# Weighted Spearman Rank correlation proof

I have been searching for alternate measures to compute spearman rank correlation that incorporates some sort of weighting in the calculation. I came across this paper by David C. Blest on weighted spearman rank correlation where he mentions the following result -

$$\sum_{k=1}^{n}\sum_{i=1}^{k} (q_i - i) = \sum_{i=1}^{n}(n+1-i)(q_i - i) = 1/2\sum_{i=1}^{n}d_i^2$$

where, $$d_i = q_i - i, 1\leqslant q_i , i \leqslant n.$$

$$q_i$$ (scheme-1) and $$i$$ (scheme-2, where $$i$$ is monotonically increasing ) are two schemes of ranking (no ties) a given set of $$n$$ elements . Therefore, the rank tuple of the $$i^{th}$$ element is given by $$(q_i,i)$$.

The relation $$\sum_{k=1}^{n}\sum_{i=1}^{k} (q_i - i) = \sum_{i=1}^{n}(n+1-i)(q_i - i)$$ is pretty straightforward to prove, but I can't prove how $$1/2\sum_{i=1}^{n}d_i^2$$ is equal to the other two. It would be great if somebody can help me work this out.

P.S Any other metric to compute weighted rank correlation is also welcome.

• (1) There is no third equality, which makes it difficult to know what you are asking. (2) Please provide details of the ranking. In particular, are ties allowed? How is $q_i$ related to $i$?
– whuber
Commented Jul 8, 2023 at 4:00
• Thanks for suggestions. I have edited . Commented Jul 8, 2023 at 6:02
• Use $\sum i^p=\sum q_i^p$ for all $p$ (you need the cases $p=1,2$).
– whuber
Commented Jul 8, 2023 at 11:07

$$\begin{eqnarray*} \sum_{i=1}^n (n+1-i)(q_i-i)&=&\sum_{i=1}^n (n+1-q_i+q_i-i)(q_i-i)\\ &=& \sum_{i=1}^n (q_i-i)^2+\sum_{i=1}^n (n+1-q_i)(q_i-i). \end{eqnarray*}$$ Now using $$\sum q_i^2=\sum i^2$$ and therefore $$\sum q_i(q_i-i)=-\sum i(q_i-i)$$, you get $$\sum_{i=1}^n (n+1-q_i)(q_i-i)=-\sum_{i=1}^n (n+1-i)(q_i-i),$$ and $$\begin{eqnarray*} 2\sum_{i=1}^n (n+1-i)(q_i-i)=\sum_{i=1}^n d_i^2. \end{eqnarray*}$$