)
|
||
|
= -\frac{2}{n\tilde{s}}\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}), \\
|
||
|
\d\log|\widetilde{\mat{\Delta}}_1| &=\tr(\widetilde{\mat{\Delta}}_1^{-1}\d\widetilde{\mat{\Delta}}_1) = \frac{2}{n}\sum_{i = 1}^n \tr(\widetilde{\mat{\Delta}}_1^{-1}\t{\mat{R}_i}\d\mat{R}_i)
|
||
|
= -\frac{2}{n}\sum_{i = 1}^n \tr(\t{\mat{f}_{y_i}}\t{\mat{\beta}}\mat{R}_i\widetilde{\mat{\Delta}}_1^{-1}\d\mat{\alpha} + \mat{f}_{y_i}\t{\mat{\alpha}}\widetilde{\mat{\Delta}}_1^{-1}\t{\mat{R}_i}\d\mat{\beta}), \\
|
||
|
\d\log|\widetilde{\mat{\Delta}}_2| &=\tr(\widetilde{\mat{\Delta}}_2^{-1}\d\widetilde{\mat{\Delta}}_2) = \frac{2}{n}\sum_{i = 1}^n \tr(\t{\mat{R}_i}\widetilde{\mat{\Delta}}_2^{-1}\d\mat{R}_i)
|
||
|
= -\frac{2}{n}\sum_{i = 1}^n \tr(\t{\mat{f}_{y_i}}\t{\mat{\beta}}\widetilde{\mat{\Delta}}_2^{-1}\mat{R}_i\d\mat{\alpha} + \mat{f}_{y_i}\t{\mat{\alpha}}\t{\mat{R}_i}\widetilde{\mat{\Delta}}_2^{-1}\d\mat{\beta})
|
||
|
\end{align*}
|
||
|
as well as
|
||
|
\begin{displaymath}
|
||
|
\d\,\tilde{s}\sum_{i = 1}^n \tr(\widetilde{\mat{\Delta}}_1^{-1}\t{\mat{R}_i}\widetilde{\mat{\Delta}}_2^{-1}\mat{R}_i)
|
||
|
= (\d\tilde{s})\sum_{i = 1}^n \tr(\widetilde{\mat{\Delta}}_1^{-1}\t{\mat{R}_i}\widetilde{\mat{\Delta}}_2^{-1}\mat{R}_i)
|
||
|
+ \tilde{s}\, \d \sum_{i = 1}^n \tr(\widetilde{\mat{\Delta}}_1^{-1}\t{\mat{R}_i}\widetilde{\mat{\Delta}}_2^{-1}\mat{R}_i).
|
||
|
\end{displaymath}
|
||
|
We have
|
||
|
\begin{displaymath}
|
||
|
\d\tilde{s} = -\frac{2}{n}\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})
|
||
|
\end{displaymath}
|
||
|
and the remaining term
|
||
|
\begin{align*}
|
||
|
\d\sum_{i = 1}^n\tr(\widetilde{\mat{\Delta}}_1^{-1}\t{\mat{R}_i}\widetilde{\mat{\Delta}}_2^{-1}\mat{R}_i)
|
||
|
= 2\sum_{i = 1}^n \tr(&\t{\mat{f}_{y_i}}\t{\mat{\beta }}(\mat{R}_i \widetilde{\mat{\Delta}}_1^{-1}\mat{S}_1\widetilde{\mat{\Delta}}_1^{-1} + \widetilde{\mat{\Delta}}_2^{-1}\mat{S}_2\widetilde{\mat{\Delta}}_2^{-1} \mat{R}_i - \widetilde{\mat{\Delta}}_2^{-1} \mat{R}_i \widetilde{\mat{\Delta}}_1^{-1})\d\mat{\alpha} \\
|
||
|
+\,&\mat{f}_{y_i} \t{\mat{\alpha}}(\widetilde{\mat{\Delta}}_1^{-1}\mat{S}_1\widetilde{\mat{\Delta}}_1^{-1}\t{\mat{R}_i} + \t{\mat{R}_i}\widetilde{\mat{\Delta}}_2^{-1}\mat{S}_2\widetilde{\mat{\Delta}}_2^{-1} - \widetilde{\mat{\Delta}}_1^{-1}\t{\mat{R}_i}\widetilde{\mat{\Delta}}_2^{-1})\d\mat{\beta }).
