# Kendall correlation between $X$ and $X^2$

Let $X \sim N(0, 1)$. Can you confirm that the Kendall's correlation coefficient between $X$ and $X^2$ is equal to $0$. The way I interpret this, is that Kendall's tau only measures monotonic dependence.

• Also Pearson's correlation is zero. The reason is the symmetry of your density around 0: for positive x, you have a positive dependency between x and its square. For negative x,the relation is negative but of equal strength. – Michael M May 28 '14 at 5:46
• @Mayer: Thank you for the intuition. Then, we agree on the fact that tau = 0 because Kendall's tau only measures monotonic dependence... isnt'it ? – user7064 May 28 '14 at 5:49
• Not fully. Symmetry of the density around 0 will lead all reasonable measures of correlation (also such that consider non-monotone relationships) to say 'no dependency'. But how to prove this claim? – Michael M May 28 '14 at 6:02
• @whuber: Coming from U-statistics theory, Kendall's tau is estimating the probability of drawing a concordant pair of points from the bivariate distribution minus the probability of a discordant pair. Couldn't this theoretical version of Kendall's tau be computed for the distribution of $(X, X^2)$? – Michael M May 28 '14 at 19:19
• @Michael: That looks like a very nice way to interpret the question. Your argument immediately proves the value is $0$ for any distribution of $X$ that is symmetric about $0$. – whuber May 28 '14 at 19:23

The population version of Kendall's $\tau$ (probability of concordance minus the probability of discordance) between $Z_1$ and $Z_2$ is given by \begin{align*} \tau(Z_1,Z_2) &= P\left[ (Z_1 - Z_1^*) (Z_2 - Z_2^*) > 0 \right] - P\left[ (Z_1 - Z_1^*) (Z_2 - Z_2^*) < 0 \right]\\ &= 2\ P\left[ (Z_1 - Z_1^*) (Z_2 - Z_2^*) > 0 \right] - 1 \end{align*} where $(Z_1,Z_2)$ and $(Z_1^*,Z_2^*)$ are iid distributed.
Now remark that $x < y$ and $x^2 < y^2$ if and only if $|x| < y$, which you can see by noting that $$x^2 < y^2\ \Leftrightarrow\ |x| < |y|,$$ and evaluating the three possibilities \begin{align*} x < y < 0 &: \quad x^2 \not< y^2 \quad \text{and} \quad |x| \not< y\\ x < 0 < y &: \quad x^2 < y^2 \ \Leftrightarrow \ |x| < y\\ 0 < x < y &: \quad x^2 < y^2 \quad \text{and} \quad |x| < y.\\ \end{align*}
We're all set to prove the result. Letting $Y \sim N(0,1)$ be an independent copy of $X$, we get \begin{align*} P\left[ (X - Y) (X^2 - Y^2) > 0 \right] &= P\left[ X > Y \quad \text{and} \quad X^2 > Y^2 \right] + P\left[ X < Y \quad \text{and} \quad X^2 < Y^2 \right]\\ &= P\left[\ |Y| < X \right] + P\left[\ |X| < Y \right]\\ &= 2\ P\left[\ |X| < Y \right]. \end{align*} That you can compute in many ways, (here I skip some obvious steps) \begin{align*} P\left[\ |X| < Y \right] &= (.5)\ P\left[\ |X| < Y\ |\ Y > 0 \right]\\ &= (.5)\ P\left[\ X < Y\ |\ Y > 0, X>0 \right]\\ &= (.5)(.5) = .25\\ \end{align*}
You therefore get $\tau(X,X^2) = 2*2*(.25) - 1 = 0$. We never use the normality assumption... Only that the distribution is symmetric about 0.