d_{H-1}$ are sufficiently large (Fig. \[fig1\]), and that each element of the $i$-th input training pattern ${\mbox{\boldmath $X$}}_i$ takes a value between $-\alpha$ and $\alpha$, that is, ${\mbox{\boldmath $X$}}_i \in {\mbox{\boldmath $I$}}^{d_x}$ for any $i$ where ${\mbox{\boldmath $I$}}=[-\alpha, \alpha]$ and $\alpha$ is a positive real number. Then we propose the following weight initialization scheme.

[**Even Initialization**]{} [ *Elements $\{w_n\}_{n=1}^{d_{k-1}}$ of the weight vector ${\mbox{\boldmath $w$}}= (w_1 \cdots w_{d_{k-1}})^T$ of any hidden neuron in the $k$-th hidden layer are initialized independently and identically according to a probability distribution whose probability density function $f_k: {\mbox{\boldmath $J$}}_k \rightarrow [0, +\infty)$ is an even function where ${\mbox{\boldmath $J$}}_k = [-1/d_{k-1}, 1/d_{k-1}]$, that is, $f_k(-x) = f_k(x)$ for any $x \in {\mbox{\boldmath $J$}}_k$ $( k = 1, \cdots, H$).* ]{} [$\Box$]{}

We designate it as [*even initialization*]{}.

[**Remarks**]{} (a) The expectation of each weight is zero because the probability density function is an even function. (b) The probability distribution in the even initialization can be a normal distribution with zero mean or a uniform distribution. (c) For the weight vectors ${\mbox{\boldmath $w$}}_1=(w_{11} \cdots w_{1M})^T, {\mbox{\boldmath $w$}}_2=(w_{21} \cdots w_{2N})^T$ of any two hidden neurons, the probability distribution which $w_{1k}$ obeys can be different from that which 