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

Question

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

Tried apply a few inequality such as Schwartz and Jensen, but none seem to be working. If not entire solution some useful inequality would also work as a suggestion

Answer 1

All reduces to the so very useful formula $$ \mathbb{E}[f(X,Y)] = \mathbb{E}\big[ \mathbb{E}[f(X,Y) \mid X]\big] $$ and the so-called "take out what is known" rule. Note that thanks to this formula first of all we see that: $$ \mathbb{E}[XY] = \mathbb{E}[X] $$ so that $\operatorname{Var}(XY) \ge \operatorname{Var}(X)$ is equivalent to showing that $$\mathbb{E}[X^2Y^2]\ge \mathbb{E}[X^2]$$ Now an application of jensen's inequality together with the above formula delivers the result.

Indeed $$ \mathbb{E}[X^2Y^2] =\mathbb{E}\big[ X^2\mathbb{E}[Y^2 \mid X ] \big] $$ and now $\mathbb{E}[Y^2 \mid X ] \ge \mathbb{E}[Y \mid X ]^2 = 1.$

Answer 2

$\newcommand{\v}{\operatorname{var}} \newcommand{\e}{\operatorname{E}}$ \begin{align} \v(XY) & = \e(\v(XY\mid X))+\v(\e(XY\mid X)) \\[10pt] & = \e(X^2\v(Y\mid X)) + \v(X\e(Y\mid X)) \\[10pt] & = \e(X^2) + (\text{a nonnegative number}) \\[10pt] & \ge \e(X^2) \ge \e(X^2) - (\e(X))^2 = \v(X). \end{align} The inequality is strict unless the variance in the second term is $0,$ i.e. unless $\e(XY\mid X)$ does not depend on $X.$