Equivalent Conditions for Conditional Independence

Question

We have random variable $V$ and two independent random vectors $U_1$ and $U_2$. How to prove that $$ E(V\mid U_1,U_2)=E(V\mid U_1) \Leftrightarrow \forall t_1, E(Ve^{i\langle t_1,U_1\rangle}\mid U_2) = E(Ve^{i\langle t_1,U_1\rangle}) $$

Answer 1

If you have suppose that $E(V\mid U_1,U_2)=E(V\mid U_1)$ then by interverting $U_1$ with $U_2$ we get $E(V\mid U_2,U_1)=E(V\mid U_2).$

In the other hand, $Ve^{i\langle t_1,U_1\rangle}$ and $U_2$ are independent since $U_1$ and $U_2$ are independent.

Then \begin{align*} E(Ve^{i\langle t_1,U_1\rangle}\mid U_2)=& e^{i\langle t_1,U_1\rangle}E(V\mid U_2)\\ =&e^{i\langle t_1,U_1\rangle}E(V\mid U_1,U_2)\\ =&e^{i\langle t_1,U_1\rangle}E(V\mid U_1)\\ =&E( e^{i\langle t_1,U_1\rangle} V\mid U_1)\\ =E(Ve^{i\langle t_1,U_1\rangle}) \end{align*} since $e^{i\langle t_1,U_1\rangle}$ is $\sigma(U_1)$ measurable.

Answer 2

$\Rightarrow$: \begin{align*} E(V\mid U_1,U_2)=&E(V\mid U_1) \\ \Rightarrow E(Ve^{i\langle t_1,U_1\rangle}\mid U_2) =& E_{U_1}(E_{V}(Ve^{i\langle t_1,U_1\rangle}\mid U_2, U_1)\mid U_2) \\ =& E_{U_1}(E_{V}(V\mid U_2, U_1)e^{i\langle t_1,U_1\rangle}\mid U_2) \\ =& E_{U_1}(E_{V}(V\mid U_1)e^{i\langle t_1,U_1\rangle}\mid U_2) \\ =& E_{U_1}(E_{V}(V\mid U_1)e^{i\langle t_1,U_1\rangle}) \\ =& E(Ve^{i\langle t_1,U_1\rangle}) \end{align*}

$\Leftarrow$: $$\forall t_1, E(Ve^{i\langle t_1,U_1\rangle}\mid U_2) = E(Ve^{i\langle t_1,U_1\rangle})$$ LHS: \begin{align*} & E(Ve^{i\langle t_1,U_1\rangle}\mid U_2) \\ =& E_{U_1}(E_{V}(Ve^{i\langle t_1,U_1\rangle}\mid U_2, U_1)\mid U_2) \\ =& E_{U_1}(E_{V}(V\mid U_2, U_1)e^{i\langle t_1,U_1\rangle}\mid U_2) \\ =& \int E(V\mid U_1=u_1,U_2) e^{i\langle t_1,U_1\rangle} dF_{U_1}(u_1) \quad\quad (U_1, U_2\text{ independent}) \quad\quad (1) \end{align*} RHS: \begin{align*} & E(Ve^{i\langle t_1,U_1\rangle}) \\ =& E(E(Ve^{i\langle t_1,U_1\rangle}\mid U_1)) \\ =& E(E(V\mid U_1)e^{i\langle t_1,U_1\rangle}) \\ =& \int E(V\mid U_1=u_1) e^{i\langle t_1,U_1\rangle} dF_{U_1}(u_1) \quad\quad (2) \end{align*}

We can see (1) and (2) as Fourier transformation, which is a one-to-one map. Since (1)=(2) holds for any $t_1$, we can say the original functions before Fourier transformation are the same, hence, $E(V\mid U_1,U_2)=E(V\mid U_1)$.