The equality about conditional expectation Let $(U_i,V_i)$ be i.i.d samples from a bivariate distribution. Show that $$\mathbb{E}[(1-U^2_1)\cdot(1-U^2_{2})\cdot|V_1-V_2|]=\mathbb{E}\left\{\mathbb{E}[(1-U^2_1)|V_1]\cdot\mathbb{E}[(1-U^2_2)|V_2]\cdot|V_1-V_2| \right\}$$
I have no ideal how to start.Are there some properties of conditional expectation I should use?
Could you give me some hints?

Let $Z_i = 1-U_i^2$, so that $(Z_i, V_i), i=1,2$ are i.i.d. samples form a bivariate distribution.
Let $\hat Z_i = \mathbb E [Z_i | V_i]$ and $W = |V_1 - V_2|$. We want to show:
$$
\mathbb E(Z_1 Z_2 W) = \mathbb E \bigl[ \hat Z_1 \hat Z_2 W \bigr]
$$
Let $V = (V_1, V_2)$. Then,
\begin{align}
\mathbb E [\mathbb E(Z_1 Z_2 W | V) ] = 
\mathbb E [\mathbb E(Z_1 Z_2 | V) W  ].
\end{align}
We just need to argue that given $V$, $Z_1$ and $Z_2$ are independent, since then $\mathbb E(Z_1 Z_2 | V) = \mathbb E (Z_1 | V) \mathbb E (Z_2 | V) = \hat Z_1 \hat Z_2$.
Why $Z_1$ and $Z_2$ are independent given $V$? You can use the approach in @Xi'an's answer. Alternatively, it is clear from the graphical model of the problem
$$
Z_1 - V_1 \\
Z_2 - V_2
$$
Conditioning on $(V_1,V_2)$ is like coloring nodes $V_1$ and $V_2$ and asking whether there a path between $Z_1$ and $Z_2$ through uncolored nodes? The answer is no, hence they are independent independent given $V$.
