9 6 7 Gaussian Copula Graphical Models

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Let $X=(X_1,\ldots,X_p)$ a non-Gaussian random variable. Let $Z=(Z_1,\ldots,Z_p)$ following a Gaussian distribution $\mathcal{N}(0,\Sigma^Z)$ and $p$ increasing differentiable functions $f_a$, $a\in\{1,\ldots,p\}$, such that $(X_1,\ldots X_p)=(f_1(Z_1),\ldots,f_p(Z_p))$. We assume that $\Sigma^Z$ is non-singular and $\Sigma^Z_{aa}=1$ for all $a\in\{1,\ldots,p\}$.

1. We note $f_X$ (resp. $f_Z$) the probability density function of $X$ (resp. $Z$). For all $A\in\mathcal{B}\left(\mathbb{R}^p\right)$,

(1)

\begin{align} \int_{\phi(A)} f_X(x_1,\ldots,x_p)~\text{d}x_1\ldots\text{d}x_p ~=~ \int_{A} f_X\left(f_1(z_1),\ldots,f_p(z_p)\right)f'(z_1)\ldots f'(z_p)~\text{d}z_1\ldots~\text{d}z_p \end{align}

where $\phi:(x_1,\ldots,x_p)\mapsto\left(f_1(x_1),\ldots,f_p(x_p)\right)$ is a diffeomorphism as the $f_i$ are increasing and differentiable.
Yet, for all $A\in\mathcal{B}\left(\mathbb{R}^p\right)$,

(2)

\begin{array} {rcl} \int_A f_Z(z_1,\ldots,z_p)~\text{d}z_1\ldots\text{d}z_p &=& \mathbb{P}[ Z_1,\ldots,Z_p\in A ] ~=~ \mathbb{P}[ f_1(Z_1),\ldots,f_p(Z_p)\in\phi(A) ] \\ &=& \mathbb{P}[ X_1,\ldots,X_p\in\phi(A) ] ~=~ \int_{\phi(A)} f_X(x_1,\ldots,x_p)~\text{d}x_1\ldots\text{d}x_p \end{array}

So, by unicity of the density function, for all $z=(z_1,\ldots,z_p)\in\mathbb{R}^p$,

(3)

\begin{align} f_Z(z_1,\ldots,z_p)~=~f_X\left(f_1(z_1),\ldots,f_p(z_p)\right)f'_1(z_1)\ldots f'_p(z_p) \end{align}

Let $a$ be in $\{1,\ldots,p\}$ and us show that $Z_a\coprod\{Z_b~\vert~b\notin\text{cl}(a)\}~\vert~\{Z_c~\vert~c\in\text{ne}(a)\}$,
i.e, for all $b\notin\text{cl}(a)$ and $c\in\text{ne}(a)$, $f_Z(z_a,z_b,z_c)~=~\frac{f_Z(z_a,z_c)\times f_Z(z_b,z_c)}{f_Z(z_c)}$.
Yet, while $\mathscr{G}^X_*$ is a graphical model for $X$,

(4)

\begin{array} {rcl} f_Z(z_a,z_a,z_c) &=& f_X\left(f_a(z_a),f_b(z_b),f_c(z_c)\right)\times f'_a(z_a)f'_b(z_b)f'_p(z_p)\\ &=& f_X(x_a,x_b,x_c)\times f'_a(z_a)f'_b(z_b)f'_c(z_c)\\ &=& \frac{f_X(x_a,x_c)\times f_X(x_b,x_c)}{f_X(x_c)}\times f'_a(z_a)f'_b(z_b)f'_c(z_c)\\ &=& \frac{\left[f_X\left(f_a(z_a),f_c(z_c)\right)\times f'_a(z_a)f'_c(z_c)\right]\times\left[f_X\left(f_b(z_b),f_c(z_c)\right)\times f'_b(z_b)f'_c(z_c)\right]}{f_X\left(f_c(z_c)\right)\times f'_c(z_c)}\\ &=& \frac{f_Z(z_a,z_c)\times f_Z(z_b,z_c)}{f_Z(z_c)} \end{array}

and $\mathscr{G}^X_*$ is a graphical model for $Z$.

Throughout the rest of the exercise we will note $\mathscr{G}_*$ the minimal graph for either variable $X$ and $Z$. In particular, for $a\neq b$, there is an edge between $a$ and $b$ in $\mathscr{G}_*$ if and only if $K^Z_{ab}\neq0$.

