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| Boilerpipe Text | 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? |
| Markdown | [Skip to main content](https://math.stackexchange.com/questions/4099490/karhunen-loeve-expansion-of-non-centered-processes#content)
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# [Karhunen-Loeve expansion of non-centered processes](https://math.stackexchange.com/questions/4099490/karhunen-loeve-expansion-of-non-centered-processes)
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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")
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\$\\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?
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- \$\\begingroup\$ isn't \$R\_X\$ the same as \$R\_Y\$? \$\\endgroup\$
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Commented Apr 25, 2022 at 7:54
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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. \$\$
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