Conditional Granger causality and partitioned Granger causality: differences and similarities

Abstract

Neural information modeling and analysis often requires a measurement of the mutual influence among many signals. A common technique is the conditional Granger causality (cGC) which measures the influence of one time series on another time series in the presence of a third. Geweke has translated this condition into the frequency domain and has explored the mathematical relationships between the time and frequency domain expressions. Chen has observed that in practice, the expressions may return (meaningless) negative numbers, and has proposed an alternative which is based on a partitioned matrix scheme, which we call partitioned Granger causality (pGC). There has been some confusion in the literature about the relationship between cGC and pGC; some authors treat them as essentially identical measures, while others have noted that some properties (such as the relationship between the time and frequency domain expressions) do not hold for the pGC. This paper presents a series of matrix equalities that simplify the calculation of the pGC. In this simplified expression, the essential differences and similarities between the cGC and the pGC become clear; in essence, the pGC is dependent on only a subset of the parameters in the model estimation, and the noise residuals (which are uncorrelated in the cGC) need not be uncorrelated in the pGC. The mathematical results are illustrated with a simulation, and the measures are applied to an EEG dataset.

Appendices

Appendix 1: Proof of Lemma 1

Proof

Equation (13) holds because

$$\begin{aligned}&\left( \begin{array}{ccc} \mathbf A (f) &{}\quad \mathbf 0 &{}\quad \mathbf B (f) \\ \mathbf 0 &{}\quad \mathbf I (f) &{}\quad \mathbf 0 \\ \mathbf C (f) &{}\quad \mathbf 0 &{}\quad \mathbf D (f) \\ \end{array} \right) \left( \begin{array}{ccc} \mathbf A ^{\prime }(f) &{}\quad \mathbf 0 &{}\quad \mathbf B ^{\prime }(f) \\ \mathbf 0 &{}\quad \mathbf I (f) &{}\quad \mathbf 0 \\ \mathbf C ^{\prime }(f) &{}\quad \mathbf 0 &{}\quad \mathbf D '(f) \\ \end{array} \right) \nonumber \\&\quad =\left( \begin{array}{ccc} \mathbf A ^{\prime }(f) &{}\quad \mathbf 0 &{}\quad \mathbf B ^{\prime }(f) \\ \mathbf 0 &{}\quad \mathbf I (f) &{}\quad \mathbf 0 \\ \mathbf C ^{\prime }(f) &{}\quad \mathbf 0 &{}\quad \mathbf D ^{\prime }(f) \\ \end{array} \right) \left( \begin{array}{ccc} \mathbf A (f) &{}\quad \mathbf 0 &{}\quad \mathbf B (f) \\ \mathbf 0 &{}\quad \mathbf I (f) &{}\quad \mathbf 0 \\ \mathbf C (f) &{}\quad \mathbf 0 &{}\quad \mathbf D (f) \\ \end{array} \right) \nonumber \\&\quad =\mathbf I . \end{aligned}$$

(33)

$\square$

Appendix 2: Proof of Lemma 2

Proof

X(f) and Y(f) calculated from (26) are equal to X(f) and Y(f) calculated from (25).$\square$

Appendix 3: Proof of Lemma 3

Proof

If $k=n$, then an invertible matrix $\mathbf B (f)$ can be chosen; hence, $\mathbf A (f)\mathbf B (f)\mathbf B ^{-1}(f)=\mathbf C (f)\mathbf B (f)\mathbf B ^{-1}(f)$. Therefore, $\mathbf A (f)=\mathbf C (f)$ (Kolman and Hill 2008).

However, if $k\ne n$, then we can define $\mathbf B _{i,j}(f)$ whose elements are zero except for one element in row i and column j, which is one. In that case, $\mathbf A (f)\mathbf B _{i,j}(f)=\mathbf C (f)\mathbf B _{i,j}(f)$. Therefore, the ith column of $\mathbf A (f)$ equals the ith column of $\mathbf C (f)$. Since this holds for all $i \in \{ 1,\dots , n \}, \mathbf A (f)=\mathbf C (f)$.$\square$

