state. Thus, half of the total number of remaining singular fade states lie inside the circle $\gamma=1$ and the other half lies outside it. This along with the fact that singular fade states are periodic, implies the number of singular fade states for the wedge $ [0,\pi /M]$, is given by $$\begin{aligned}
\nonumber
\frac{1}{M}\left( \frac{\frac{M^3}{4}-\frac{M^2}{2}}{2}+M\right)  =\frac{M^2}{8}-\frac{M}{4}+1.\end{aligned}$$ Also from [@Vijay], it is clear that these fade states lie along $\theta =0$ and $\theta = \pi /M$ lines.

We denote this set of all singular fade states lying in the wedge $ [0, \pi/M] $ by $ \mathcal{H}_W $. Let $N_W = \vert \mathcal{H}_W \vert $.

Observe from , that the distance between two points in $ \mathcal{S}_{\text{eff}} $ is a function of $ \gamma $ and $ \theta $. Let $ \vert \triangle s_j \vert = \vert s_j-s'_j \vert  $ and $ \phi_j = \angle (s_j-s'_j) $ for $ j=1,\,2 $. Now from , the distance between the elements of the pair $ \lbrace (s_1,s_2),(s'_1,s'_2) \rbrace  \in  \mathcal{S}^2_{\text{eff}} $ is

[$$\begin{aligned}
\label{eq:distclassex}
d^2_{(s_1,s_2)\leftrightarrow (s'_1,s'_2)}= \vert \triangle s_1 \vert ^2 + \gamma ^2 \vert \triangle s_2 \vert ^2 + 2 \gamma \vert \triangle s_1 \triangle s_2 \vert \cos (\theta + \phi_2 -\phi_1 