1$. Therefore, $$\begin{aligned}
\label{eqn4-10}
& & P \left( [Z_i]_{(j,p)} = 1 \ \Big\vert \ w_{j_1 j_0} = \lambda_1, 
\cdots, w_{j_{H+1} j_H } = \lambda_{H+1} \right) \nonumber\\
&\fallingdotseq& \frac{1}{2^H} \hspace*{1cm} \mbox{(from Eq. (\ref{eqn4-6}))}\nonumber\\
&=& P \left( [Z_i]_{(j,p)} = 1 \right). \hspace*{1cm} \mbox{(from Theorem \ref{thm1})}\end{aligned}$$ This completes the proof.

\[thm3\] For any training pattern $i$, any output neuron $j$, and any path $p$ from an input neuron to the output neuron $j$ in the Kawaguchi model initialized by the even initialization method, random variable $[Z_i]_{(j,p)}$ is independent of the input training singnal ${\mbox{\boldmath $X$}}_i$. [$\Box$]{}

[*Proof*]{}.   Take training pattern $i$, output neuron $j$, path $p$ from an input neuron to output neuron $j$ of the deep nonlinear neural network in the Kawaguchi model initialized by the even initialization method, and ${\mbox{\boldmath $\mu$}}\in {\mbox{\boldmath $R$}}^{d_x}$ arbitrarily and fix them. Then, in the same mode of the proof of Theorem \[thm1\], it is apparent that $$\label{eqn4-12}
P\left([Z_i]_{(j,p)} = 1 \ \big\vert \ {\mbox{\boldmath $X$}}_i = {\mbox{\boldmath $\mu$}}\right) = \frac{1}{2^H}.$$ Therefore, it follows that $$\begin{aligned}
\label{eqn4-13}
P\left([Z_i]_{(j,p)} = 1 \right) &=& \frac{1}{2^H}  \hspace*{1.0cm} (\mbox{from Theorem} \ref{thm1} ) \nonumber\\
&=& P\left([Z_i]_{(j,p)} = 1 \ \ \big\vert \ \ {\mbox{\boldmath $X$}}_i = {\mbox{\boldmath $\mu$}}\right) \hspace*{0.5cm} (\mbox{from Eq.} 