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Reallocated singl_function (elt_size=4) to 100 items from 50. Database file #1: Divilkovskiy2024SourceSpace_en-blx.bib Database file #2: references.bib -Biblatex version: 3.17 +Biblatex version: 3.19 Reallocated wiz_functions (elt_size=4) to 9000 items from 6000. Reallocated singl_function (elt_size=4) to 100 items from 50. You've used 18 entries, 6399 wiz_defined-function locations, - 1313 strings with 13350 characters, -and the built_in function-call counts, 158071 in all, are: -= -- 13348 + 1313 strings with 13352 characters, +and the built_in function-call counts, 158083 in all, are: += -- 13349 > -- 2002 < -- 370 + -- 5464 - -- 822 -* -- 10751 -:= -- 19918 +* -- 10752 +:= -- 19920 add.period$ -- 0 call.type$ -- 18 change.case$ -- 130 chr.to.int$ -- 84 cite$ -- 34 duplicate$ -- 17847 -empty$ -- 12986 +empty$ -- 12987 format.name$ -- 1362 -if$ -- 30564 +if$ -- 30567 int.to.chr$ -- 0 int.to.str$ -- 36 missing$ -- 0 @@ -51,9 +51,9 @@ pop$ -- 6640 preamble$ -- 1 purify$ -- 286 quote$ -- 0 -skip$ -- 2910 +skip$ -- 2911 stack$ -- 0 -substring$ -- 24299 +substring$ -- 24302 swap$ -- 2544 text.length$ -- 430 text.prefix$ -- 0 diff --git a/paper/Divilkovskiy2024SourceSpace_en.pdf b/paper/Divilkovskiy2024SourceSpace_en.pdf index afb6e88..296bffc 100644 Binary files a/paper/Divilkovskiy2024SourceSpace_en.pdf and b/paper/Divilkovskiy2024SourceSpace_en.pdf differ diff --git a/paper/Divilkovskiy2024SourceSpace_en.synctex.gz b/paper/Divilkovskiy2024SourceSpace_en.synctex.gz index a380417..84c2f8d 100644 Binary files a/paper/Divilkovskiy2024SourceSpace_en.synctex.gz and b/paper/Divilkovskiy2024SourceSpace_en.synctex.gz differ diff --git a/paper/Divilkovskiy2024SourceSpace_en.tex b/paper/Divilkovskiy2024SourceSpace_en.tex index 9f6d075..6293ac4 100644 --- a/paper/Divilkovskiy2024SourceSpace_en.tex +++ b/paper/Divilkovskiy2024SourceSpace_en.tex @@ -16,7 +16,7 @@ \title{Calculating the value of a multivariate time series at a point by the predicted matrix of pairwise correlation} \author{% - Maxim Divilkovskiy\footnote{\textbf{Huawei ???, sizzziy@yandex.ru ??? RU}}, % + Maxim Divilkovskiy\footnote{Forecsys, mdivilkovskij@gmail.com}, % Konstantin Yakovlev\footnote{Forecsys, bicst808@gmail.com}, % Vadim Strijov\footnote{Forecsys, strijov@gmail.com}, % } @@ -48,7 +48,7 @@ \section{Point-wise prediction of a set of time series} A set of $d$ time series is given by $t$ vectors: \[ -\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_t], \text{for each } k: \mathbf{x}_k \in \mathbb{R}^d, +[\mathbf{x}_1, \mathbf{x}_2, \ldots, \mathbf{x}_t], \text{for each } k: \mathbf{x}_k \in \mathbb{R}^d, \] where $\mathbf{x}_{t_i, k} \in \mathbb{R}$ is the value of the series with index $k$ at the time moment $t_i$. One has to predict $\mathbf{x}_{t+1}$ the set of all time series at the time moment $t+1$. The matrix $\mathbf{x}$ of size $t \times d$ is treated as a \emph{multidimensional} time series, treating the value at a point as an element of space $\mathbb{R}^d$. @@ -146,18 +146,21 @@ \section{Pairwise correlation between time series} \textbf{Proof.