Abstract
In this paper, we present a method of estimating the partial power spectrum of a bivariate point process. The method is based on the simple and cross-periodograms which are smoothed by applying a simple averaging weighting scheme in order to improve the properties of the estimate. The asymptotic distribution of the estimate is shown to be approximately normal when the amount of smoothing is large. An illustrative example is presented from the field of Neurophysiology when the complex system of the muscle spindle is affected by a static gamma motoneuron.
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Acknowledgements
The authors are most grateful to Professor Ken Linsday and to Professor Jay Rosenberg for their kindness to provide the Neurophysiological data set.
Funding
This work was supported by European Regional Development Fund-Project “SINGING PLANT” \((No.\,CZ.02.1.01/0.0/0.0/16\_026/0008446)\) which received a financial contribution from the Ministry of Education, Youths and Sports of the Czech Republic in the form of special support through the National Program for Sustainability II funds.
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Appendix
Appendix
1.1 Proofs of Theorems
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Proof of Theorem 1 Let \(g: \; R^3\rightarrow R\) be a differentiable function with \(g(x,y,z)=\frac{xy}{z},\; z \ne 0\). By setting \(x=f_{kl}(\lambda ), \; y=f_{lk}(\lambda )\) and \(z=f_{ll}(\lambda )\), and using the multivariate delta method, we get that [27]:
$$\begin{aligned} g(\hat{x},\hat{y},\hat{z})-g(x,y,z)=\frac{f_{kl}^{(T)}(\lambda ) f_{lk}^{(T)}(\lambda )}{f_{ll}^{(T)}(\lambda )}-\frac{f_{kl}(\lambda )f_{lk}(\lambda )}{f_{ll}(\lambda )}, \end{aligned}$$which can be expressed as follows:
$$\begin{aligned}&g(\hat{x},\hat{y},\hat{z})-g(x,y,z)=\frac{\partial g}{\partial x}(\hat{x}-x)+\frac{\partial g}{\partial y}(\hat{y}-y)+ \frac{\partial g}{\partial z}(\hat{z}-z)\\&\quad =\frac{f_{lk}(\lambda )}{f_{ll}(\lambda )} \lbrace f_{kl}^{(T)}(\lambda )-f_{kl}(\lambda ) \rbrace +\frac{f_{kl}(\lambda )}{f_{ll}(\lambda )} \lbrace f_{lk}^{(T)}(\lambda )-f_{lk}(\lambda ) \rbrace \\&\qquad -\frac{f_{kl}(\lambda )f_{lk}(\lambda )}{f_{ll}^2(\lambda )} \lbrace f_{ll}^{(T)}(\lambda )-f_{ll}(\lambda ) \rbrace . \end{aligned}$$On using definitions (3) and (13), we find