2.i. Let $\tilde{Z}$ be an independent copy of $Z$ and define $\tilde{X}~:=~\left(f_1(\tilde{Z_1}),\ldots,f_p(\tilde{Z_p})\right)$.
Let $a,b\in\{1,\ldots,p\}$ and $\tau_{ab}~:=~\mathbb{E}\left[\text{sign}\left((X_a-\tilde{X}_a)(X_b-\tilde{X}_b)\right)\right]$.

By definition, $\tau_{ab}~=~\mathbb{E}\left[\text{sign}\left((f_a(Z_a)-f_a(\tilde{Z}_a))(f_b(Z_b)-f_b(\tilde{Z}_b))\right)\right]$. Or, by growing of $f_a$ and $f_b$,

(5)

\begin{align} & \left(f_a(Z_a)-f_a(\tilde{Z_a})\right)\left(f_b(Z_b)-f_b(\tilde{Z}_b)\right)>0\\ & \quad\quad\quad\quad\iff~\left\{\begin{array}{r} f_a(Z_a)-f_a(\tilde{Z}_a)>0\\ f_b(Z_b)-f_b(\tilde{Z}_b)>0\end{array}\right. \quad \text{or}\quad \left\{\begin{array}{r} f_a(Z_a)-f_a(\tilde{Z}_a)<0\\ f_b(Z_b)-f_b(\tilde{Z}_b)<0\end{array}\right. \\ & \quad\quad\quad\quad\iff \left\{\begin{array}{r} Z_a-\tilde{Z}_a>0\\ Z_b-\tilde{Z}_b>0\end{array}\right. \quad \text{or}\quad \left\{\begin{array}{r} Z_a-\tilde{Z}_a<0\\ Z_b-\tilde{Z}_b<0\end{array}\right. \\ & \quad\quad\quad\quad\iff \left(Z_a-\tilde{Z}_a\right)\left(Z_b-\tilde{Z}_b\right)>0 \end{align}

Similarly, $\left(f_a(Z_a)-f_a(\tilde{Z_a})\right)\left(f_b(Z_b)-f_b(\tilde{Z}_b)\right)\leqslant0~\iff~\left(Z_a-\tilde{Z}_a\right)\left(Z_b-\tilde{Z}_b\right)\leqslant0$ and thus,

(6)

\begin{align} \text{sign}\left((f_a(Z_a)-f_a(\tilde{Z}_a))(f_b(Z_b)-f_b(\tilde{Z}_b))\right)~=~\text{sign}\left((Z_a-\tilde{Z}_a)(Z_b-\tilde{Z}_b)\right) \end{align}

and

(7)

\begin{align} \tau_{ab}=\mathbb{E}\left[\text{sign}\left((Z_a-\tilde{Z}_a)(Z_b-\tilde{Z}_b)\right)\right] \end{align}

ii. Since $Z_a$ and $\tilde{Z}_a$ are independent with a variance of $1$,

(8)

\begin{array} {rcl} \text{Var}~(Z_a-\tilde{Z}_a) &=& \text{Var}~(Z_a)+\text{Var}~(\tilde{Z}_a)\\ &=& 1+1~=~2 \end{array}

Let $\zeta_a=\frac{1}{\sqrt{2}}\left(Z_a-\tilde{Z}_a\right)$ and $\zeta_b=\frac{1}{\sqrt{2}}\left(Z_b-\tilde{Z}_b\right)$. By independence of $Z_a$ and $\tilde{Z}_a$ (resp. $Z_b$ and $\tilde{Z}_b$), $\zeta_a$ (resp. $\zeta_b$) is a normal distribution with a variance of $1$. Moreover, since $\tilde{Z}$ is an independent copy of $Z$, $\zeta$ and $-\zeta$ follow the same law. So, we have,