Appendix 4: Proof of Lemma 4

Proof

Since $(\mathbf AB )^{-1}=\mathbf B ^{-1}\mathbf A ^{-1}$, (21) can be rewritten as

$$\begin{aligned} \mathbf G _\mathrm{sub}^{-1}(f)=\left( \begin{array}{cc} \mathbf I _m &{}\quad \mathbf 0 \\ -\varvec{\Gamma }(f) &{}\quad \mathbf I _k \\ \end{array} \right) \left( \begin{array}{cc} \mathbf H _{\textit{XX}}(f) &{}\quad \mathbf H _{\textit{XZ}}(f) \\ \mathbf H _{\textit{ZX}}(f) &{}\quad \mathbf H _{\textit{ZZ}}(f) \\ \end{array} \right) ^{-1}. \end{aligned}$$

(34)

Express the inverse $\left( {\begin{array}{cc} \mathbf H _{\textit{XX}}(f) &{}\quad \mathbf H _{\textit{XZ}}(f) \\ \mathbf H _{\textit{ZX}}(f) &{}\quad \mathbf H _{\textit{ZZ}}(f) \\ \end{array}} \right) ^{-1}$ in form as $\left( \begin{array}{cc} \mathbf L (f) &{}\quad \mathbf M (f) \\ \mathbf N (f) &{}\quad \mathbf K (f) \\ \end{array} \right)$, then

$$\begin{aligned} \mathbf G _\mathrm{sub}^{-1}(f)&=\left( \begin{array}{c@{\quad }c} \mathbf I _m &{} \mathbf 0 \\ -\varvec{\Gamma }(f) &{} \mathbf I _k \\ \end{array} \right) \left( \begin{array}{cc} \mathbf L (f) &{}\quad \mathbf M (f) \\ \mathbf N (f) &{}\quad \mathbf K (f) \\ \end{array} \right) \nonumber \\&= \left( \begin{array}{cc} \mathbf L (f) &{}\quad \mathbf M (f) \\ -\varvec{\Gamma }(f)\mathbf L (f)+\mathbf N (f) &{}\quad -\varvec{\Gamma }(f)\mathbf M (f)+\mathbf K (f) \nonumber \\ \end{array} \right) \\&=\left( \begin{array}{cc} \mathbf G _{\textit{XX}}(f) &{}\quad \mathbf G _{\textit{XZ}}(f) \\ \mathbf G _{\textit{ZX}}(f) &{}\quad \mathbf G _{\textit{ZZ}}(f) \\ \end{array} \right) ^{-1}. \end{aligned}$$

(35)

Moreover, according to (13) in Lemma 1

$$\begin{aligned} \mathbf G (f)&=\left( \begin{array}{ccc} \mathbf G _{\textit{XX}}(f) &{}\quad \mathbf 0 &{}\quad \mathbf G _{\textit{XZ}}(f)\\ \mathbf 0 &{}\quad \mathbf I &{}\quad \mathbf 0 \\ \mathbf G _{\textit{ZX}}(f) &{}\quad \mathbf 0 &{}\quad \mathbf G _{\textit{ZZ}}(f) \end{array} \right) ^{-1}\nonumber \\&= \left( \begin{array}{ccc} \mathbf L (f) &{}\quad \mathbf 0 &{}\quad \mathbf M (f)\\ \mathbf 0 &{}\quad \mathbf I &{}\quad \mathbf 0 \\ -\varvec{\Gamma } \mathbf (f) L(f) +\mathbf N(f) &{}\quad \mathbf 0 &{}\quad -\varvec{\Gamma } \mathbf (f) M(f) +\mathbf K(f) \\ \end{array} \right) . \end{aligned}$$

(36)

