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The typical form of Karhunen-Loeve expansion is on a detrended stochastic process. E.g., let $Y(t)$ be a stochastic process on $[0,T]$ , and let $X(t) = Y(t)-\mathbb{E}Y(t)$ with a continuous covariance function $R_X(s,t)$ , then there exists a series of orthonormal functions $\phi_k(t)$ on $[0,T]$ and independent zero-mean and normalized random variables $Z_k$ such that $$ X(t) = \sum_{k=1}^{\infty} Z_k \sqrt{\lambda_k} \phi_k(t) $$ where $\lambda_k$ and $\phi_k(t)$ are the solutions of the following eigensystems equation: $$ \int_0^T R_X(s,t)\phi_k(t)dt=\lambda_k\phi_k(s) $$ My question is, if we do not subtract the mean from the process, and just focus on the raw autocorrelation function $R_Y(s,t)$ , as $R_Y(s,t)$ is also positive definite symmetric functions in $s,t$ , it has a spectral decomposition (Mercer's theorem) $$ R_Y(s,t) = \sum_k \mu_k \psi_k(s)\psi_k(t) $$ Can we expand the original process $Y(t)$ in $\psi(t)$ ? And does this decomposition enjoy some similar properties of the original K-L decomposition, such as the optimal compaction of energies of the signal in the eigenfunctions? $$ Y(t)=\sum_k \tilde{Z}_k \sqrt{\mu_k} \psi_k(t) $$ The empirical reason is that sometimes we are interested in the information contained in the mean as well, not only in the fluctuation around the mean. While the original K-L expansion optimally compacts the total variance into the first few eigenfunctions $\phi_k(t)$ , the direct decomposition of $Y(t)$ may optimally compact the total fluctuation around $0$ , i.e., the raw energy of the signal: $\mathbb{E}\int_0^T |Y(t)|^2dt$ into $\psi_k(t)$ . Has it been shown to be true? And if true, is the result trivial in the sense that $\psi_k(t)$ and $\phi_k(t)$ are connected in some simple way?
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[Skip to main content](https://math.stackexchange.com/questions/4099490/karhunen-loeve-expansion-of-non-centered-processes#content) #### Stack Exchange Network Stack Exchange network consists of 183 Q\&A communities including [Stack Overflow](https://stackoverflow.com/), the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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[Learn more about Teams](https://stackoverflow.co/teams/) # [Karhunen-Loeve expansion of non-centered processes](https://math.stackexchange.com/questions/4099490/karhunen-loeve-expansion-of-non-centered-processes) [Ask Question](https://math.stackexchange.com/questions/ask) Asked 4 years, 4 months ago Modified [3 years, 3 months ago](https://math.stackexchange.com/questions/4099490/karhunen-loeve-expansion-of-non-centered-processes?lastactivity "2022-04-26 07:52:24Z") Viewed 445 times 7 \$\\begingroup\$ The typical form of Karhunen-Loeve expansion is on a detrended stochastic process. E.g., let \$Y(t)\$ be a stochastic process on \$\[0,T\]\$, and let \$X(t) = Y(t)-\\mathbb{E}Y(t)\$ with a continuous covariance function \$R\_X(s,t)\$, then there exists a series of orthonormal functions \$\\phi\_k(t)\$ on \$\[0,T\]\$ and independent zero-mean and normalized random variables \$Z\_k\$ such that \$\$ X(t) = \\sum\_{k=1}^{\\infty} Z\_k \\sqrt{\\lambda\_k} \\phi\_k(t) \$\$ where \$\\lambda\_k\$ and \$\\phi\_k(t)\$ are the solutions of the following eigensystems equation: \$\$ \\int\_0^T R\_X(s,t)\\phi\_k(t)dt=\\lambda\_k\\phi\_k(s) \$\$ My question is, if we do not subtract the mean from the process, and just focus on the raw autocorrelation function \$R\_Y(s,t)\$, as \$R\_Y(s,t)\$ is also positive definite symmetric functions in \$s,t\$, it has a spectral decomposition (Mercer's theorem) \$\$ R\_Y(s,t) = \\sum\_k \\mu\_k \\psi\_k(s)\\psi\_k(t) \$\$ Can we expand the original