(9)

\begin{array} {rcl} \tau_{ab} &=& \mathbb{E}\left[ \text{sign}\left( \frac{Z_a-\tilde{Z}_a}{\sqrt{2}}\times\frac{Z_b-\tilde{Z}_b}{\sqrt{2}} \right) \right] ~=~ \mathbb{E}\left[ \text{sign}\left(\zeta_a\zeta_b\right) \right]\\ &=& \mathbb{E}\left[ \textbf{1}_{\zeta_a\zeta_b>0} - \textbf{1}_{\zeta_a\zeta_b\leqslant 0} \right] ~=~ \mathbb{P}\left[\zeta_a\zeta_b>0\right]-\mathbb{P}\left[\zeta_a\zeta_b\leqslant 0\right] \\ &=& \mathbb{P}\left[\zeta_a>0,~\zeta_b>0\right] + \mathbb{P}\left[\zeta_a<0,~\zeta_b<0\right] - \mathbb{P}\left[\zeta_a\geqslant0,~\zeta_b\leqslant0\right] - \mathbb{P}\left[\zeta_a\leqslant0,~\zeta_b\geqslant0\right] \\ &=& \mathbb{P}\left[\zeta_a>0,~\zeta_b>0\right] - \mathbb{P}\left[\overline{\zeta_a>0,~\zeta_b>0}\right] + 2~\mathbb{P}\left[\zeta_a<0,~\zeta_b<0\right]\\ &=& \mathbb{P}\left[\zeta_a>0,~\zeta_b>0\right] - (1 - \mathbb{P}\left[\zeta_a>0,~\zeta_b>0\right]) + 2~\mathbb{P}\left[\zeta_a>0,~\zeta_b>0\right] \\ &=& 4~\mathbb{P}\left[\zeta_a>0,~\zeta_b>0\right] - 1 \end{array}

Since $\mathbb{P}\left[\zeta_a>0,~\zeta_b>0\right] = \frac{1}{4}\left(1+\frac{2}{\pi}\arcsin(\Sigma^Z_{ab})\right)$, we have the result :

(10)

\begin{align} \tau_{ab} ~=~ \mathbb{E}\left[\text{sign}\left((Z_a-\tilde{Z}_a)(Z_b-\tilde{Z}_b)\right)\right] ~=~ \frac{2}{\pi}\arcsin\left(\Sigma^Z_{ab}\right) \end{align}

A) Kendall's tau

1. We define $F:\mathbb{R}^{2n}\rightarrow[-1,1]$ by

(11)

\begin{align} F\left( (x_1,y_1),\ldots,(x_n,y_n) \right) := \frac{2}{n(n-1)}\sum_{i<j} \text{sign}\left( (x_i-x_j)(y_i-y_j) \right) \end{align}

We first remark that

(12)

\begin{align} -1~\leqslant~\frac{2}{n(n-1)}\sum_{i<j}(-1)~\leqslant~F\left( (x_1,y_1),\ldots,(x_n,y_n) \right)~\leqslant~\frac{2}{n(n-1)}\sum_{i<j}1~\leqslant~1 \end{align}

and $F$ is well-defined.

For all $i\in\{1,\ldots,n\}$ and $x_1,\ldots,x_n,y_1,\ldots,y_n,x'_i,y'_i\in\mathbb{R}$, with an evident notation,

(13)

\begin{array} {rcl} \Delta_i &:=& {\displaystyle \left\lvert F\left( (x_1,y_1),\ldots,(x_i,y_i),\ldots,(x_n,y_n) \right) - F\left( (x_1,y_1),\ldots,(x'_i,y'_i),\ldots,(x_n,y_n) \right) \right\rvert}\\ &=& {\displaystyle \frac{2}{n(n-1)} \left\lvert \sum_{1\leqslant k<l\leqslant n} \underbrace{\text{sign}\left( (x_k-x_l)(y_k-y_l) \right) - \text{sign}\left( (x'_k-x'_l)(y'_k-y'_l) \right)}_{:= s_{k,l}} \right\rvert} \end{array}

Yet,

• if $l<i$, $x'_l=x_l,~y'_l=y_l$ and for all $k<l$, $s_{k,l}=0$ ;
• if $l=i$, $s_{.,i} = \sum_{k=1}^{i-1} \text{sign}\left( (x_k-x_i)(y_k-y_i) \right) - \text{sign}\left( (x_k-x'_i)(y_k-y'_i) \right)$ ;
• if $l>i$, $s_{k,l}=0$ except if $k=i$ :
  $s_{i,.} = \sum_{l=i+1}^n \text{sign}\left( (x_i-x_l)(y_i-y_l) \right) - \text{sign}\left( (x'_i-x_l)(y'_i-y_l) \right)$.