|
||
|
\end{align*}
|
||
|
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}) \\
|
||
|
&\hspace{3em} - \frac{n p}{2}\Big(-\frac{2}{n}\Big)\sum_{i = 1}^n \tr(\t{\mat{f}_{y_i}}\t{\mat{\beta}}\mat{R}_i\widetilde{\mat{\Delta}}_1^{-1}\d\mat{\alpha} + \mat{f}_{y_i}\t{\mat{\alpha}}\widetilde{\mat{\Delta}}_1^{-1}\t{\mat{R}_i}\d\mat{\beta}) \\
|
||
|
&\hspace{3em} - \frac{n q}{2}\Big(-\frac{2}{n}\Big)\sum_{i = 1}^n \tr(\t{\mat{f}_{y_i}}\t{\mat{\beta}}\widetilde{\mat{\Delta}}_2^{-1}\mat{R}_i\d\mat{\alpha} + \mat{f}_{y_i}\t{\mat{\alpha}}\t{\mat{R}_i}\widetilde{\mat{\Delta}}_2^{-1}\d\mat{\beta}) \\
|
||
|
&\hspace{3em} -\frac{1}{2}\Big(-\frac{2}{n}\Big)\Big(\sum_{i = 1}^n \tr(\widetilde{\mat{\Delta}}_1^{-1}\t{\mat{R}_i}\widetilde{\mat{\Delta}}_2^{-1}\mat{R}_i)\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}) \\
|
||
|
&\hspace{3em} -\frac{\tilde{s}}{2}2\sum_{i = 1}^n \tr\!\Big(\t{\mat{f}_{y_i}}\t{\mat{\beta }}(\mat{R}_i \widetilde{\mat{\Delta}}_1^{-1}\mat{S}_1\widetilde{\mat{\Delta}}_1^{-1} + \widetilde{\mat{\Delta}}_2^{-1}\mat{S}_2\widetilde{\mat{\Delta}}_2^{-1} \mat{R}_i - \widetilde{\mat{\Delta}}_2^{-1} \mat{R}_i \widetilde{\mat{\Delta}}_1^{-1})\d\mat{\alpha} \\
|
||
|
&\hspace{3em} \hspace{4.7em} + \mat{f}_{y_i} \t{\mat{\alpha}}(\widetilde{\mat{\Delta}}_1^{-1}\mat{S}_1\widetilde{\mat{\Delta}}_1^{-1}\t{\mat{R}_i} + \t{\mat{R}_i}\widetilde{\mat{\Delta}}_2^{-1}\mat{S}_2\widetilde{\mat{\Delta}}_2^{-1} - \widetilde{\mat{\Delta}}_1^{-1}\t{\mat{R}_i}\widetilde{\mat{\Delta}}_2^{-1})\d\mat{\beta }\Big) \\
|
||
|
%
|
||
|
&= \sum_{i = 1}^n \tr\bigg(\t{\mat{f}_{y_i}}\t{\mat{\beta}}\Big(
|
||
|
-p q \tilde{s}^{-1} \mat{R}_i + p \mat{R}_i\widetilde{\mat{\Delta}}_1^{-1} + q \widetilde{\mat{\Delta}}_2^{-1}\mat{R}_i + \tr(\widetilde{\mat{\Delta}}_1^{-1}\mat{S}_1)\mat{R}_i \\
|
||
|
&\hspace{3em} \hspace{4.7em} - \tilde{s}(\mat{R}_i \widetilde{\mat{\Delta}}_1^{-1}\mat{S}_1\widetilde{\mat{\Delta}}_1^{-1} + \widetilde{\mat{\Delta}}_2^{-1}\mat{S}_2\widetilde{\mat{\Delta}}_2^{-1} \mat{R}_i - \widetilde{\mat{\Delta}}_2^{-1} \mat{R}_i \widetilde{\mat{\Delta}}_1^{-1})
|
||
|
\Big)\d\mat{\alpha}\bigg) \\
|
||
|
&\hspace{3em}+ \sum_{i = 1}^n \tr\bigg(\mat{f}_{y_i}\t{\mat{\alpha}}\Big(
|
||
|