} Let us denote by $\mathbf{\Sigma}$ the true matrix at the moment of time $T$, and $\hat{\mathbf{\Sigma}}$ the predicted one. By construction, ${\mathbf{\Sigma}} = \frac{1}{T} \sum_{k=1}^{T} (\mathbf{a}^T_k - \boldsymbol{\mu}_T)(\mathbf{a}^T_k - \boldsymbol{\mu}_T)^\intercal\texttt{}$. The matrix $\mathbf{A}$ is a transposed matrix $\mathbf{X}$ of time series, where the first dimension is the number of the series, not the moment of time as in the case of $\mathbf{X}$. Then, consider what the elements of the matrices $\mathbf{\Sigma}$ and $\hat{\mathbf{\Sigma}}$ are equal to. \begin{align*} - \mathbf{\Sigma}_{ij} &= \frac{1}{T}\sum_{k=1}^{T}(\mathbf{a}_{ik} - \boldsymbol{\mu}_i)(\mathbf{a}_{jk}-\boldsymbol{\mu}_j),\\ - \hat{\mathbf{\Sigma}}_{ij} &= \frac{1}{T}\sum_{k=1}^{T-1}(\mathbf{a}_{ik} - \hat{\boldsymbol{\mu}}_i)(\mathbf{a}_{jk}-\hat{\boldsymbol{\mu}}_j) + (\mathbf{y}_i - \hat{\boldsymbol{\mu}}_i)(\mathbf{y}_j - \hat{\boldsymbol{\mu}}_j). + &\mathbf{\Sigma}_{ij} = \frac{1}{T}\sum_{k=1}^{T}(\mathbf{a}_{ik} - \boldsymbol{\mu}_i)(\mathbf{a}_{jk}-\boldsymbol{\mu}_j),\\ + &\text{separating last term,}\\ + &\hat{\mathbf{\Sigma}}_{ij} = \frac{1}{T}\sum_{k=1}^{T-1}(\mathbf{a}_{ik} - \hat{\boldsymbol{\mu}}_i)(\mathbf{a}_{jk}-\hat{\boldsymbol{\mu}}_j) + (\mathbf{y}_i - \hat{\boldsymbol{\mu}}_i)(\mathbf{y}_j - \hat{\boldsymbol{\mu}}_j). \end{align*} Since we minimise the norm of the difference, both matrices are equal to each other. Consider the equality of diagonal elements. \begin{gather*} \text{For any } i, j \text{ such that } i = j \text{ the following is true: } \mathbf{\Sigma}_{ii} = \hat{\mathbf{\Sigma}}_{ii},\\ \sum_{k=1}^{T}(\mathbf{a}_{ik} - \boldsymbol{\mu}_i)(\mathbf{a}_{ik}-\boldsymbol{\mu}_i) = \sum_{k=1}^{T-1}(\mathbf{a}_{ik} - \hat{\boldsymbol{\mu}}_i)(\mathbf{a}_{ik}-\hat{\boldsymbol{\mu}}_i) + (\mathbf{y}_i - \hat{\boldsymbol{\mu}}_i)(\mathbf{y}_i - \hat{\boldsymbol{\mu}}_i),\\ + \text{since } \boldsymbol{\mu}_i \text{ and } \hat{\boldsymbol{\mu}}_i \text{ are equal up to } (T-1) \text{-th coordinate,}\\ (\mathbf{a}_{iT}-\boldsymbol{\mu}_i)^2 = (\mathbf{y}_i-\hat{\boldsymbol{\mu}}_i)^2.\\ \text{Consider } \hat{\boldsymbol{\mu}}_i \text{ and } \hat{\boldsymbol{\mu}}:\\ \hat{\boldsymbol{\mu}}_i = \frac{1}{T}\sum_{k=1}^{T-1}\mathbf{a}_{ik} + \frac{1}{T}\mathbf{y}_i,\\ - \boldsymbol{\mu}_i = \frac{1}{T}\sum_{k=1}^{T}\mathbf{a}_{ik},\\ + \boldsymbol{\mu}_i = \frac{1}{T}\sum_{k=1}^{T}\mathbf{a}_{ik}.\\ + \text{Substitute the expressions for } \hat{\boldsymbol{\mu}} \text{ into } (\mathbf{a}_{iT}-\boldsymbol{\mu}_i)^2 = (\mathbf{y}_i-\hat{\boldsymbol{\mu}}_i)^2:\\ \left[ \begin{array}{ll} \frac{T-1}{T}\mathbf{y}_i-\frac{1}{T}\sum_{k=1}^{T-1}\mathbf{a}_{ik}=\mathbf{a}_{iT}-\boldsymbol{\mu}_i @@ -165,6 +168,7 @@ \section{Pairwise correlation between time series} \frac{T-1}{T}\mathbf{y}_i-\frac{1}{T}\sum_{k=1}^{T-1}\mathbf{a}_{ik}=\boldsymbol{\mu}_i-\mathbf{a}_{iT} \end{array}, \right .