So,

(14)

\begin{array} {rcl} \Delta_i &=& {\displaystyle \frac{2}{n(n-1)} \left\lvert \sum_{k=1}^{i-1} \text{sign}\left( (x_k-x_i)(y_k-y_i) \right) - \text{sign}\left( (x_k-x'_i)(y_k-y'_i) \right) \right. }\\ && {\displaystyle \left. \quad\quad\quad\quad\quad + \sum_{l=i+1}^n \text{sign}\left( (x_i-x_l)(y_i-y_l) \right) - \text{sign}\left( (x'_i-x_l)(y'_i-y_l) \right) \right\rvert} \\ &\leqslant& {\displaystyle \frac{2}{n(n-1)} \left( \sum_{k=1}^{i-1} \left\lvert \text{sign}\left( (x_k-x_i)(y_k-y_i) \right) - \text{sign}\left( (x_k-x'_i)(y_k-y'_i) \right) \right\rvert \right. }\\ && {\displaystyle \left. \quad\quad\quad\quad\quad + \sum_{l=i+1}^n \left\lvert \text{sign}\left( (x_i-x_l)(y_i-y_l) \right) - \text{sign}\left( (x'_i-x_l)(y'_i-y_l) \right) \right\rvert \right) }\\ &\leqslant& {\displaystyle \frac{2}{n(n-1)} \left( \sum_{k=1}^{i-1} 2 + \sum_{l=i+1}^n 2 \right) = \frac{4}{n(n-1)} ( i-1 + n-(i+1)+1 ) ~=~ \frac{4}{n} } \end{array}

2. Let $\hat{\tau}_{ab}~:=~F\left( (X_a^{(1)},X_b^{(1)}),\ldots,(X_a^{(n)},X_b^{(n)}) \right)~=~F\left(X_a,X_b\right)$. Then,

(15)

\begin{array} {rcl} \mathbb{E}\left[ \hat{\tau}_{ab} \right] &=& \frac{2}{n(n-1)} \sum_{1\leqslant i<j\leqslant n} \mathbb{E}\left[\text{sign}\left( (X_a^{(i)}-X_a^{(j)})(X_b^{(i)}-X_b^{(j)}) \right)\right] \\ &=& \frac{2}{n(n-1)} \sum_{1\leqslant i<j\leqslant n} \tau_{ab} \end{array}

Since ${\displaystyle \#\left\{ (i,j)\in\{1,\ldots,n\} ~\vert~ i<j \right\} ~=~ \sum_{k=2}^n k-1 ~=~ \frac{n(n-1)}{2}}$, we find that $\mathbb{E}\left[\hat{\tau}_{ab}\right]~=~\tau_{ab}$. Similarly, $\mathbb{E}\left[-\hat{\tau}_{ab}\right]~=~-\tau_{ab}$.

Last, as ${\displaystyle \sum_{i=1}^n \left(\frac{4}{n}\right)^2~=~\frac{16}{n} }$, from McDiarmind concentration inequality, for all $a<b$ and $t>0$, we have :

(16)

\begin{array} {rcl} {\displaystyle \mathbb{P}\left[ \left\lvert \hat{\tau}_{ab}-\tau_{ab} \right\rvert > t \right]} &\leqslant& {\displaystyle \mathbb{P}\left[ \hat{\tau}_{ab}>\tau_{ab}+t \right] + \mathbb{P}\left[ -\hat{\tau}_{ab}>-\tau_{ab}+t \right]} \\ &\leqslant& {\displaystyle 2 \text{e}^{ -2t^2\times\frac{n}{16} } ~=~ 2 \text{e}^{ \frac{-nt^2}{8} }} \end{array}

By definition,

(17)

\begin{align} \left\{\begin{array}{rcl} \hat{\Sigma}^Z_{ab} &=& \sin\left(\frac{\pi}{2}\hat{\tau}_{ab}\right)~=~\sin\left(\frac{\pi}{2}F(X_a,X_b)\right)~:=~G(X_a,X_b)\\ \Sigma^Z_{ab} &=& \mathbb{E}\left[G(X_a,X_b)\right] \end{array}\right. \end{align}

For brevity, we note $(X'_a,X'_b)$ for $\left( (X_a^{(1)},X_b^{(1)}),\ldots,(X_a^{(i')},X_b^{(i')}),\ldots(X_a^{(n)},X_b^{(n)})\right)$ and we have for all $a<b$,

