construct random variable with a fixed level of Spearman Coefficient to another

Question

In my quest to understand all things Spearman, consider the following problem:

Given random variable $x$ with known variance, $\sigma^2$, and $p \in [-1,1]$, one can construct a random variable $y$ such that the Pearson correlation coefficient of the variables $x$ and $y$ is exactly $p$. (Let $y = p x + \sqrt{(1-p^2)} z,$ where $z$ is a random variable independent of $x$ with variance $\sigma^2$.)

I am wondering if there is an analogue for the Spearman Rank Correlation Coefficient. I am defining the population Spearman Correlation Coefficient of the jointly distributed $(x,y)$ as $$E[sign((x_i - x_j)(y_i - y_j))].$$ (this is the subject of another of my questions here on M.O.)

It seems like I would need to know more than just the variance of $x$ to construct $y$, perhaps the entire CDF.

Answer 1

You don't even need $\sigma^2$ to construct such a variable.

Let $Z$ be $+1$ with probability $q$, and $-1$ with probability $1-q$, and independent of $X$.

Let $Y = e^XZ$. This squashes X to the positive reals preserving order, and then may change the sign.

$y_1$ and $y_2$ have the same ordering as $x_1$ and $x_2$ when the greater value isn't negated. That is, if $x_1 \gt x_2$, then $sign((x_1 - x_2)(y_1-y_2)) = sign(y_1)$. If $x_1\lt x_2$, then $sign((x_1 - x_2)(y_1-y_2)) = sign(y_2)$. So, $E[sign((x_1 - x_2)(y_1-y_2))] = E[Z].$

Choose $Z$ to have average value $p$ (set $q=\frac{(p+1)}2$), and then $X$ and $Y$ have Spearman Rank Correlation Coefficient $p$.

Actually, there is a little ambiguity (to me) about whether you allow $x_1 = x_2$, which I ignored above. If under your definition, the rank correlation of $X$ with itself is $\alpha$, then the rank correlation of $X$ with $Y$ is $\alpha p$, and you can get any value in $[-\alpha,\alpha]$.