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.
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1$\begingroup$ What have you tried and where are you stuck? $\endgroup$– whuber ♦Commented Nov 12, 2022 at 16:10
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$\begingroup$ I split the variance into the form of expectation, but I don't know how to relate it to the previous conditional expectation $\endgroup$– wysngCommented Nov 12, 2022 at 16:14
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2$\begingroup$ Please add your attempts and approach to the question and not as a comment. Elaborate your attempts clearly so that the community might be able to steer away any confusion or so. $\endgroup$– User1865345Commented Nov 12, 2022 at 16:30
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2 Answers
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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.
answered Nov 12, 2022 at 21:15
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Using the law of total variance, $$\operatorname{Var}(X_1X_2) =E\underbrace{\operatorname{Var}(X_1X_2|X_1)}_{\ge 0}+\operatorname{Var}E(X_1X_2|X_1) \ge\operatorname{Var}(X_1\underbrace{E(X_2|X_1)}_{=1}) =\operatorname{Var}X_1. $$
answered Nov 13, 2022 at 12:53