$$\begin{aligned} f_{kk.l}^{(T)}(\lambda )-f_{kk.l}(\lambda )= & {} \lbrace f_{kk}^{(T)}(\lambda )-f_{kk}(\lambda ) \rbrace - \frac{f_{lk}(\lambda )}{f_{ll}(\lambda )} \lbrace f_{kl}^{(T)}(\lambda )-f_{kl}(\lambda ) \rbrace \nonumber \\&-\frac{f_{kl}(\lambda )}{f_{ll}(\lambda )}\lbrace f_{lk}^{(T)}(\lambda )-f_{lk}(\lambda ) \rbrace + \frac{f_{kl}(\lambda )f_{lk}(\lambda )}{f_{ll}^2(\lambda )} \lbrace f_{ll}^{(T)}(\lambda )-f_{ll}(\lambda ) \rbrace . \end{aligned}$$(17)It now follows from (17) that:
$$\begin{aligned} E \lbrace f_{kk.l}^{(T)}(\lambda ) \rbrace \simeq f_{kk.l}(\lambda ), \end{aligned}$$since \(E \lbrace f_{kk}^{(T)}(\lambda ) \rbrace \simeq f_{kk}(\lambda )\) and \(E \lbrace f_{kl}^{(T)}(\lambda ) \rbrace \simeq f_{kl}(\lambda ), \; (k \ne l)\). The covariance between \(f_{kk.l}^{(T)}(\lambda )\) and \(f_{kk.l}^{(T)}(\mu )\) is obtained as follows:
$$\begin{aligned} \mathrm {Cov} \lbrace f_{kk.l}^{(T)}(\lambda )-f_{kk.l}(\lambda ),f_{kk.l}^{(T)}(\mu )-f_{kk.l}(\mu ) \rbrace =\mathrm {Cov} \lbrace f_{kk.l}^{(T)}(\lambda ),f_{kk.l}^{(T)}(\mu ) \rbrace \end{aligned}$$and
$$\begin{aligned}&\mathrm {Cov} \lbrace f_{kk.l}^{(T)}(\lambda ),f_{kk.l}^{(T)}(\mu ) \rbrace =\mathrm {Cov} \lbrace f_{kk}^{(T)}(\lambda ),f_{kk}^{(T)}(\mu ) \rbrace - \frac{f_{lk}(-\mu )}{f_{ll}(\mu )} \mathrm {Cov} \lbrace f_{kk}^{(T)}(\lambda ),f_{kl}^{(T)}(\mu ) \rbrace \\&\quad -\frac{f_{kl}(-\mu )}{f_{ll}(\mu )} \mathrm {Cov} \lbrace f_{kk}^{(T)}(\lambda ),f_{lk}^{(T)}(\mu ) \rbrace + \frac{\mid f_{lk}(\mu ) \mid ^2}{f_{ll}^2 (\mu )}\mathrm {Cov} \lbrace f_{kk}^{(T)}(\lambda ),f_{ll}^{(T)}(\mu ) \rbrace \\&\quad - \frac{f_{lk}(\lambda )}{f_{ll}(\lambda )} \mathrm {Cov} \lbrace f_{kl}^{(T)}(\lambda ),f_{kk}^{(T)}(\mu ) \rbrace + \frac{f_{lk}(\lambda )}{f_{ll}(\lambda )} \frac{f_{lk}(-\mu )}{f_{ll}(\mu )}\mathrm {Cov} \lbrace f_{kl}^{(T)}(\lambda ),f_{kl}^{(T)}(\mu ) \rbrace \\&\quad +\frac{f_{lk}(\lambda )}{f_{ll}(\lambda )} \frac{f_{kl}(-\mu )}{f_{ll}(\mu )}\mathrm {Cov} \lbrace f_{kl}^{(T)}(\lambda ),f_{lk}^{(T)}(\mu ) \rbrace - \frac{f_{lk}(\lambda )}{f_{ll}(\lambda )} \frac{\mid f_{kl}(\mu ) \mid ^2}{f_{ll}^2 (\mu )} \mathrm {Cov} \lbrace f_{kl}^{(T)}(\lambda ),f_{ll}^{(T)}(\mu ) \rbrace \\&\quad - \frac{f_{kl}(\lambda )}{f_{ll}(\lambda )} \mathrm {Cov} \lbrace f_{lk}^{(T)}(\lambda ),f_{kk}^{(T)}(\mu ) \rbrace + \frac{f_{kl}(\lambda )}{f_{ll}(\lambda )} \frac{f_{lk}(-\mu )}{f_{ll}(\mu )} \mathrm {Cov} \lbrace f_{lk}^{(T)}(\lambda ),f_{kl}^{(T)}(\mu ) \rbrace \\&\quad + \frac{f_{kl}(\lambda )}{f_{ll}(\lambda )} \frac{f_{kl}(-\mu )}{f_{ll}(\mu )} \mathrm {Cov} \lbrace f_{lk}^{(T)}(\lambda ),f_{lk}^{(T)}(\mu ) \rbrace -\frac{f_{kl}(\lambda )}{f_{ll}(\lambda )} \frac{\mid