A Proof of Corollary 1
Since both \(\overline{D}_{P}(r)\) and \(\underline{D}_{P}(r)\) are piecewise linear function of r, all we have to do to prove that they are equal at some interval is to prove that the vertices are the same. First, let us note that
$$\begin{aligned} r_1&= 0\end{aligned}$$
(65)
$$\begin{aligned} r_2&= \frac{(N-1)P \alpha _1}{ N\left( \sum _{k=2}^K \left( {\begin{array}{c}K\\ k\end{array}}\right) \alpha _k -\sum _{k=2}^{K-P} \left( {\begin{array}{c}K-P\\ k\end{array}}\right) \alpha _k \right) + (N-1)P \alpha _1} \end{aligned}$$
(66)
Then, we note that
$$\begin{aligned} \overline{D}_{P}(r_2)&= \frac{N \left( \sum _{k=2}^K \left( {\begin{array}{c}K\\ k\end{array}}\right) \alpha _k \right) }{ N\left( \sum _{k=2}^K \left( {\begin{array}{c}K\\ k\end{array}}\right) \alpha _k -\sum _{k=2}^{K-P} \left( {\begin{array}{c}K-P\\ k\end{array}}\right) \alpha _k \right) + (N-1)P \alpha _1}\end{aligned}$$
(67)
$$\begin{aligned}&= \frac{1}{(1-\frac{1}{N})U} \sum _{i=1}^{P} \gamma _i r_i^{K-P} \left[ (N-1)(N^{\frac{K}{P}}-1) -\frac{PK(N-1)}{\gamma _i} \right] \end{aligned}$$
(68)
where \(U = N\left( \sum _{k=2}^K \left( {\begin{array}{c}K\\ k\end{array}}\right) \alpha _k -\sum _{k=2}^{K-P} \left( {\begin{array}{c}K-P\\ k\end{array}}\right) \alpha _k \right) + (N-1)P \alpha _1\), \(\alpha _k=\sum _{i=1}^{P} \gamma _i r_i^{K-P-k}\).
Further, we note from (20), by choosing \(r = r_2\), we have
$$\begin{aligned} \underline{D}_{P}(r_2)&\ge \sum _{i=0}^{ \frac{K}{P} - 1} \frac{1}{N^{i}} \left[ 1-r_{2} \left( \frac{K}{P} - i \right) \right] \end{aligned}$$
(69)
$$\begin{aligned}&= \frac{1-(\frac{1}{N})^{\frac{K}{P}}}{1-\frac{1}{N}} - \frac{(N-1)P \alpha _1}{U} \left[ \frac{ \frac{K}{P}- \frac{1}{N}(1 - (\frac{1}{N})^{\frac{K}{P}})}{(1-\frac{1}{N})^{2}}\right] \end{aligned}$$
(70)
$$\begin{aligned}&= \frac{1}{(1-\frac{1}{N})U} \left[ \left( 1-\frac{1}{N^{\frac{K}{P}}}\right) U -(N-1)P\alpha _1 \left( \frac{ \frac{K}{P}- \frac{1}{N}(1 - (\frac{1}{N})^{\frac{K}{P}})}{(1-\frac{1}{N})}\right) \right] \end{aligned}$$
(71)
$$\begin{aligned}&= \frac{1}{(1-\frac{1}{N})U} \left[ (N-1)P\alpha _1 \left( \left( 1+\frac{1}{N-1} \right) \left( 1-\frac{1}{N^{\frac{K}{P}}} \right) - \frac{K}{P} \right) \right] \end{aligned}$$
(72)
$$\begin{aligned}&+ \frac{1}{(1-\frac{1}{N})U} \left[ N \left( 1-\frac{1}{N^{\frac{K}{P}}}\right) \sum _{i=1}^{P} \gamma _i r_i^{K-P} \left( N^{\frac{K}{P}} - N^{\frac{K}{P} - 1} -\frac{P}{\gamma _i} \right) \right] \end{aligned}$$
(73)
$$\begin{aligned}&= \frac{1}{(1-\frac{1}{N})U} \sum _{i=1}^{P} \gamma _i r_i^{K-P} \left[ (N-1)(N^{\frac{K}{P}}-1) -\frac{PK(N-1)}{\gamma _i} \right] \end{aligned}$$
(74)
$$\begin{aligned}&= \overline{D}_{P}(r_2) \end{aligned}$$
(75)
Thus, since \( \underline{D}_{P}(r_2) \le \overline{D}_{P}(r_2)\) by definition, (75) implies \( \underline{D}_{P}(r_2) = \overline{D}_{P}(r_2)\). We also note that \( \underline{D}_{P}(r_1) = \overline{D}_{P}(r_1)\). Since both \(\overline{D}_{P}(r)\) and \(\underline{D}_{P}(r)\) are piecewise linear function of r, and since \( \underline{D}_{P}(r_2) = \overline{D}_{P}(r_2) \) and \( \underline{D}_{P}(r_1) = \overline{D}_{P}(r_1)\), we have \( \underline{D}_{P}(r) = \overline{D}_{P}(r) = D^{*}_{P}(r)\) for \(r_1 \le r \le r_{2}\). Thus we complete the proof of corollary 1.