\\[1ex] + \text{Rewriting } \boldsymbol{\mu} \text{,}\\ \left[ \begin{array}{ll} \frac{T-1}{T}\mathbf{y}_i = \frac{T-1}{T}\mathbf{a}_{iT} @@ -188,15 +192,17 @@ \section{Pairwise correlation between time series} \end{align*} $$ \blacksquare $$ +Figure \ref{fig:fig2} shows an example of a function $||\hat{\mathbf{\Sigma}}_{t+1} - \bar{\mathbf{\Sigma}}_{t+1}||_2^2$ with two minimum for certain time series. + \textbf{Corollary. (A trivial method for obtaining a pair of possible answers.)} This theorem shows that using pairwise correlation as a distance function gives at most \emph{two} different answers when reconstructing. Moreover, having obtained one, we can explicitly find the second one. Then, to find both possible answers, it is proposed to apply any non-convex optimisation method to find at least one of the minimum of the function. Therefore with the formula above we are able to find another minimum. \begin{figure}[!htbp] \centering \begin{center} \includegraphics[width=0.8\textwidth]{NonConvex.eps} - \label{fig:fig5} \end{center} \caption{Function $||\hat{\mathbf{\Sigma}}_{t+1} - \bar{\mathbf{\Sigma}}_{t+1}||_2^2$ for following series: $(1, 3)$ and $(2, 4)$. Minumums: (3; 4) is desired and (-1; 0) is alternative.} + \label{fig:fig2} \end{figure} @@ -213,7 +219,7 @@ \section{Pairwise correlation between time series} \mathbf{\Sigma}_t = \frac{1}{t} \sum_{i=1}^{t} \mathbf{x}_i \mathbf{x}_i^\intercal - \frac{1}{t} \left( \sum_{i=1}^{t} \mathbf{x}_i\right) \boldsymbol{\mu}_t^\intercal - \boldsymbol{\mu}_t \frac{1}{t} \left( \sum_{i=1}^{t} \mathbf{x}_i^\intercal\right) + \boldsymbol{\mu}_t \boldsymbol{\mu}_t^\intercal =\\= \frac{1}{t} \sum_{i=1}^{t} \mathbf{x}_i \mathbf{x}_i^\intercal - \boldsymbol{\mu}_t \boldsymbol{\mu}_t^\intercal \Rightarrow\\ \sum_{i=1}^{t} \mathbf{x}_i \mathbf{x}_i^\intercal = t \mathbf{\Sigma}_t + t \boldsymbol{\mu}_t \boldsymbol{\mu}_t^\intercal. \end{gather*} - \item Express $\mathbf{\Sigma_{t+1}}$ through $\mathbf{\Sigma_t}$ using the previous expressions: + \item Express $\mathbf{\Sigma_{t+1}}$ with $\mathbf{\Sigma_t}$ using the previous expressions: \begin{gather*} \mathbf{\Sigma}_{t+1} = \frac{1}{t+1} \left(\sum_{i=1}^{t} \mathbf{x}_i \mathbf{x}_i^\intercal + \mathbf{x}_{t+1} \mathbf{x}_{t+1}^\intercal \right) - \boldsymbol{\mu}_{t+1} \boldsymbol{\mu}_{t+1}^\intercal = \\ = \frac{t}{t+1}\mathbf{\Sigma}_t + \frac{t}{t+1}\boldsymbol{\mu}_{t} \boldsymbol{\mu}_{t}^\intercal + \frac{1}{t+1} \mathbf{x}_{t+1} \mathbf{x}_{t+1}^\intercal - \frac{1}{(t+1)^2} (t \boldsymbol{\mu}_t + \mathbf{x}_{t+1})(t \boldsymbol{\mu}_t + \mathbf{x}_{t+1})^\intercal =\\ @@ -271,6 +277,7 @@ \section{Algorithm for reconstructing time series values in case of accurate pre In real data, the probability of matching sets of answers is 0, just as it is in synthetic data with random noise added. \end{enumerate} +\ref{fig:fig3} shows the example of reconstructed time series with accurate predicted matrix. As you can see in the graph, the reconstruction is