f_{kl}(\mu ) \mid ^2}{f_{ll}^2 (\mu )} \mathrm {Cov} \lbrace f_{lk}^{(T)}(\lambda ),f_{ll}^{(T)}(\mu ) \rbrace \\&\quad + \frac{\mid f_{kl}(\lambda ) \mid ^2}{f_{ll}^2(\lambda )} \mathrm {Cov} \lbrace f_{ll}^{(T)}(\lambda ),f_{kk}^{(T)}(\mu ) \rbrace - \frac{\mid f_{kl}(\lambda ) \mid ^2}{f_{ll}^2(\lambda )} \frac{f_{lk}(-\mu )}{f_{ll}(\mu )} \mathrm {Cov} \lbrace f_{ll}^{(T)}(\lambda ),f_{kl}^{(T)}(\mu ) \rbrace \\&\quad -\frac{\mid f_{kl}(\lambda ) \mid ^2}{f_{ll}^2(\lambda )} \frac{f_{kl}(-\mu )}{f_{ll}(\mu )} \mathrm {Cov} \lbrace f_{ll}^{(T)}(\lambda ),f_{lk}^{(T)}(\mu ) \rbrace \\&\quad + \frac{\mid f_{kl}(\lambda ) \mid ^2}{f_{ll}^2(\lambda )} \frac{\mid f_{kl}(\mu ) \mid ^2}{f_{ll}^2(\mu )} \mathrm {Cov} \lbrace f_{ll}^{(T)}(\lambda ),f_{ll}^{(T)}(\mu ) \rbrace , \end{aligned}$$since \(f_{lk}(\lambda )=f_{kl}(-\lambda )\), \(\mid f_{lk}(\lambda ) \mid ^2 = \mid f_{kl}(\lambda ) \mid ^2\) and
$$\begin{aligned} \mathrm {Cov} \lbrace f_{kl}^{(T)} (\lambda ), a(\lambda )f_{kl}^{(T)}(\mu ) \rbrace = \overline{a(\lambda )} \mathrm {Cov} \lbrace f_{kl}^{(T)} (\lambda ), f_{kl}^{(T)}(\mu ) \rbrace , \end{aligned}$$where \(\overline{a(\lambda )}\) is the conjugate function of the complex function \(a(\lambda )\). If we substitute in the above relation the covariances from (12), we obtain the required result. If we set \(\lambda =\mu\) in the \(\mathrm {Cov} \lbrace f_{kk.l}^{(T)} (\lambda ), f_{kk.l}^{(T)}(\mu ) \rbrace\), we get
$$\begin{aligned}&\mathrm {Var} \lbrace f_{kk.l}^{(T)}(\lambda ) \rbrace \simeq \frac{S}{T} \frac{C_p}{2m+1} \left[f_{kk}^2 (\lambda )-\frac{f_{kl}(\lambda )f_{kl}(-\lambda )f_{kk}(\lambda )}{f_{ll}(\lambda )}\right.\\&\left. \qquad -\frac{\mid f_{kl}(\lambda ) \mid ^2}{f_{ll}(\lambda )}f_{kk}(\lambda )+\frac{\mid f_{kl}(\lambda ) \mid ^2 f_{kl}(\lambda ) f_{kl}(-\lambda )}{f_{ll}^2(\lambda )}-\frac{\mid f_{kl}(\lambda ) \mid ^2 f_{kk}(\lambda )}{f_{ll}(\lambda )}\right.\\\\&\left.\qquad +\frac{\mid f_{kl}(\lambda ) \mid ^2 f_{kk}(\lambda )}{f_{ll}(\lambda )}+\frac{\mid f_{kl}(\lambda ) \mid ^2 \mid f_{kl}(\lambda ) \mid ^2}{f_{ll}^2(\lambda )}-\frac{\mid f_{kl}(\lambda ) \mid ^2 \mid f_{lk}(\lambda ) \mid ^2}{f_{ll}^2(\lambda )}\right.\\&\left.\qquad -\frac{\mid f_{lk}(\lambda ) \mid ^2 f_{kk}(\lambda )}{f_{ll}(\lambda )}+ \frac{\mid f_{kl}(\lambda ) \mid ^2 \mid f_{kl}(\lambda ) \mid ^2}{f_{ll}^2(\lambda )}+\frac{f_{kl}(\lambda )f_{kl}(-\lambda )f_{kk}(\lambda )}{f_{ll}(\lambda )}\right.