-p q \tilde{s}^{-1} \t{\mat{R}_i} + p \widetilde{\mat{\Delta}}_1^{-1}\t{\mat{R}_i} + q \t{\mat{R}_i}\widetilde{\mat{\Delta}}_2^{-1} + \tr(\widetilde{\mat{\Delta}}_1^{-1}\mat{S}_1)\t{\mat{R}_i} \\
|
||
|
&\hspace{3em}\hspace{3em} \hspace{4.7em} - \tilde{s}(\widetilde{\mat{\Delta}}_1^{-1}\mat{S}_1\widetilde{\mat{\Delta}}_1^{-1}\t{\mat{R}_i} + \t{\mat{R}_i}\widetilde{\mat{\Delta}}_2^{-1}\mat{S}_2\widetilde{\mat{\Delta}}_2^{-1} - \widetilde{\mat{\Delta}}_1^{-1} \t{\mat{R}_i} \widetilde{\mat{\Delta}}_2^{-1})
|
||
|
\Big)\d\mat{\beta}\bigg).
|
||
|
\end{align*}
|
||
|
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
|
||
|
+ (q\mat{I}_p - \tilde{s}\widetilde{\mat{\Delta}}_2^{-1}\mat{S}_2)\widetilde{\mat{\Delta}}_2^{-1}\mat{R}_i
|
||
|
+ \mat{R}_i\widetilde{\mat{\Delta}}_1^{-1}(p\mat{I}_q - \tilde{s}\mat{S}_1\widetilde{\mat{\Delta}}_1^{-1})
|
||
|
+ \tilde{s}\widetilde{\mat{\Delta}}_2^{-1}\mat{R}_i\widetilde{\mat{\Delta}}_1^{-1}
|
||
|
\end{displaymath}
|
||
|
and with $\mathcal{G}$ the order 3 tensor stacking the $\mat{G}_i$'s such that the first mode indexes the observation
|
||
|
\begin{displaymath}
|
||
|
\ten{G}
|
||
|
= (\tr(\widetilde{\mat{\Delta}}_1^{-1}\mat{S}_1) - p q \tilde{s}^{-1})\ten{R}
|
||
|
+ \ten{R}\ttm[2](q\mat{I}_p - \tilde{s}\widetilde{\mat{\Delta}}_2^{-1}\mat{S}_2)\widetilde{\mat{\Delta}}_2^{-1}
|
||
|
+ \ten{R}\ttm[3](p\mat{I}_q - \tilde{s}\widetilde{\mat{\Delta}}_1^{-1}\mat{S}_1)\widetilde{\mat{\Delta}}_1^{-1}
|
||
|
+ \tilde{s}\ten{R}\ttm[2]\widetilde{\mat{\Delta}}_2^{-1}\ttm[3]\widetilde{\mat{\Delta}}_1^{-1}
|
||
|
\end{displaymath}
|
||
|
This leads to the following form of the differential of $\tilde{l}$ given by
|
||
|
\begin{displaymath}
|
||
|
\d\tilde{l}(\mat{\alpha}, \mat{\beta})
|
||
|
= \sum_{i = 1}^n \tr(\t{\mat{f}_{y_i}}\t{\mat{\beta}}\mat{G}_i\d\mat{\alpha})
|
||
|
+ \sum_{i = 1}^n \tr(\mat{f}_{y_i}\t{\mat{\alpha}}\t{\mat{G}_i}\d\mat{\beta})
|
||
|
\end{displaymath}
|
||
|
and therefore the gradients
|
||
|
\begin{align*}
|
||
|
\nabla_{\mat{\alpha}}\tilde{l}(\mat{\alpha}, \mat{\beta}) &= \sum_{i = 1}^n \t{\mat{G}_i}\mat{\beta}\mat{f}_{y_i}
|
||
|
= \ten{G}_{(3)}\t{(\ten{F}\ttm[2]\beta)_{(3)}}, \\
|
||
|
\nabla_{\mat{\beta}} \tilde{l}(\mat{\alpha}, \mat{\beta}) &= \sum_{i = 1}^n \mat{G}_i\mat{\alpha}\t{\mat{f}_{y_i}}
|
||
|
= \ten{G}_{(2)}\t{(\ten{F}\ttm[3]\beta)_{(2)}}.