(18)

\begin{array} {rcl} {\displaystyle \left\lvert G\left(X_a,X_b\right) - G\left(X'_a,X'_b\right) \right\rvert} &=& {\displaystyle \left\lvert \sin\left(\frac{\pi}{2}F(X_a,X_b)\right) - \sin\left(\frac{\pi}{2}F(X'_a,X'_b)\right) \right\rvert} \\ &=& {\displaystyle 2 ~\left\lvert \cos\left( \frac{\pi}{4}\left(F(X_a,X_b)+F(X'_a,X'_b)\right) \right) \right.}\\ && {\displaystyle \left. \quad\quad\quad \times \sin\left( \frac{\pi}{4}\left(F(X_a,X_b)-F(X'_a,X'_b)\right) \right) \right\rvert} \\ &\leqslant& {\displaystyle 2\times\frac{\pi}{4}\left\lvert F(X_a,X_b)-F(X'_a,X'_b)\right\rvert ~\leqslant~ \frac{\pi}{2}\times\frac{4}{n}} \end{array}

Similarly, for all $a<b$ and $t>0$, ${\displaystyle \sum_{i=1}^n \left(\frac{\pi}{2}\times\frac{4}{n}\right)^2~=~\sum_{i=1}^n\left(\frac{2\pi}{n}\right)^2~=~\frac{4\pi^2}{n} }$ and

(19)

\begin{align} \mathbb{P}\left[ \left\lvert \hat{\Sigma}^Z_{ab}-\Sigma^Z_{ab} \right\rvert > t \right] \leqslant 2 \text{e}^{ -2t^2\times\frac{n}{4\pi^2} } ~=~ 2 \text{e}^{ \frac{-nt^2}{2\pi^2} } \end{align}

3. By using the symmetry of $\hat{\Sigma}^Z$ and $\Sigma^Z$, we know that

(20)

\begin{align} \mathbb{P}\left[ \left\lvert \hat{\Sigma}^Z-\Sigma^Z\right\rvert_{\infty} >t \right] = \mathbb{P}\left[\max_{a\leqslant b}\left\lvert\hat{\Sigma}^Z_{ab}-\Sigma^Z_{ab}\right\rvert>t\right] \end{align}

In fact,

(21)

\begin{align} \hat{\tau}_{aa} ~=~ \frac{2}{n(n-1)}\sum_{1\leqslant i<j\leqslant n}\text{sign}\left(\left(X_a^{(i)}-X_a^{(j)}\right)^2\right) ~=~1 \quad\text{and}\quad \hat{\Sigma}^Z_{aa} ~=~ \sin\left(\frac{\pi}{2}\hat{\tau}_{aa}\right) ~=~1 \end{align}

So,

(22)

\begin{array} {rcl} {\displaystyle \mathbb{P}\left[ \left\lvert \hat{\Sigma}^Z-\Sigma^Z\right\rvert_{\infty} >t \right]} &=& {\displaystyle \mathbb{P}\left[\max_{a<b}\left\lvert\hat{\Sigma}^Z_{ab}-\Sigma^Z_{ab}\right\rvert>t\right] ~=~ \mathbb{P}\left[\bigcup_{a<b}\left\lvert\hat{\Sigma}^Z_{ab}-\Sigma^Z_{ab}\right\rvert>t\right]} \\ &=& {\displaystyle \sum_{1\leqslant a<b\leqslant p} \mathbb{P}\left[ \left\lvert \hat{\Sigma}_{ab}-\Sigma_{ab} \right\rvert > t \right]} \\ &\leqslant& {\displaystyle \frac{p(p-1)}{2}\times 2\text{e}^{ \frac{-nt^2}{2\pi^2} }} ~\leqslant~ {\displaystyle p^2\text{e}^{ \frac{-nt^2}{2\pi^2} }} \end{array}

In particular, for $t=2\pi\sqrt{\frac{2\log p}{n}}$,

(23)

\begin{align} -\frac{nt^2}{2\pi^2} ~=~ -\frac{n}{2\pi^2}\times4\pi^2\left(\frac{2\log p}{n}\right) ~=~ -4\log p \end{align}

and therefore

(24)

\begin{align} \mathbb{P}\left[ \left\lvert \hat{\Sigma}^Z-\Sigma^Z\right\rvert_{\infty} > 2\pi\sqrt{\frac{2\log p}{n}} \right] \leqslant p^2\times\frac{1}{p^4} ~=~ \frac{1}{p^2} \end{align}

