How to prove $\text{var}(X_1X_2)\geq \text{var}(X_1)$ when $E(X_2|X_1)=1$ How to prove that $\text{var}(X_1X_2)\geq \text{var}(X_1)$ when $E(X_2|X_1)=1$ ?
I just try $\text{var}(X_1X_2)=E(X_1^2X_2^2)-E(X_1X_2)^2$ and $\text{var}(X_1)=E(X_1^2)-E(X_1)^2$,
but then I do not know how to proceed.
 A: Using the law of total variance,
$$\operatorname{Var}(X_1X_2)
  =E\underbrace{\operatorname{Var}(X_1X_2|X_1)}_{\ge 0}+\operatorname{Var}E(X_1X_2|X_1)
 \ge\operatorname{Var}(X_1\underbrace{E(X_2|X_1)}_{=1})
=\operatorname{Var}X_1.
$$
A: First note by the law of iterative expectation and assumption $E(X_2|X_1) = 1$, we have
\begin{align}
E(X_1X_2) = E[E(X_1X_2|X_1)] = E[X_1E(X_2|X_1)] = E(X_1).
\end{align}
Therefore,
\begin{align}
& \operatorname{Var}(X_1X_2) = E[(X_1X_2 - E(X_1X_2))^2] = E[(X_1X_2 - E(X_1))^2] \\
=& E[(X_1X_2 - X_1 + X_1 - E(X_1))^2] \\
=& E[(X_1X_2 - X_1)^2] + 2E[(X_1X_2 - X_1)(X_1 - E(X_1))] + \operatorname{Var}(X_1).
\tag{1}
\end{align}
The inequality $\operatorname{Var}(X_1X_2) \geq \operatorname{Var}(X_1)$ thus holds if we can prove the middle term in $(1)$ is $0$.  Indeed, applying the law of iterative expectation and $E(X_2|X_1) = 1$ again yields
\begin{align}
& E[(X_1X_2 - X_1)(X_1 - E(X_1))] = E[E[(X_1X_2 - X_1)(X_1 - E(X_1))|X_1]] \\
=& E[(X_1 - E(X_1))E[X_1X_2 - X_1 | X_1]] \\
=& E[(X_1 - E(X_1))(E(X_1X_2|X_1) - X_1)] \\
=& E[(X_1 - E(X_1))(X_1 - X_1)] = 0.
\end{align}
This completes the proof.