|
||
|
\end{align*}
|
||
|
|
||
|
\todo{check the tensor version of the gradient!!!}
|
||
|
|
||
|
\newpage
|
||
|
|
||
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||
|
%%% Numerical Examples %%%
|
||
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||
|
\section{Numerical Examples}
|
||
|
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
|
||
|
\begin{displaymath}
|
||
|
\mat{X} = \mat{\beta}\mat{f}_y \t{\mat{\alpha}} + \mat{\epsilon}, \qquad \vec{\mat{\epsilon}} \sim \mathbb{N}_{p q}(\mat{0}, \mat{\Delta})
|
||
|
\end{displaymath}
|
||
|
where $\mat{\Delta}_{i j} = 0.5^{|i - j|}$ for $i, j = 1, ..., p q$.
|
||
|
|
||
|
\begin{figure}
|
||
|
\centering
|
||
|
\includegraphics{loss_Ex01.png}
|
||
|
\end{figure}
|
||
|
\begin{figure}
|
||
|
\centering
|
||
|
\includegraphics{estimates_Ex01.png}
|
||
|
\end{figure}
|
||
|
\begin{figure}
|
||
|
\centering
|
||
|
\includegraphics{Delta_Ex01.png}
|
||
|
\end{figure}
|
||
|
\begin{figure}
|
||
|
\centering
|
||
|
\includegraphics{hist_Ex01.png}
|
||
|
\end{figure}
|
||
|
|
||
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||
|
%%% Bib and Index %%%
|
||
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||
|
\printindex
|
||
|
\nocite{*}
|
||
|
\printbibliography
|
||
|
|
||
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
||
|
%%% Appendix %%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\appendix
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\section{Matrix Differential Rules}
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Let $\mat A$ be a square matrix (and invertible if needed) and $|.|$ stands for the determinant
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\begin{align*}
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\d\log\mat A &= \frac{1}{|\mat A|}\d\mat{A} \\
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\d|\mat A| &= |\mat A|\tr \mat{A}^{-1}\d\mat A \\
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\d\log|\mat A| &= \tr\mat{A}^{-1}\d\mat A \\
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\d\mat{X}^{-1} &= -\mat{X}^{-1}(\d\mat{X})\mat{X}^{-1}
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\end{align*}
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\section{Useful Matrix Identities}
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In this section we summarize a few useful matrix identities, for more details see for example \cite{MatrixAlgebra-AbadirMagnus2005}.
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For two matrices $\mat A$ of dimensions $q\times r$ and $\mat B$ of dimensions $p\times k$ holds
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\begin{equation}\label{eq:vecKron}
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\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).
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\end{equation}
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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
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\begin{displaymath}
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(\mat A + \mat a\t{\mat b})^{-1} = \mat{A}^{-1} - \frac{1}{1 + \t{\mat b}A^{-1}\mat a}\mat{A}^{-1}\mat{a}\t{\mat{b}}\mat{A}^{-1}
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\end{displaymath}
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as well as
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\begin{displaymath}
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\det(\mat A + \mat a\t{\mat b}) = \det(\mat A)(1 + \t{\mat b}{\mat A}^{-1}\mat a)
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\end{displaymath}
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which even holds in the case $\t{\mat b}A^{-1}\mat a = -1$. This is known as Sylvester's determinant theorem.
