In "A new coefficient of correlation" (Chatterjee, 2019), Chatterjee defined an asymptotically unbiased estimator for the following index, which is zero for independent random variables $X,Y$ and one if $Y$ is a deterministic function of $X$: $$\xi(X,Y) = \frac{\int \mbox{Var}\left(E(1_{\\{Y\geq t\\}}|X)\right) d\mu(t) }{\int \mbox{Var}(1_{\\{Y\geq t\\}}) d\mu(t)}$$

where $\mu$ is the (unconditional) probability distribution of $Y$. The paper does not say so, but I guess that the variance in the numerator is taken with respect to $X$, not with respect to $t$, because otherwise the outer integral with respect to $t$ would make no sense.

As I am not very familiar with this notation, I would first like to ask whether my following understanding is correct:

  1. As $1_{\\{Y\geq t\\}}$ can only be one or zero, its expectaion value under the condition of $X$ is simply the conditional probability $P(Y\geq t|X)$.
  2. The variance in the denominator is the variance of a Bernoulli variable and thus $pq$, or $P(Y\geq t)(1-P(Y\geq t))$.
  3. The outer integrals are irrelevant for the property that $\xi$ is zero for independence of $X$ and $Y$ and one for strict dependency. They are just applied to make the value independent of $t$.

If my above interpretations are correct, the index $\xi$ can equivalently be written in terms of the unconditional distributions $\lambda$ for $X$ and $\mu$ for $Y$ as $$\xi(X,Y) = \frac{\int\int \Big( P(Y\geq t|X=x)-m(t)\Big)^2 d\lambda(x) d\mu(t)}{\int P(Y\geq t) \Big(1-P(Y\geq t)\Big) d\mu(t)} \quad\mbox{with}\quad m(t)=\int P(Y\geq t|X=x) d\lambda(x) = P(Y\geq t)$$

My second question is, what is the value of $\xi$ for the very simple model $Y=f(X) + \varepsilon$ where $X$ is uniformly or normally distributed, $f$ is a deterministic function, and $\varepsilon$ is a normally distributed noise with zero mean and variance $\sigma^2$. A simple Monte Carlo estimate can be achieved by simulating data and use Chatterjee's estimator, but I wonder whether the integrals can be solved in closed form in this simple situation.

  • $\begingroup$ I'm also lost with this notation. It's a shame that no one answered this. $\endgroup$
    – Juan
    Nov 30, 2023 at 20:56
  • $\begingroup$ @Juan In the meantime, I could verify that my interpretation is correct by computing the integrals numerically for a variety of models and comparing this true $\xi$ with the value of xicor() for a large data set generated according to the respective model. I am currently doing a number of tests with xicor for these models and can keep you informed about the results, if you are interested. $\endgroup$
    – cdalitz
    Dec 1, 2023 at 14:40
  • $\begingroup$ Oh! Well I think this us way above my level of understanding. In any case, it's always nice to share your own research, if you're feeling like it. $\endgroup$
    – Juan
    Dec 1, 2023 at 16:51
  • $\begingroup$ @Juan Meanhwile, we have written down our results. Please have a look at section 2 of this paper for a detailed explanation of writing Cahtterjee's $\xi$ in terms of conditional probabilities and how it is related to a similar quantity be Dette et al: arxiv.org/abs/2312.15496 $\endgroup$
    – cdalitz
    Dec 27, 2023 at 13:22

1 Answer 1


In order to mark this question as answered: Yes, the formulation with conditional probabilities is equivalent to Chatterjee's original formulation.

Interestingly, for continuous $Y$, some of the integrals can be readily computed because of $d\mu(t)=-P'(Y\geq t)\,dt$ where the prime denotes derivation with respect to $t$. This yields \begin{align} \int P(Y\!\geq\! t)^2 d\mu(t) &= -\frac{1}{3}P(Y\!\geq\! t)^3\Big|_{-\infty}^{\infty} = \frac{1}{3} \\ \int P(Y\!\geq\! t) d\mu(t) &= -\frac{1}{2}P(Y\!\geq\! t)^2\Big|_{-\infty}^{\infty} = \frac{1}{2} \end{align} which results in $$\xi(X,Y) = 6\int \!\!\!\int\! P(Y\!\geq\! t|X\!=\!x)^2 d\lambda(x)\, d\mu(t) - 2$$ Moreover, for continuous $Y$, the double integral can also be rewritten as \begin{align} \int \!\!\!\int\! & P(Y\!\geq\! t|X\!=\!x)^2 d\lambda(x)\, d\mu(t) \\ &= \int \!\!\!\int\! \Big(1-P(Y\!\leq\! t|X\!=\!x)\Big)^2 d\lambda(x)\, d\mu(t) \\ &= 1 - 2\int \!\!\!\int\! P(Y\!\leq\! t|X\!=\!x) d\lambda(x)\, d\mu(t) \\ &\quad\quad + \int \!\!\!\int\! P(Y\!\leq\! t|X\!=\!x)^2 d\lambda(x)\, d\mu(t) \\ &= 1 - 2 \int P(Y\!\leq\! t)\, d\mu(t) + \int \!\!\!\int\! P(Y\!\leq\! t|X\!=\!x)^2 d\lambda(x)\, d\mu(t) \\ &= \int \!\!\!\int\! P(Y\!\leq\! t|X\!=\!x)^2 d\lambda(x)\, d\mu(t) \end{align} which results in the following expression for $\xi$ that was already proposed in 2010 by Dette, Sieburg & Stoimenov as a measure for correlation between continuous random variables: $$\xi(X,Y) = 6\int \!\!\!\int\! P(Y\!\leq\! t|X\!=\!x)^2 d\lambda(x)\, d\mu(t) - 2$$ R code for computing $\xi$ for some models of $X$ and $Y$ can be found in the appendix of https://arxiv.org/abs/2312.15496.


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