# Are there bounds on the Spearman correlation of a sum of two variables?

Given $n$-vectors $x, y_1, y_2$ such that the Spearman correlation coefficient of $x$ and $y_i$ is $\rho_i = \rho(x,y_i)$, are there known bounds on the Spearman coefficient of $x$ with $y_1 + y_2$, in terms of the $\rho_i$ (and $n$, presumably)? That is, can one find (non-trivial) functions $l(\rho_1,\rho_2,n), u(\rho_1,\rho_2,n)$ such that $$l(\rho_1,\rho_2,n) \le \rho(x,y_1+y_2) \le u(\rho_1,\rho_2,n)$$

edit: per @whuber's example in the comment, it appears that in the general case, only the trivial bounds $l = -1, u = 1$ can be made. Thus, I would like to further impose the constraint:

• $y_1, y_2$ are permutations of the integers $1 \ldots n$.
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Only knowing $\rho_{1}, \rho_{2}$, the interval containing $\rho(x, y_{1} + y_{2})$ must include $\rho_{1}$ and $\rho_{2}$: for each $y_{1}, y_{2}$ could have very small values (while having any rank-order), and thus simply "jitter" the values in $y_{1}$ when added to $y_{1}$. Thus the rank-order of $y_{1}$ wouldn't be affected. I don't know if the interval can exceed the $\rho_{i}$. –  caracal Dec 22 '10 at 20:36
@caracal Good observations. The interval definitely can be wider than the $\rho_i$: just consider the case where both correlations are zero. The correlation with the sum can easily be nonzero--it can range all the way from -1 to 1. E.g., x = (1,2,3,4,5); y1 = (3,-10,2,10,1); y2 = (-8,9,-2,-9,4); y1+y2 = (-5,-1,0,1,5) has $\rho_1=\rho_2=0$ but $\rho=1$. –  whuber Dec 22 '10 at 20:57
@whuber: this seems to imply only trivial bounds exist (i.e. $l = -1, u = 1$). Perhaps I have to throw another constraint at the problem. –  shabbychef Dec 22 '10 at 21:13
@shabbychef No, you have posted a nice problem: it's not trivial. In case $\rho_1 = \rho_2 = 1$, for instance, the only possibility is $\rho = 1$. I suspect the bounds are nontrivial except when $\rho_1 = \rho_2 = 0$; they must get narrower as $\rho_1$ and $\rho_2$ approach $\pm 1$. –  whuber Dec 22 '10 at 21:21
Here’s another pathological case. Suppose that $x = y_1$ and $y_1 = -y_2$. Then $\rho(x, y_1 + y_2) = 0$, but $\rho_1 = 1$ and $\rho_2 = −1$. It might be enlightening to think about a simpler, probabilistic version of the problem. Let $X$, $Y_1$, and $Y_2$ be random variables, each with marginally Uniform distributions. Now let $G$ be the CDF of $Y_1 + Y_2$. What can we say about $Cov(X, G(Y_1 + Y_2))$ based on $Cov(X,Y_1)$ and $Cov(X,Y_2)$? –  vqv Dec 22 '10 at 23:56
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Spearman's rank correlation is just the Pearson product-moment correlation between the ranks of the variables. Shabbychef's extra constraint means that $y_1$ and $y_2$ are the same as their ranks and that there are no ties, so they have equal standard deviation $\sigma_y$ (say). If we also replace x by its ranks, the problem becomes the equivalent problem for the Pearson product-moment correlation.
By definition of the Pearson product-moment correlation, \begin{align} \rho(x,y_1+y_2) &= \frac{\operatorname{Cov}(x,y_1+y_2)} {\sigma_x \sqrt{\operatorname{Var}(y_1+y_2)}} \\ &= \frac{\operatorname{Cov}(x,y_1) + \operatorname{Cov}(x,y_2)} {\sigma_x \sqrt{\operatorname{Var}(y_1)+\operatorname{Var}(y_2) + 2\operatorname{Cov}(y_1,y_2)}} \\ &= \frac{\rho_1\sigma_x\sigma_y + \rho_2\sigma_x\sigma_y} {\sigma_x \sqrt{2\sigma_y^2 + 2\sigma_y^2\rho(y_1,y_2)}} \\ &= \frac{\rho_1 + \rho_2} {\sqrt{2}\left(1+\rho(y_1,y_2)\right)^{1/2}}. \\ \end{align} For any set of three variables, if we know two of their three correlations we can put bounds on the third correlation (see e.g. Vos 2009, or from the formula for partial correlation): $$\rho_1\rho_2 - \sqrt{1-\rho_1^2}\sqrt{1-\rho_2^2} \leq \rho(y_1,y_2) \leq \rho_1\rho_2 + \sqrt{1-\rho_1^2}\sqrt{1-\rho_2^2}$$ Therefore $$\frac{\rho_1 + \rho_2} {\sqrt{2}\left(1+\rho_1\rho_2 + \sqrt{1-\rho_1^2}\sqrt{1-\rho_2^2}\right)^{1/2}} \leq \rho(x,y_1+y_2) \leq \frac{\rho_1 + \rho_2} {\sqrt{2}\left(1+\rho_1\rho_2 - \sqrt{1-\rho_1^2}\sqrt{1-\rho_2^2}\right)^{1/2}}$$ if $\rho_1 + \rho_2 \geq 0$; if $\rho_1 + \rho_2 \le 0$ you need to switch the bounds around.
@vqv but if $y_1$ and $y_2$ are permutations of the integers $1\ldots n$ then they are exactly the same as their ranks. –  onestop Dec 23 '10 at 20:52
The ranked values of $y_1 + y_2$ are in general a nonlinear function of $y_1 + y_2$ — even if $y_1$ and $y_2$ are each a permutation of the integers $1,\ldots,n$. Here’s an example: $y_1 = (1,2,3,4)$ and $y_2 = (2,3,1,4)$. Then $y_1+y_2 = (3,5,4,8)$ and $rank(y_1+y_2) = (1,3,2,4)$. Plot $y_1+y_2$ against $rank(y_1+y_2)$ and you’ll see that there is no linear relationship between the two. The above assertion that $\rho(x,y_1+y_2) = Cov(x,y_1+y_2) / \cdots$ is in general false, even under the assumption that $y_1$ and $y_2$ are permutations of the integers. –  vqv Dec 26 '10 at 6:16