process \$Y(t)\$ in \$\\psi(t)\$? And does this decomposition enjoy some similar properties of the original K-L decomposition, such as the optimal compaction of energies of the signal in the eigenfunctions? \$\$ Y(t)=\\sum\_k \\tilde{Z}\_k \\sqrt{\\mu\_k} \\psi\_k(t) \$\$ The empirical reason is that sometimes we are interested in the information contained in the mean as well, not only in the fluctuation around the mean. While the original K-L expansion optimally compacts the total **variance** into the first few eigenfunctions \$\\phi\_k(t)\$, the direct decomposition of \$Y(t)\$ may optimally compact the total fluctuation around \$0\$, i.e., the raw energy of the signal: \$\\mathbb{E}\\int\_0^T \|Y(t)\|^2dt\$ into \$\\psi\_k(t)\$. Has it been shown to be true? And if true, is the result trivial in the sense that \$\\psi\_k(t)\$ and \$\\phi\_k(t)\$ are connected in some simple way? - [probability](https://math.stackexchange.com/questions/tagged/probability "show questions tagged 'probability'") - [stochastic-processes](https://math.stackexchange.com/questions/tagged/stochastic-processes "show questions tagged 'stochastic-processes'") - [spectral-theory](https://math.stackexchange.com/questions/tagged/spectral-theory "show questions tagged 'spectral-theory'") - [signal-processing](https://math.stackexchange.com/questions/tagged/signal-processing "show questions tagged 'signal-processing'") [Share](https://math.stackexchange.com/q/4099490 "Short permalink to this question") Cite Follow asked Apr 12, 2021 at 15:52 [![yixianshuiesuan's user avatar](https://lh3.googleusercontent.com/-PODzdRDtHDM/AAAAAAAAAAI/AAAAAAAAAfc/n3fC4ei3Sxw/s64-rj/photo.jpg)](https://math.stackexchange.com/users/525589/yixianshuiesuan) [yixianshuiesuan](https://math.stackexchange.com/users/525589/yixianshuiesuan)yixianshuiesuan 9755 bronze badges \$\\endgroup\$ 1 - \$\\begingroup\$ isn't \$R\_X\$ the same as \$R\_Y\$? \$\\endgroup\$ – [G. Gare](https://math.stackexchange.com/users/568973/g-gare "1,555 reputation") Commented Apr 25, 2022 at 7:54 [Add a comment](https://math.stackexchange.com/questions/4099490/karhunen-loeve-expansion-of-non-centered-processes "Use comments to ask for more information or suggest improvements. Avoid answering questions in comments.") \| ## 1 Answer 1 Sorted by: [Reset to default](https://math.stackexchange.com/questions/4099490/karhunen-loeve-expansion-of-non-centered-processes?answertab=scoredesc#tab-top) 2 \$\\begingroup\$ It is trivial to see that \\begin{align} R\_X(t,s)&=E\[(X(t) - E\[X(t)\])(X(s)-E\[X(s)\])\] \\\\ &= E\[(Y(t) - E\[Y(t)\])(Y(s)-E\[Y(s)\])\] = R\_Y(t,s). \\end{align} Since both Mercer's theorem and the KL decomposition use an eigendecomposition of \$R\_X\$ (and thus \$R\_Y\$), we have that \$\\psi\_k\$ and \$\\phi\_k\$ are actually the same vectors. In order to find a decomposition of the non-centered process \$Y(t)\$ it suffices to notice that \$\\phi\_k\$ is a complete orthonormal basis of some Lebesgue space on \$\[0,T\]\$ (e.g. \$L^2\$) - in most cases - and therefore denoting \$y(t) = E\[Y(t)\]\$ there exist scalars \$\\{y\_k\\}\$ such that \$\$ y(t) = \\sum\_{k} y\_k \\sqrt{\\lambda\_k} \\phi\_k(t). \$\$ You can then write the process \$Y(t)\$ as \$\$ Y(t) = X(t) + y(t) =\\sum\_k (Z\_k+y\_k)\\sqrt{\\lambda\_k}\\phi\_k. \$\$ [Share](https://math.stackexchange.com/a/4436519 "Short permalink to this answer") Cite Follow answered Apr 26, 2022 at 7:52 [![G. Gare's user avatar](https://www.gravatar.com/avatar/7ca20256620a9ffb1230c159137af53e?s=64&d=identicon&r=PG&f=y&so-version=2)](https://math.stackexchange.com/users/568973/g-gare) [G. Gare](https://math.stackexchange.com/users/568973/g-gare)G. 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