\\&\left. \qquad - \frac{\mid f_{kl}(\lambda ) \mid ^2 f_{kl}(\lambda )f_{kl}(-\lambda )}{f_{ll}^2(\lambda )}+\frac{\mid f_{kl}(\lambda ) \mid ^2 f_{lk}(\lambda )f_{lk}(-\lambda )}{f_{ll}^2(\lambda )}-\frac{\mid f_{kl}(\lambda ) \mid ^2 \mid f_{kl}(\lambda ) \mid ^2}{f_{ll}^2(\lambda )}\right.\\&\qquad \left. -\frac{\mid f_{kl}(\lambda ) \mid ^2 f_{lk}(\lambda )f_{lk}(-\lambda )}{f_{ll}^2(\lambda )}+\frac{\mid f_{kl}(\lambda ) \mid ^2 \mid f_{kl}(\lambda ) \mid ^2}{f_{ll}^2(\lambda )} \right]\\&\quad =\frac{S}{T}\frac{C_p}{2m+1} \left [f_{kk}^2(\lambda )-\frac{2\mid f_{kl}(\lambda ) \mid ^2 f_{kk}(\lambda )}{f_{ll}(\lambda )}+\frac{\mid f_{kl}(\lambda ) \mid ^2 \mid f_{kl}(\lambda ) \mid ^2}{f_{ll}^2(\lambda )}\right]\\&\quad =\frac{S}{T}\frac{C_p}{2m+1} f_{kk}^2(\lambda )\left[1-2 \mid R_{kl}(\lambda ) \mid ^2 + \mid R_{kl}(\lambda ) \mid ^4 \right]\\&\quad =\frac{S}{T}\frac{C_p}{2m+1}\left [f_{kk}(\lambda )(1-\mid R_{kl}(\lambda ) \mid ^2)\right]^2=\frac{S}{T}\frac{C_p}{2m+1}f_{kk.l}^2(\lambda ). \end{aligned}$$ -
Proof of Proposition 1 The distribution of \(f_{kk}^{(T)}(\lambda _j)\) is a multiple of a \(\chi ^2\)-distribution with \(2(2m+1)\) degrees of freedom. Moreover, the variates \(\lbrace f_{kk}^{(T)}(\lambda _j) \rbrace ,\, j=1,2,\ldots ,J\) are asymptotically independent distributed. For large m, the variates \(\lbrace f_{kk}^{(T)}(\lambda _j) \rbrace ,\, j=1,2,\ldots ,J\) are asymptotically independent normally distributed. Also, the distribution of \(f_{kl}^{(T)}(\lambda _j)\) follows a Wishart distribution, \(\frac{1}{2m+1}W_1^C \lbrace 2m+1, f_{kl}(\lambda _j) \rbrace\). The variates \(\lbrace f_{kl}^{(T)}(\lambda _j) \rbrace ,\, j=1,2,\ldots ,J\) are asymptotically independent distributed. For large m the Wishart distribution tends to a normal distribution. For more details refer to [12].
For large m, the estimate of PPS approximated by (17) is the sum of 4 independent normal distributions, therefore tends to a normal distribution with mean and variance obtained in Theorem 1.
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Proof of Proposition 2: This follows in the same way as the logarithm to base 10 of the estimate of the PS [12].
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Michailidis, G.E., Spyroglou, I.I., Zaridis, D. et al. A Method of Estimating the Partial Power Spectrum of a Bivariate Point Process and an Application to a Neurophysiological Data Set. J Stat Theory Pract 14, 36 (2020). https://doi.org/10.1007/s42519-020-00105-8
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DOI: https://doi.org/10.1007/s42519-020-00105-8