Using (24), where $\mathbf Q (f)=\mathbf G (f) \mathbf H (f)$

$$\begin{aligned} \mathbf Q (f)&=\left( \begin{array}{ccc} \mathbf Q _{\textit{XX}}(f) &{}\quad \mathbf Q _{\textit{XY}}(f) &{}\quad \mathbf Q _{\textit{XZ}}(f)\\ \mathbf Q _{\textit{YX}}(f) &{}\quad \mathbf Q _{\textit{YY}}(f) &{}\quad \mathbf Q _{YZ}(f)\\ \mathbf Q _{\textit{ZX}}(f) &{}\quad \mathbf Q _{\textit{ZY}}(f)&{}\quad \mathbf Q _{\textit{ZZ}}(f) \end{array} \right) \nonumber \\&=\left( \begin{array}{ccc} \mathbf L (f) &{}\quad \mathbf 0 &{}\quad \mathbf M (f)\\ \mathbf 0 &{}\quad \mathbf I (f) &{}\quad \mathbf 0 \\ -\varvec{\Gamma } \mathbf (f) L(f) +\mathbf N(f) &{}\quad \mathbf 0 &{}\quad -\varvec{\Gamma } \mathbf (f) M(f) +\mathbf K(f) \\ \end{array} \right) \nonumber \\&\quad \times \left( \begin{array}{ccc} \mathbf H _{\textit{XX}}(f)&{}\quad \mathbf H _{\textit{XY}}(f)&{}\quad \mathbf H _{\textit{XZ}}(f)\\ \mathbf H _{\textit{YX}}(f) &{}\quad \mathbf H _{\textit{YY}}(f) &{}\quad \mathbf H _{YZ}(f)\\ \mathbf H _{\textit{ZX}}(f) &{}\quad \mathbf H _{\textit{ZY}}(f) &{}\quad \mathbf H _{\textit{ZZ}}(f)\\ \end{array} \right) \nonumber \\&= \left( \begin{array}{ccc} \mathbf L (f)\mathbf H _{\textit{XX}}(f)+\mathbf M (f)\mathbf H _{\textit{ZX}}(f) &{}\quad \mathbf Q _{\textit{XY}}(f) &{}\quad \mathbf Q _{\textit{XZ}}(f)\\ \mathbf Q _{\textit{YX}}(f) &{}\quad \mathbf Q _{\textit{YY}}(f) &{}\quad \mathbf Q _{YZ}(f)\\ \mathbf Q _{\textit{ZX}}(f) &{}\quad \mathbf Q _{\textit{ZY}}(f) &{}\quad \mathbf Q _{\textit{ZZ}}(f)\\ \end{array} \right) . \end{aligned}$$

(37)

So, $\mathbf Q _{\textit{XX}}(f)=\mathbf L (f)\mathbf H _{\textit{XX}}(f)+\mathbf M (f)\mathbf H _{\textit{ZX}}(f)$. On the other hand, since the matrix $\left( \begin{array}{cc} \mathbf L (f) &{}\quad \mathbf M (f) \\ \mathbf N (f) &{}\quad \mathbf K (f) \\ \end{array} \right)$ is the inverse of $\left( {\begin{array}{cc} \mathbf H _{\textit{XX}}(f) &{}\quad \mathbf H _{\textit{XZ}}(f) \\ \mathbf H _{\textit{ZX}}(f) &{}\quad \mathbf H _{\textit{ZZ}}(f) \\ \end{array}} \right)$, then

$$\begin{aligned} \left( \begin{array}{cc} \mathbf L (f) &{}\quad \mathbf M (f) \\ \mathbf N (f) &{}\quad \mathbf K (f) \\ \end{array} \right) \left( \begin{array}{cc} \mathbf H _{\textit{XX}}(f) &{}\quad \mathbf H _{\textit{XZ}}(f) \\ \mathbf H _{\textit{ZX}}(f) &{}\quad \mathbf H _{\textit{ZZ}}(f) \\ \end{array} \right) =\left( \begin{array}{cc} \mathbf I &{}\quad \mathbf 0 \\ \mathbf 0 &{}\quad \mathbf I \\ \end{array} \right) . \end{aligned}$$

(38)

Therefore, $\mathbf L (f)\mathbf H _{\textit{XX}}(f)+\mathbf M (f)\mathbf H _{\textit{ZX}}(f)=\mathbf I$, and consequently $\mathbf Q _{\textit{XX}}(f)=\mathbf I$.$\square$

Appendix 5: Proof of Lemma 5

Proof

According to (37),

$$\begin{aligned} \mathbf Q _{\textit{XZ}}(f)=\mathbf L (f)\mathbf H _{\textit{XZ}}(f)+\mathbf M (f)\mathbf H _{\textit{ZZ}}(f), \end{aligned}$$

(39)

on the other hand from (38), it can be seen that $\mathbf L (f)\mathbf H _{\textit{XZ}}(f)+\mathbf M (f)\mathbf H _{\textit{ZZ}}(f)=\mathbf 0$; therefore, $\mathbf Q _{\textit{XZ}}(f)=\mathbf 0$.$\square$

