a simple "paradox" Let $x$ be a random vector uniformly distributed on the unit sphere $S^2$ in $\mathbb{R}^3$.
Let $b$ be any fixed vector on the unit sphere. 
Goal: Compute the variance of the random variable $|b^\top x|$.
I have proved that 
$\mathbb{E}[\operatorname{sign}(b^\top x) x] = \frac{2}{3} \cdot b$. A little intuition about this
result: $\mathbb{E}[\operatorname{sign}(b^\top x) x]$ is colinear with $b$, 
because the set $\left\{\operatorname{sign}(b^\top y) y, \, y \in S^2\right\}$ is the hemisphere defined by the direction $b$. The factor $\frac{2}{3}$ is the average height
of the unit hemisphere in $\mathbb{R}^3$. Consequently ,
\begin{align}
\mathbb{E}[|b^\top x|]=\mathbb{E}[b^\top \operatorname{sign}(b^\top x) x] = \frac{2}{3}. 
\end{align}
Now, it is known that the covariance matrix of $x$ is $\frac{1}{3} I$, where
$I$ is the $3 \times 3$ identity matrix. Then
\begin{align}
\operatorname{Var}[|b^\top x|] &= \mathbb{E}[|b^\top x|^2] - \left(\mathbb{E}[|b^\top x|]\right)^2 \\
&= \mathbb{E}[(b^\top x)^2] - \left(\frac{2}{3}\right)^2 \\
&=b^\top \mathbb{E}[x x^\top] b - \frac{4}{9} \\
&= b^\top \left(\frac{1}{3} I\right) b - \frac{4}{9} \\
&= \frac{1}{3} b^\top b - \frac{4}{9} \\
&= \frac{1}{3} - \frac{4}{9} \\
&= -\frac{1}{9}.
\end{align} But this is impossible because the variance is always non-negative.
Question: Where is error?

The average height of the unit hemisphere in $\mathbb R^3$ is $1/2$, not $2/3$. In fact, the height of the points on $S^2$ is uniformly distributed.