B) Graph estimation

1. We note like in the exercice 7.6.6 ${\displaystyle \lvert K\rvert_{1,\infty}=\max_b\sum_a\vert K_{ab}\rvert}$. We also note $K^Z:={\Sigma^Z}^{-1}$ and $\hat{K}^Z:=\text{argmin}\left\{ \left\lvert B\right\rvert_{1,\infty} ~\vert~ B\in\mathscr{M}_p\mathbb{R},~\left\lvert\hat{\Sigma}B-I_n\right\rvert_{\infty}\leqslant\lambda \right\}$ where $\lambda\in\mathbb{R}+$. Last, we define a graph $\hat{\mathscr{G}}$ by setting an edge between $a$ and $b$ if and only if both $\hat{K}^Z_{ab}$ and $\hat{K}^Z_{ba}$ are non-zero.

According to the question $A.3$ of the exercise cited above, if $\lambda$ fulfils $\lambda\geqslant\left\lvert K^Z\right\rvert_{1,\infty}\left\lvert\hat{\Sigma}^Z-\Sigma^Z\right\rvert_{\infty}$ then $\left\lvert\hat{K}^Z-K^Z\right\rvert_{\infty}\leqslant2\lambda\left\lvert K^Z\right\rvert_{1,\infty}$.
In particular, for $\lambda=2\pi\sqrt{\frac{2 \log p}{n}}\left\lvert K^Z\right\rvert_{1,\infty}$, the condition is equivalent to $2\pi\sqrt{\frac{2\log p}{n}}\geqslant\left\lvert\hat{\Sigma}^Z-\Sigma^Z\right\rvert_{\infty}$ because $\Sigma^Z_{aa}=1$ for all $a$ and so $K^Z$ is non-zero.
This condition is verified with probability at least $1-\frac{1}{p^2}$ (question $A.3$) and in that case,

(25)

\begin{align} \left\lvert\hat{K}^Z-K^Z\right\vert_{\infty}~\leqslant~4\pi\sqrt{\frac{2\log p}{n}}\left\lvert K^Z\right\rvert_{1,\infty}^2 \end{align}

2. We assume now that all entries $K^Z_{ab}$ of $K^Z$ are such that

(26)

\begin{align} K^Z_{ab}=0 \quad\text{or}\quad \left\vert K^Z_{ab}\right\rvert>4\pi\sqrt{\frac{2\log p}{n}}\left\lvert K^Z\right\rvert_{1,\infty}^2 \end{align}

By construction, for $a\neq b$, there is an edge between $a$ and $b$ in

• $\mathscr{G}_*$ if and only if $K^Z_{ab}\neq 0$,
• $\hat{\mathscr{G}}$ if and only if $\hat{K}_{ab}\neq0$ and $\hat{K}_{ba}\neq0$.

Firstly, we remark that

(27)

\begin{align} \left\lvert\hat{K}^Z-K^Z\right\vert_{\infty}~\leqslant~4\pi\sqrt{\frac{2\log p}{n}}\left\lvert K^Z\right\rvert_{1,\infty}^2 ~\implies~ \forall a,b\in\{1,\ldots,p\},\quad\left\lvert\hat{K}^Z_{ab}-K^Z_{ab}\right\vert~\leqslant~4\pi\sqrt{\frac{2\log p}{n}}\left\lvert K^Z\right\rvert_{1,\infty}^2 \end{align}

We want to show that $\mathscr{G}_*\subset\hat{\mathscr{G}}$, i.e that if $K^Z_{ab}\neq0$ then $\hat{K}^Z_{ab}\neq0$ and $\hat{K}^Z_{ba}\neq0$. We proceed by contrapositive : let us assume that it exists $a,b\in\{1,\ldots,p\}$ $\hat{K}_{ab}=0$. Then,

(28)

\begin{align} \left\lvert K^Z_{ab}\right\rvert~=~\left\lvert\hat{K}^Z_{ab}-K^Z_{ab}\right\rvert~\leqslant~4\pi\sqrt{\frac{2\log p}{n}}\left\lvert K^Z\right\rvert_{1,\infty}^2 \end{align}

and so $K^Z_{ab}=0$ as otherwise $4\pi\sqrt{\frac{2\log p}{n}}\left\lvert K^Z\right\rvert_{1,\infty}^2<4\pi\sqrt{\frac{2\log p}{n}}\left\lvert K^Z\right\rvert_{1,\infty}^2$, which is impossible.