Appendix 6: Proof of Lemma 6

Proof

Given (37), $\mathbf Q _{\textit{XY}}(f)=\mathbf L (f)\mathbf H _{\textit{XY}}(f)+\mathbf M (f)\mathbf H _{\textit{ZY}}(f)$. Moreover, by using (38), $\bar{E}_X(f)$ in (23), i.e.,

$$\begin{aligned} \left( \begin{array}{c} \bar{E}_X(f)\\ \bar{E}_Z(f) \end{array}\right)&=\left( \begin{array}{c} E_X(f)\\ E_Z(f) \end{array}\right) +\left( \begin{array}{cc} \mathbf H _{\textit{XX}}(f) &{}\quad \mathbf H _{\textit{XZ}}(f)\\ \mathbf H _{\textit{ZX}}(f) &{}\quad \mathbf H _{\textit{ZZ}}(f)\\ \end{array} \right) ^{-1}\nonumber \\&\quad \times \left( \begin{array}{c} \mathbf H _{\textit{XY}}(f)\\ \mathbf H _{\textit{ZY}}(f) \end{array}\right) E_Y(f), \end{aligned}$$

(40)

can be written as

$$\begin{aligned} \left( \begin{array}{c} \bar{E}_X(f)\\ \bar{E}_Z(f) \end{array}\right)&=\left( \begin{array}{c} E_X(f)\\ E_Z(f) \end{array}\right) +\left( \begin{array}{cc} \mathbf L (f) &{}\quad \mathbf M (f)\\ \mathbf N (f) &{}\quad \mathbf K (f)\\ \end{array} \right) \nonumber \\&\quad \times \left( \begin{array}{c} \mathbf H _{\textit{XY}}(f)\\ \mathbf H _{\textit{ZY}}(f) \end{array}\right) E_Y(f). \end{aligned}$$

(41)

So, $\bar{E}_X(f)=E_X(f)+\mathbf Q _{\textit{XY}}(f)E_Y(f)$, where $\mathbf Q _{\textit{XY}}(f)$ has been calculated in (37). Based on Lemma 2, (22) can be written as

$$\begin{aligned} \left( \begin{array}{c} X(f)\\ T(f)\\ Z(f) \end{array}\right) =\left( \begin{array}{ccc} \mathbf H _{\textit{XX}}(f) &{}\quad \mathbf H _{\textit{XY}}(f) &{}\quad \mathbf H _{\textit{XZ}}(f) \\ \mathbf H _{\textit{YX}}(f) &{}\quad \mathbf H _{\textit{YY}}(f) &{}\quad \mathbf H _{YZ}(f) \\ \mathbf H _{\textit{ZX}}(f) &{}\quad \mathbf H _{\textit{ZY}}(f) &{}\quad \mathbf H _{\textit{ZZ}}(f) \end{array} \right) \left( \begin{array}{c} \bar{E}_X(f)\\ 0 \\ \bar{E}_Z(f) \end{array}\right) , \end{aligned}$$

(42)

where based on (41), $\left( \begin{array}{c} \bar{E}_X(f)\\ 0 \\ \bar{E}_Z(f) \end{array}\right)$ is

$$\begin{aligned} \left( \begin{array}{c} \bar{E}_X(f)\\ 0 \\ \bar{E}_Z(f)\end{array}\right)&=\left( \begin{array}{c} E_X(f)\\ 0 \\ E_Z(f) \end{array}\right) \nonumber \\&\quad +\,\left( \begin{array}{ccc} \mathbf H _{\textit{XX}}(f) &{}\quad \mathbf H _{\textit{XY}}(f) &{}\quad \mathbf H _{\textit{XZ}}(f) \\ \mathbf H _{\textit{YX}}(f) &{}\quad \mathbf H _{\textit{YY}}(f) &{}\quad \mathbf H _{YZ}(f) \\ \mathbf H _{\textit{ZX}}(f) &{}\quad \mathbf H _{\textit{ZY}}(f) &{}\quad \mathbf H _{\textit{ZZ}}(f) \end{array} \right) ^{-1}\nonumber \\&\quad \times \left( \begin{array}{c} \mathbf H _{\textit{XY}}(f)\\ \gamma (f) \\ \mathbf H _{\textit{ZY}}(f) \end{array}\right) E_Y(f), \end{aligned}$$

