construct random variable with a fixed level of Spearman Coefficient to another - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T05:11:39Z http://mathoverflow.net/feeds/question/8930 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/8930/construct-random-variable-with-a-fixed-level-of-spearman-coefficient-to-another construct random variable with a fixed level of Spearman Coefficient to another Steven Pav 2009-12-15T01:25:35Z 2010-01-16T16:43:48Z <p>In my quest to understand all things Spearman, consider the following problem:</p> <p>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 $\sigma^2$.) </p> <p>I am wondering if there is an analogue for the Spearman Rank Correlation Coefficient. I am defining the <em>population</em> 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.)</p> <p>It seems like I would need to know more than just the variance of $x$ to construct $y$, perhaps the entire CDF.</p> http://mathoverflow.net/questions/8930/construct-random-variable-with-a-fixed-level-of-spearman-coefficient-to-another/11995#11995 Answer by Douglas Zare for construct random variable with a fixed level of Spearman Coefficient to another Douglas Zare 2010-01-16T16:43:48Z 2010-01-16T16:43:48Z <p>You don't even need $\sigma^2$ to construct such a variable.</p> <p>Let $Z$ be $+1$ with probability $q$, and $-1$ with probability $1-q$, and independent of $X$.</p> <p>Let $Y = e^XZ$. This squashes X to the positive reals preserving order, and then may change the sign.</p> <p>$y_1$ 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, $E[sign((x_1 - x_2)(y_1-y_2))] = E[Z].$</p> <p>Choose $Z$ to have average value $p$ (set $q=\frac{(p+1)}2$), and then $X$ and $Y$ have Spearman Rank Correlation Coefficient $p$.</p> <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 $[-\alpha,\alpha]$.</p>