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.
 A: 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.
