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Given that $E[Y\mid X] = 1$, show that $\operatorname{Var}(XY)\geqslant\operatorname{Var}(X)$.

So I tried to expand $\operatorname{Var}(XY) = \operatorname{E}(X^2 Y^2) - 1$ and was stuck here.

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  • $\begingroup$ So, $Y=1+Z$ where $E(Z\mid X)=0$. In particular, $E(XY)=E(X)+E(XZ)=E(X)+E(XE(Z\mid X))$ hence $E(XY)=E(X)$ and $E(X^2Y^2)=E(X^2(1+Z)^2)=E(X^2)+2E(X^2Z)+E(X^2Z^2)$ hence $E(X^2Y^2)\geqslant E(X^2)$ as soon as $E(X^2Z)=0$, right? Now, ... $\endgroup$
    – Did
    Oct 28, 2017 at 15:04

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We have $$\operatorname{Var}[XY] = \operatorname{E}[X^2 Y^2] - \operatorname{E}[XY]^2.$$ Try to tackle each of these two terms by first conditioning on $X$ and then taking expectation. For a full proof, hover below.

Write $\operatorname{E}[XY] = \operatorname{E}[\operatorname{E}[XY \mid X]] = \operatorname{E}[X \operatorname{E}[Y \mid X] ] = \operatorname{E}[X]$ since $\operatorname{E}[Y \mid X] = 1$. Lastly, we have $$\operatorname{E}[X^2Y^2] = \operatorname{E}[ X^2 \operatorname{E}[Y^2 \mid X]] \geq \operatorname{E}[ X^2 \operatorname{E}[Y\mid X]^2 ] = \operatorname{E}[X^2]$$ where we use Jensen's inequality for conditional expectation. Thus $$\operatorname{Var}[XY] \geq \operatorname{E}[X^2] - \operatorname{E}[X]^2.$$

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  • $\begingroup$ You guys are awesome! $\endgroup$ Oct 29, 2017 at 8:20
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$\newcommand{\v}{\operatorname{var}}\newcommand{\e}{\operatorname{E}}$By the law of total variance, you have \begin{align} \v(XY) & = \v(\e(XY\mid X)) + \e(\v(XY\mid X)) \\[10pt] & = \v(X\e(Y\mid X)) + \e(X^2\v(Y\mid X)) \\[10pt] & = \v(X) + \e(X^2\v(Y\mid X)) \\[10pt] & \ge \v(X). \end{align}

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