(43)

or

$$\begin{aligned} \left( \begin{array}{c} \bar{E}_X(f)\\ 0 \\ \bar{E}_Z(f) \end{array}\right)&=\left( \begin{array}{c} E_X(f)\\ 0 \\ E_Z(f) \end{array}\right) \nonumber \\&\quad +\,\left( \begin{array}{ccc} \mathbf A _{\textit{XZ}}(f) &{}\quad \mathbf A _{\textit{XY}}(f) &{}\quad \mathbf A _{\textit{XZ}}(f) \\ \mathbf A _{\textit{YX}}(f) &{}\quad \mathbf A _{\textit{YY}}(f) &{}\quad \mathbf A _{YZ}(f) \\ \mathbf A _{\textit{ZX}}(f) &{}\quad \mathbf A _{\textit{ZY}}(f) &{}\quad \mathbf A _{\textit{ZZ}}(f) \end{array} \right) \nonumber \\&\quad \times \left( \begin{array}{c} \mathbf H _{\textit{XY}}(f)\\ \gamma (f) \\ \mathbf H _{\textit{ZY}}(f) \end{array}\right) E_Y(f). \end{aligned}$$

(44)

Equation (43) holds because, after substituting it into (42), X(f) and Z(f) must satisfy (22). $\gamma (f)$ should be defined such that

$$\begin{aligned} \mathbf A _{\textit{YX}}(f)\mathbf H _{\textit{XY}}(f)+\mathbf A _{\textit{YY}}(f)\gamma (f)+\mathbf A _{YZ}(f)\mathbf H _{\textit{ZY}}(f)=\mathbf 0 . \end{aligned}$$

(45)

On the other hand as $\mathbf H ^{-1}(f)=\mathbf A (f)$, so

$$\begin{aligned} \mathbf A _{\textit{YX}}(f)\mathbf H _{\textit{XY}}(f)+\mathbf A _{\textit{YY}}(f)\mathbf H _{\textit{YY}}(f)+\mathbf A _{YZ}(f)\mathbf H _{\textit{ZY}}(f)=\mathbf I . \end{aligned}$$

(46)

So, by adding $\mathbf A _{\textit{YY}}(f)\mathbf H _{\textit{YY}}(f)$ to both sides of (45)

$$\begin{aligned}&\mathbf A _{\textit{YX}}(f)\mathbf H _{\textit{XY}}(f)+\mathbf A _{\textit{YY}}(f)\mathbf H _{\textit{YY}}(f)+\mathbf A _{\textit{YY}}(f)\gamma (f)\nonumber \\&\qquad +\,\mathbf A _{YZ}(f)\mathbf H _{\textit{ZY}}(f)=\mathbf A _{\textit{YY}}(f)\mathbf H _{\textit{YY}}(f). \end{aligned}$$

(47)

By substituting (46) into (47)

$$\begin{aligned} \mathbf I +\mathbf A _{\textit{YY}}(f)\gamma (f)=\mathbf A _{\textit{YY}}(f)\mathbf H _{\textit{YY}}(f). \end{aligned}$$

(48)

Therefore,

$$\begin{aligned} \mathbf A _{\textit{YY}}(f)\gamma (f)= & {} \mathbf A _{\textit{YY}}(f)\mathbf H _{\textit{YY}}(f)-\mathbf I ,\nonumber \\ \gamma (f)= & {} \mathbf A _{\textit{YY}}^{-1}(f)(\mathbf A _{\textit{YY}}(f)\mathbf H _{\textit{YY}}(f)-\mathbf I ),\nonumber \\ \gamma (f)= & {} \mathbf H _{\textit{YY}}(f)-\mathbf A _{\textit{YY}}^{-1}(f). \end{aligned}$$

(49)

Replacing (49) into (44) gives

$$\begin{aligned} \left( \begin{array}{c} \bar{E}_X(f)\\ 0 \\ \bar{E}_Z(f) \end{array}\right)&=\left( \begin{array}{c} E_X(f)\\ 0 \\ E_Z(f) \end{array}\right) \nonumber \\&\quad +\,\left( \begin{array}{ccc} \mathbf A _{\textit{XZ}}(f) &{}\quad \mathbf A _{\textit{XY}}(f) &{}\quad \mathbf A _{\textit{XZ}}(f) \\ \mathbf A _{\textit{YX}}(f) &{}\quad \mathbf A _{\textit{YY}}(f) &{}\quad \mathbf A _{YZ}(f) \\ \mathbf A _{\textit{ZX}}(f) &{}\quad \mathbf A _{\textit{ZY}}(f) &{}\quad \mathbf A _{\textit{ZZ}}(f) \end{array} \right) \nonumber \\&\quad \times \left( \begin{array}{c} \mathbf H _{\textit{XY}}(f)\\ \mathbf H _{\textit{YY}}(f)-\mathbf A _{\textit{YY}}^{-1}(f) \\ \mathbf H _{\textit{ZY}}(f) \end{array}\right) . \end{aligned}$$

