, $$\begin{aligned} \label{eqn0-0} P\Big( [Z_i]_{(j,p)} = 0 \Big) &=& \left(\frac{0}{0}\right)^{H-0} \cdot \frac{0}{0} \nonumber\\ &=& \frac{0}{0^H}, \end{aligned}$$ which means that the case of $H$ indeed holds. Th