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\section{Commutation Matrix and Permutation Identities}
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\begin{center}
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Note: In this section we use 0-indexing for the sake of simplicity!
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\end{center}
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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.
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Let $\mat A$ be an arbitrary $p\times q$ matrix. The permutation matrix $\mat K_{p, q}$ satisfies
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\begin{displaymath}
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\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
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\end{displaymath}
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where $\pi_{p, q}$ is a permutation of the indices $i = 0, ..., p q - 1$ such that
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\begin{displaymath}
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\pi_{p, q}(i + j p) = j + i q, \quad\text{for }i = 0, ..., p - 1; j = 0, ..., q - 1.
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\end{displaymath}
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\begin{table}[!htp]
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\centering
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\begin{minipage}{0.8\textwidth}
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\centering
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\begin{tabular}{l c l}
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$\mat{K}_{p, q}$ & $\hat{=}$ & $\pi_{p, q}(i + j p) = j + i q$ \\
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$\mat{I}_r\kron\mat{K}_{p, q}$ & $\hat{=}$ & $\tilde{\pi}_{p, q, r}(i + j p + k p q) = j + i q + k p q$ \\
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$\mat{K}_{p, q}\kron\mat{I}_r$ & $\hat{=}$ & $\hat{\pi}_{p, q, r}(i + j p + k p q) = r(j + i q) + k$
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\end{tabular}
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\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 $k = 0, ..., r - 1$.}
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\end{minipage}
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\end{table}
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\section{Matrix and Tensor Operations}
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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}
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$\hada$ 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}
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\todo{ resolve confusion between Khatri-Rao, Column-wise Kronecker / Khatri-Rao, Row-wise Kronecker / Khatri-Rao, Face-Splitting Product, .... Yes, its a mess.}
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\paragraph{Kronecker Product $\kron$:}
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\paragraph{Khatri-Rao Product $\hada$:}
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\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
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\begin{displaymath}
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\t{(\mat{A}\odot_{t}\mat{B})} = \t{\mat{A}}\hada\t{\mat{B}}
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\end{displaymath}
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\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 $j = \sum_{k = 1, k\neq n}^m i_k\prod_{l = 1, l\neq n}^{k - 1}t_l$.
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\todo{ give an example!}
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\paragraph{$n$-mode Tensor Product $\ttm[n]$:}
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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
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\begin{displaymath}
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(\ten{T}\ttm[n] \mat{M})_{i_1, ..., i_{n-1}, j, i_{n+1}, ..., i_m} = \sum_{k = 1}^{t_n} \ten{T}_{i_1, ..., i_{n-1}, k, i_{n+1}, ..., i_m} \mat{M}_{j, k}
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\end{displaymath}
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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 $t_1\times t_2\times ... \times p\times ... \times t_m$.
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\begin{example}[Matrix Multiplication Analogs]
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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
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\begin{align*}
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(\mat{A}\ttm[1] \mat{B})_{i,j} &= \sum_{l = 1}^{t_1} \mat{A}_{l,j}\mat{B}_{i,l}
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= (\mat{B}\mat{A})_{i,j}
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& \text{for }t_1 = q, \\
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(\mat{A}\ttm[2] \mat{B})_{i,j} &= \sum_{l = 1}^{t_2} \mat{A}_{i,l}\mat{B}_{j,l}
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= (\mat{A}\t{\mat{B}})_{i,j} = \t{(\mat{B}\t{\mat{A}})}_{i,j}
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& \text{for }t_2 = q.
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\end{align*}
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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.
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\end{example}
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\begin{example}[Order Three Analogs]
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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
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\begin{align*}
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(\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, \\
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(\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, \\
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(\mat{A}\ttm[3]\mat{B})_{i,j,k} &= \sum_{l = 1}^{t_3} \mat{A}_{i,j,l}\mat{B}_{k,l} \equiv \t{(\mat{B}\t{\mat{A}_{i,:,:}})}_{j,k} & \text{for }t_3 = q.
