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?

The joint distribution of $(U_1,U_2,V_2,V_2)$ factorises as
$$f(u_1,v_1)f(u_2,v_2)$$ by assumption.
(i) Since
$$(U_1,U_2)|(V_1,V_2) \sim p(u_1,u_2|v_1,v_2)\propto f(u_1,v_1)f(u_2,v_2)$$
the factorisation shows that $U_1$ and $U_2$ are independent given $(V_1,V_2)$, hence
$$\mathbb E[(1 - U_1^2)(1 - U_2^2)|V_1, V_2] = \mathbb E[(1 - U_1^2)|V_1, V_2]\mathbb E[(1 - U_2^2)|V_1, V_2]$$
(ii) Since $(i=1,2)$
$$U_i|(V_1,V_2) \sim q(u_i|v_1,v_2)\propto f(u_i,v_i)\int f(u_{3-i},v_{3-i})\,\text du_{3-i}\propto f(u_i,v_i)$$
$U_i$ is independent of $V_{3-i}$ given $V_i$, hence
$$\mathbb E[(1-U_i^2)|V_1,V_2]=\mathbb E[(1-U_i^2)|V_i]$$
