<s>[INST] Prove the equivalence of the following two formulas for Spearman correlation From wikipedia, Spearman's rank correlation is calculated by converting variables $X_i$ and $Y_i$ into ranked variables $x_i$ and $y_i$, and then calculating Pearson's correlation between the ranked variables:

However, the article goes on to state that if there are no ties amongst the variables $X_i$ and $Y_i$, the above formula is equivalent to

where $d_i = y_i - x_i$, the difference in ranks.
Can someone give a proof of this please? I don't have access to the textbooks referenced by the wikipedia article.
 [/INST] $ \rho = \frac{\sum_i(x_i-\bar{x})(y_i-\bar{y})}{\sqrt{\sum_i (x_i-\bar{x})^2 \sum_i(y_i-\bar{y})^2}}$
Since there are no ties, the $x$'s and $y$'s both consist of the integers from $1$ to $n$ inclusive.
Hence we can rewrite the denominator:
$\frac{\sum_i(x_i-\bar{x})(y_i-\bar{y})}{\sum_i (x_i-\bar{x})^2}$
But the denominator is just a function of $n$:
$\sum_i (x_i-\bar{x})^2 = \sum_i x_i^2 - n\bar{x}^2 \\
\quad= \frac{n(n + 1)(2n + 1)}{6} - n(\frac{(n + 1)}{2})^2\\
\quad= n(n + 1)(\frac{(2n + 1)}{6} - \frac{(n + 1)}{4})\\
\quad= n(n + 1)(\frac{(8n + 4-6n-6)}{24})\\
\quad= n(n + 1)(\frac{(n -1)}{12})\\
\quad= \frac{n(n^2 - 1)}{12}$
Now let's look at the numerator:
$\sum_i(x_i-\bar{x})(y_i-\bar{y})\\
\quad=\sum_i x_i(y_i-\bar{y})-\sum_i\bar{x}(y_i-\bar{y}) \\
\quad=\sum_i x_i y_i-\bar{y}\sum_i x_i-\bar{x}\sum_iy_i+n\bar{x}\bar{y} \\
\quad=\sum_i x_i y_i-n\bar{x}\bar{y} \\
\quad= \sum_i x_i y_i-n(\frac{n+1}{2})^2 \\
\quad= \sum_i  x_i y_i- \frac{n(n+1)}{12}3(n +1)  \\
\quad= \frac{n(n+1)}{12}.(-3(n +1))+\sum_i  x_i y_i   \\
\quad= \frac{n(n+1)}{12}.[(n-1) - (4n+2)] + \sum_i  x_i y_i  \\
\quad= \frac{n(n+1)(n-1)}{12} - n(n+1)(2n+1)/6 + \sum_i  x_i y_i  \\
\quad= \frac{n(n+1)(n-1)}{12} -\sum_i x_i^2+ \sum_i  x_i y_i  \\
\quad= \frac{n(n+1)(n-1)}{12} -\sum_i (x_i^2+ y_i^2)/2+ \sum_i  x_i y_i  \\
\quad= \frac{n(n+1)(n-1)}{12} - \sum_i (x_i^2 - 2x_i y_i + y_i^2) /2\\
\quad= \frac{n(n+1)(n-1)}{12} - \sum_i(x_i - y_i)^2/2\\
\quad= \frac{n(n^2-1)}{12} - \sum d_i^2/2$
Numerator/Denominator 
$= \frac{n(n+1)(n-1)/12 - \sum d_i^2/2}{n(n^2 - 1)/12}\\
\quad= {\frac {n(n^2 - 1)/12 -\sum d_i^2/2}{n(n^2 - 1)/12}}\\
\quad= 1- {\frac {6 \sum d_i^2}{n(n^2 - 1)}}\,$.
Hence
$ \rho = 1- {\frac {6 \sum d_i^2}{n(n^2 - 1)}}.$
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