(50)

The equation for $\bar{E}_{X}(f)$ from (50) can be simplified as

$$\begin{aligned} \bar{E}_{X}(f)= & {} E_{X}(f)+(\mathbf A _{\textit{XZ}}(f)\mathbf H _{\textit{XY}}(f)+ \mathbf A _{\textit{XY}}(f)(\mathbf H _{\textit{YY}}(f)\nonumber \\&-\mathbf A _{\textit{YY}}^{-1}(f))+\mathbf A _{\textit{XZ}}(f)\mathbf H _{\textit{ZY}}(f))E_Y(f)\nonumber \\= & {} E_{X}(f)+(\mathbf A _{\textit{XZ}}(f)\mathbf H _{\textit{XY}}(f)+\mathbf A _{\textit{XY}}(f)\mathbf H _{\textit{YY}}(f) \nonumber \\&-\mathbf A _{\textit{XY}}(f)\mathbf A _{\textit{YY}}^{-1}(f)+\mathbf A _{\textit{XZ}}(f)\mathbf H _{\textit{ZY}}(f))E_Y(f)\nonumber \\= & {} E_{X}(f)-\mathbf A _{\textit{XY}}(f)\mathbf A _{\textit{YY}}^{-1}(f)E_Y(f). \end{aligned}$$

(51)

On the other hand, based on (41), $\bar{E}_X(f)=E_X(f)+\mathbf Q _{\textit{XY}}(f)E_Y(f)$ for all $E_Y(f)$, so based on Lemma 3, $\mathbf Q _{\textit{XY}}(f)=-\mathbf A _{\textit{XY}}(f)\mathbf A _{\textit{YY}}^{-1}(f)$ .$\square$

Appendix 7: Proof of Theorem 1

Proof

Based on Lemmas 4–6, the equation for conditional Granger causality, i.e., (17), can be simplified as follows

$$\begin{aligned}&\mathfrak {f}_{Y\rightarrow X|Z}(f)=\ln \frac{|\varvec{\Sigma }_\varTheta |}{\left| \mathbf Q _{\textit{XX}}(f)\varvec{\Sigma }_{\textit{XX}}\mathbf Q _{\textit{XX}}^*(f)\right| }\nonumber \\&\quad =\ln \frac{\left| \mathbf Q _{\textit{XX}}(f)\varvec{\Sigma }_{\textit{XX}}\mathbf Q _{\textit{XX}}^*(f)+\mathbf Q _{\textit{XY}}(f)\varvec{\Sigma }_{\textit{YY}}\mathbf Q _{\textit{XY}}^*(f)+\mathbf Q _{\textit{XZ}}(f)\varvec{\Sigma }_{\textit{ZZ}}\mathbf Q _{\textit{XZ}}^*(f)\right| }{\left| \mathbf Q _{\textit{XX}}(f)\varvec{\Sigma }_{\textit{XX}}\mathbf Q _{\textit{XX}}^*(f)\right| }\nonumber \\&\quad =\ln \frac{\left| \varvec{\Sigma }_{\textit{XX}}+\left( -\mathbf A _{\textit{XY}}(f)\mathbf A _{\textit{YY}}^{-1}(f)\right) \varvec{\Sigma }_{\textit{YY}}\left( -\mathbf A _{\textit{XY}}(f)\mathbf A _{\textit{YY}}^{-1}(f)\right) ^*\right| }{\left| \varvec{\Sigma }_{\textit{XX}}\right| }\nonumber \\&\quad = \ln \frac{\left| \varvec{\Sigma }_{\textit{XX}}+\left( \mathbf A _{\textit{XY}}(f)\mathbf A _{\textit{YY}}^{-1}(f)\right) \varvec{\Sigma }_{\textit{YY}}\left( \mathbf A _{\textit{XY}}(f)\mathbf A _{\textit{YY}}^{-1}(f)\right) ^*\right| }{\left| \varvec{\Sigma }_{\textit{XX}}\right| }. \end{aligned}$$

(52)

$\square$