How to prove variance of Spearman's Rho How does one derive the variance of Spearman's Rho?
The result is $\frac{1}{n-1}$.
I simplified the formula into the last line shown in the picture but how does one proceed from then? ($R(X_i)$ and $R(Y_i)$ denote the rank of $X$ and $Y$, respectively). I don't know how to calculate the variance. Or is this the wrong approach?

 A: Assuming $(X_1,Y_1),(X_2,Y_2),\ldots,(X_n,Y_n)$ are i.i.d continuous random vectors, the result in the question is true only under the hypothesis $H_0$ (say) of independence of $X$ and $Y$.
Let $\text{Rank}(X_i)=R_i$ and $\text{Rank}(Y_i)=S_i$.
Define the statistic
$$L_n=\sum_{i=1}^n R_i S_i $$
Under $H_0$, the rank vectors $\boldsymbol R=(R_1,\ldots,R_n)$ and $\boldsymbol S=(S_1,\ldots,S_n)$ are independent, and both are uniformly distributed over the set of $n!$ permutations of $(1,2,\ldots,n)$.
Therefore, under $H_0$, $L_n$ can be written in the simpler form
$$L_n=\sum_{i=1}^n i S_i \tag{1}$$
Mean of $L_n$ under $H_0$ is of course
$$E_{H_0}(L_n)=n \left(\frac{n+1}{2}\right)\left(\frac{n+1}{2}\right)$$
And from $(1)$, it can be worked out that
$$\operatorname{Var}_{H_0}(L_n)=\frac1{n-1}\sum_{i=1}^n \left(i-\frac{n+1}{2}\right)^2 \sum_{i=1}^n \left(i-\frac{n+1}{2}\right)^2 \tag{2}$$
Define the standardized statistic
\begin{align}
T_n =\frac{L_n-E_{H_0}(L_n)}{\sqrt{\operatorname{Var}_{H_0}(L_n)}}
&=\sqrt{n-1}\cdot\frac{\sum_{i=1}^n \left(R_i-\frac{n+1}2 \right)\left(S_i-\frac{n+1}2 \right)}{\sqrt{\sum_{i=1}^n \left(i-\frac{n+1}2 \right)^2}\sqrt{\sum_{i=1}^n \left(i-\frac{n+1}2\right)^2}}
\\&= \sqrt{n-1}\cdot\frac{\sum_{i=1}^n \left(R_i-\frac{n+1}2 \right)\left(S_i-\frac{n+1}2 \right)}{\sqrt{\sum_{i=1}^n \left(R_i-\frac{n+1}2 \right)^2}\sqrt{\sum_{i=1}^n \left(S_i-\frac{n+1}2\right)^2}}
\end{align}
Spearman's rank correlation coefficient is therefore
$$r_S=\frac{T_n}{\sqrt{n-1}}$$
But $T_n$ has zero mean and unit variance under $H_0$, whence
$$E_{H_0}(r_S)=0$$
and
$$\operatorname{Var}_{H_0}(r_S)=\frac1{n-1}$$

To prove $(2)$, simply use
$$\operatorname{Var}_{H_0}(L_n)=\sum_{i=1}^n i^2 \operatorname{Var}_{H_0}(S_i)+\sum_{i\ne j}ij \operatorname{Cov}_{H_0}(S_i,S_j)$$
Of course,
$$\operatorname{Var}_{H_0}(S_i)=\frac1n\sum_{i=1}^n \left(i-\frac{n+1}{2}\right)^2 =\sigma^2 \quad,\,\text{say}$$
And for $i\ne j$,
$$\operatorname{Cov}_{H_0}(S_i,S_j)=\frac1{n(n-1)}\sum_{i\ne j}\left(i-\frac{n+1}{2}\right)\left(j-\frac{n+1}{2}\right)=-\frac{\sigma^2}{n-1}$$
It follows that
$$\operatorname{Var}_{H_0}(L_n)=\frac{n^2\sigma^2\sigma^2}{n-1}$$