lossless. \begin{figure}[!htbp] \centering @@ -285,6 +292,7 @@ \section{Algorithm for reconstructing time series values in case of inaccurate m Instead of two values of $T$ and $T^\prime$, we propose take $K$ values. We get $K$ matrices with some noise that came from inaccuracy in prediction. Thus, each matrix is reduced to the nearest positive semi-definite matrix. Algorithm is explained in \cite{HIGHAM1988103}. Then, for each value algorithm for obtaining the pair of possible answers is applied. + We get $K$ sets of answers: \begin{gather*} [\hat{\mathbf{x}}_{t+1}^1, \hat{\mathbf{x}}^{\prime 1}_{t+1}],\\ @@ -295,6 +303,9 @@ \section{Algorithm for reconstructing time series values in case of inaccurate m Then we propose to search through $2^K$ sets of answers and choose the set in which the diameter is minimal. The diameter of set is calculated as maximum Euclidean distance between two different points in set. It is a \emph{necessary} condition for the actual answer. In the case of an accurate prediction, the diameter of such a set will always be zero. The asymptotic complexity of this reconstruction will be $O(2^K \times K \times N)$ $+$ the complexity of the minimum search algorithm used. + +On \ref{fig:fig4} you can see an example of reconstructing time series values when the predicted matrix contains some random noise. The reconstruction is no more lossless. + \begin{figure}[!htbp] \centering \includegraphics[width=\textwidth]{NonIdealRecovery.eps} @@ -311,7 +322,7 @@ \section{Experiment} \begin{table}[!h] \def\arraystretch{2.3} \begin{center} -\caption{Loss on synthetic data} +\caption{Loss on synthetic data. As expected, the error is less on bigger $K$ value. See \ref{fig:fig5} for the example of reconstruction with $K=10$ and noise $\mathcal{N}(0, 0.05)$.} \begin{tabular}{|l||l||*{3}{c|}}\hline {Noise} &\makebox[3em]{Metric}&\makebox[3em]{$K=2$}&\makebox[3em]{$K=4$}&\makebox[3em]{$K=10$}\\\hline @@ -327,7 +338,7 @@ \section{Experiment} \centering \includegraphics[width=\textwidth]{synthetic_time_series_K10N005.eps} \caption{Synthetic data reconstruction plot at $K=10$, Additional noise $\mathcal{N}(0, 0.05)$. \textbf{MAE: 0.116, MSE: 0.025}} - \label{fig:fig6} + \label{fig:fig5} \end{figure} \paragraph{Electricity Transformer Temperature}\ @@ -338,13 +349,13 @@ \section{Experiment} \centering \includegraphics[width=\textwidth]{ETT_time_series_K10N005.eps} \caption{ETTh1 data reconstruction plot at $K=10$, Additional Noise $\mathcal{N}(0, 0.05)$. \textbf{MAE: 0.096, MSE: 0.019}} - \label{fig:fig7} + \label{fig:fig6} \end{figure} \begin{table}[!h] \def\arraystretch{2.3} \begin{center} -\caption{Loss on ETTh1 data} +\caption{Loss on ETTh1 data. The same dependence of the error on the value of $K$ as on the synthetic data can be seen. See \ref{fig:fig6} for the example of reconstruction with $K=10$ and noise $\mathcal{N}(0, 0.05)$.} \begin{tabular}{|l||l||*{3}{c|}}\hline {Noise} &\makebox[3em]{Metric}&\makebox[3em]{$K=2$}&\makebox[3em]{$K=4$}&\makebox[3em]{$K=10$}\\\hline