Algebraic Summation Problem

Question

Suppose that, $$i \in M \Rightarrow c_i =\frac{\sum_{j\in M, j\ne i}^{N}(z_j)}{[\sum_{j\in M}^{N}(z_j)]^2}$$ Given that $z_i$ > $0$, it's obvious that

$$c_i <\frac{\sum_{j\in M}^{N}(z_j)}{[\sum_{j\in M}^{N}(z_j)]^2}$$

But why is it true that?

$$\frac{\sum_{j\in M}^{N}(z_j)}{[\sum_{j\in M}^{N}(z_j)]^2}= \frac{\sum_{j\in M}^{N}(c_j)}{|M|-1} $$

where $|M|$ is the number of elements in $M$.

Answer 1

This happens because: $$ c_i = \frac{\sum_{j \in M, j \neq i}^N z_j}{[\sum_{j \in M}^N z_j]^2} = \frac{\sum_{j \in M}^N z_j}{[\sum_{j \in M}^N z_j]^2} - \frac{z_i}{[\sum_{j \in M}^N z_j]^2} $$

Now, summing over all possible $i \in M$, $$ \sum_{i \in M} c_i = \frac{|M|\sum_{j \in M}^N z_j}{[\sum_{j \in M}^N z_j]^2} - \frac{\sum_{i \in M}z_i}{[\sum_{j \in M}^N z_j]^2} = \frac{(|M| - 1)\sum_{j \in M}^N z_j}{[\sum_{j \in M}^N z_j]^2} $$

From where the statement follows by rearrangement.