Show that rank correlation lies between -1 and 1

Question

Show that Spearman's rank correlation r lies between -1 and 1.

where $r = 1 - \frac{6 \sum d_i^2}{n(n^2 - 1)}$

Answer 1

For maximum r, d should be the least. So consider the ranks (x, y) like:

x        y        d        
1        1        0
2        2        0
3        3        0
...     ...       ...
n        n        0

So, $\sum d^2 =0$ and $r=1$

For minimum r, consider the ranks like:

x        y        d        
n        1        n-1
n-1      2        n-3
n-2      3        n-4
...     ...       ...
1        n        1-n

We can write $d_i = n + 1 - 2i$

$$\begin{align} & \sum_{i=1}^n d_i^2 \\ &= \sum_{i=1}^n (n + 1 - 2i)^2 \\ &= \sum ((n + 1)^2 - 4(n + 1)i + 4i^2) \\ &= \frac n3 (n^2+1) \end{align}$$

Plugging into the formula of r, we get $r=-1$