Refactor

paper/Divilkovskiy2024SourceSpace_en.aux

@@ -41,26 +41,26 @@
 \newlabel{fig:fig5}{{\caption@xref {fig:fig5}{ on input line 196}}{10}{Pairwise correlation between time series}{figure.caption.2}{}}
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 \@writefile{lof}{\contentsline {figure}{\numberline {4}{\ignorespaces Prediction reconstruction in case of inaccurate correlation matrix $\mathbf  {\Sigma }$ prediction. In addition to the prediction error caused by noise in the correlation matrix a new type of error is added. When selecting a set, it is possible that the diameter is minimised not at the right set, as this is only a necessary condition, but not a sufficient one.\relax }}{15}{figure.caption.4}\protected@file@percent }
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paper/Divilkovskiy2024SourceSpace_en.tex

@@ -209,14 +209,14 @@ \section{Pairwise correlation between time series}
 	\item Similarly, express $\mathbf{\Sigma_t}$ through the values of the series:
 		\begin{gather*}
 		\mathbf{\Sigma}_t = \frac{1}{t} \sum_{i=1}^{t} (\mathbf{x}_i-\boldsymbol{\mu}_t)(\mathbf{x}_i-\boldsymbol{\mu}_t)^\intercal;\\
-		\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\\
+		\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:
 	\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 =\\
-		= \frac{t}{t+1}\mathbf{\Sigma}_t + \frac{t}{t+1}\boldsymbol{\mu}_{t} \boldsymbol{\mu}_{t}^\intercal + \frac{t}{(t+1)^2} \mathbf{x}_{t+1} \mathbf{x}_{t+1}^\intercal - \frac{t^2}{(t+1)^2}\boldsymbol{\mu}_{t} \boldsymbol{\mu}_{t}^\intercal - \frac{t}{(t+1)^2}\boldsymbol{\mu}_{t} \mathbf{x}_{t+1}^\intercal - \frac{t}{(t+1)^2}\mathbf{x}_{t+1} \boldsymbol{\mu}_{t}^\intercal =\\
+		= \frac{t}{t+1}\mathbf{\Sigma}_t + \frac{t}{t+1}\boldsymbol{\mu}_{t} \boldsymbol{\mu}_{t}^\intercal + \frac{t}{(t+1)^2} \mathbf{x}_{t+1} \mathbf{x}_{t+1}^\intercal -\\- \frac{t^2}{(t+1)^2}\boldsymbol{\mu}_{t} \boldsymbol{\mu}_{t}^\intercal - \frac{t}{(t+1)^2}\boldsymbol{\mu}_{t} \mathbf{x}_{t+1}^\intercal - \frac{t}{(t+1)^2}\mathbf{x}_{t+1} \boldsymbol{\mu}_{t}^\intercal =\\
 		= \frac{t}{t+1}\mathbf{\Sigma}_t + \frac{t}{t+1}\boldsymbol{\mu}_{t} \boldsymbol{\mu}_{t}^\intercal - \frac{t}{(t+1)^2} \left( -\mathbf{x}_{t+1} \mathbf{x}_{t+1}^\intercal + t\boldsymbol{\mu}_{t} \boldsymbol{\mu}_{t}^\intercal + \boldsymbol{\mu}_{t} \mathbf{x}_{t+1}^\intercal + \mathbf{x}_{t+1} \boldsymbol{\mu}_{t}^\intercal \right) =\\
 		= \frac{t}{t+1}\mathbf{\Sigma}_t + \frac{t}{t+1}\boldsymbol{\mu}_{t} \boldsymbol{\mu}_{t}^\intercal - \frac{t}{(t+1)^2} \left( -(\mathbf{x}_{t+1}-\boldsymbol{\mu}_t)(\mathbf{x}_{t+1}-\boldsymbol{\mu}_t)^\intercal + (t+1)\boldsymbol{\mu}_{t} \boldsymbol{\mu}_{t}^\intercal \right) =\\
 		= \frac{t}{t+1}\mathbf{\Sigma}_t + \frac{t}{t+1}\boldsymbol{\mu}_{t} \boldsymbol{\mu}_{t}^\intercal - \frac{t(t+1)}{(t+1)^2}\boldsymbol{\mu}_{t} \boldsymbol{\mu}_{t}^\intercal + \frac{t}{(t+1)^2}(\mathbf{x}_{t+1}-\boldsymbol{\mu}_t)(\mathbf{x}_{t+1}-\boldsymbol{\mu}_t)^\intercal =\\
@@ -226,7 +226,7 @@ \section{Pairwise correlation between time series}
 
 	\item We solve the problem of finding the minimum of the function $||\hat{\mathbf{\Sigma}}_{t+1} - \bar{\mathbf{\Sigma}}_{t+1}||_2^2$. In our case, this is analogous to the equality of this function to zero. Let us write the expression under the norm:
 	\begin{gather*}
-		\hat{\mathbf{\Sigma}}_{t+1} - \bar{\mathbf{\Sigma}}_{t+1} = \frac{t}{t+1}\mathbf{\Sigma}_t + \frac{t}{(t+1)^2}(\hat{\mathbf{x}}_{t+1}-\boldsymbol{\mu}_t)(\hat{\mathbf{x}}_{t+1}-\boldsymbol{\mu}_t)^\intercal - \frac{t}{t+1}\mathbf{\Sigma}_t + \frac{t}{(t+1)^2}(\bar{\mathbf{x}}_{t+1}-\boldsymbol{\mu}_t)(\bar{\mathbf{x}}_{t+1}-\boldsymbol{\mu}_t)^\intercal =\\
+		\hat{\mathbf{\Sigma}}_{t+1} - \bar{\mathbf{\Sigma}}_{t+1} = \frac{t}{t+1}\mathbf{\Sigma}_t + \frac{t}{(t+1)^2}(\hat{\mathbf{x}}_{t+1}-\boldsymbol{\mu}_t)(\hat{\mathbf{x}}_{t+1}-\boldsymbol{\mu}_t)^\intercal -\\- \frac{t}{t+1}\mathbf{\Sigma}_t + \frac{t}{(t+1)^2}(\bar{\mathbf{x}}_{t+1}-\boldsymbol{\mu}_t)(\bar{\mathbf{x}}_{t+1}-\boldsymbol{\mu}_t)^\intercal =\\
 		=\frac{t}{(t+1)^2}(\hat{\mathbf{x}}_{t+1}-\boldsymbol{\mu}_t)(\hat{\mathbf{x}}_{t+1}-\boldsymbol{\mu}_t)^\intercal -\frac{t}{(t+1)^2}(\bar{\mathbf{x}}_{t+1}-\boldsymbol{\mu}_t)(\bar{\mathbf{x}}_{t+1}-\boldsymbol{\mu}_t)^\intercal.
  	\end{gather*}
  	Then, the problem above is equivalent to finding the minimum of the function \[||(\hat{\mathbf{x}}_{t+1}-\boldsymbol{\mu}_t)(\hat{\mathbf{x}}_{t+1}-\boldsymbol{\mu}_t)^\intercal-(\bar{\mathbf{x}}_{t+1}-\boldsymbol{\mu}_t)(\bar{\mathbf{x}}_{t+1}-\boldsymbol{\mu}_t)^\intercal||_2^2.\]