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\end{align*}
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\end{example}
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Letting $\ten{F}$ be the $3$-tensor of dimensions $n\times k\times r$ such that $\ten{F}_{i,:,:} = \mat{f}_{y_i}$, then
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\begin{displaymath}
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\mat{\beta}\mat{f}_{y_i}\t{\mat{\alpha}} = (\ten{F}\ttm[2]\mat{\beta}\ttm[3]\mat{\alpha})_{i,:,:}
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\end{displaymath}
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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 $i = 1, ..., n$.
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Another analog way of writing this is
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\begin{displaymath}
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(\ten{F}\ttm[2]\mat{\beta}\ttm[3]\mat{\alpha})_{(1)} = \mathbb{F}_{y}(\t{\mat{\alpha}}\kron\t{\mat{\beta}})
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\end{displaymath}
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\section{Equivalencies}
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In this section we give a short summary of alternative but equivalent operations.
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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 ($\mat{A}\widehat{=}\mat{B}\ :\Leftrightarrow\ \vec{\mat{A}} = \vec{\mat{B}}$).
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Therefore, we use $\mat{A}, \mat{B}, \mat{X}, \mat{F}, \mat{R}, ...$ for matrices. 3-Tensors are written as $\ten{A}, \ten{B}, \ten{T}, \ten{X}, \ten{F}, \ten{R}, ...$.
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\begin{align*}
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\ten{T}\ttm[3]\mat{A}\ &{\widehat{=}}\ \mat{T}\t{\mat A} & \ten{T}(n, p, q)\ \widehat{=}\ \mat{T}(n p, q), \mat{A}(p, q) \\
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\ten{T}\ttm[2]\mat{B}\ &{\widehat{=}}\ \mat{B}\ten{T}_{(2)} & \ten{T}(n, p, q), \ten{T}_{(2)}(p, n q), \mat{B}(q, p)
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\end{align*}
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\section{Matrix Valued Normal Distribution}
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A random variable $\mat{X}$ of dimensions $p\times q$ is \emph{Matrix-Valued Normal Distribution}, denoted
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\begin{displaymath}
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\mat{X}\sim\mathcal{MN}_{p\times q}(\mat{\mu}, \mat{\Delta}_2, \mat{\Delta}_1),
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\end{displaymath}
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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
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\begin{displaymath}
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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).
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\end{displaymath}
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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%%% Reference Summaries %%%
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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\section{Reference Summaries}
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This section contains short summaries of the main references with each sub-section concerning one paper.
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\subsection{}
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\subsection{Generalized Tensor Decomposition With Features on Multiple Modes}
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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
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\begin{displaymath}
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\E(\ten{Y} | \mat{X}_1, ..., \mat{X}_K) = f(\ten{C}\times\{ \mat{X}_1\mat{M}_1, ..., \mat{X}_K\mat{M}_K \}),\qquad \t{\mat{M}_k}\mat{M}_k = \mat{I}_{r_k}\ \forall k\in[K].
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\end{displaymath}
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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
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\begin{displaymath}
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\ten{C}\times\{ \mat{X}_1\mat{M}_1, ..., \mat{X}_K\mat{M}_K \}
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= \ten{C}\ttm[1]\mat{X}_1\mat{M}_1\ ...\ttm[K]\mat{X}_K\mat{M}_K
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\end{displaymath}
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with $\ttm[k]$ denoting the $k$-mode tensor matrix product.
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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
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\begin{displaymath}
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l(\ten{C}, \mat{M}_1, ..., \mat{M}_K) = \langle \ten{Y}, \Theta \rangle - \sum_{i_1, ..., i_K} b(\Theta_{i_1, ..., i_K}), \qquad \Theta = \ten{C}\times\{ \mat{X}_1\mat{M}_1, ..., \mat{X}_K\mat{M}_K \}
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\end{displaymath}
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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.
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The main benefit is that this approach generalizes well to a multitude of different structured data sets.
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\todo{ how does this relate to the $\mat{X} = \mat{\mu} + \mat{\beta}\mat{f}_y\t{\mat{\alpha}} + \mat{\epsilon}$ model.}
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\end{document}
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