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Let $X$ and $Y$ be continuous random variables with finite first and second moments. Then, is it true that $Var[X]\geq Var[E(X|Y)]$?

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  • $\begingroup$ Posting this kind of question on math.stackexchange.com/ would be more appropriate. $\endgroup$
    – Christophe Leuridan
    Commented Jan 18, 2023 at 20:17
  • $\begingroup$ hi @ChristopheLeuridan, how come? what's the difference between the two? $\endgroup$
    – Adrian Leverkuhn
    Commented Jan 18, 2023 at 23:06
  • $\begingroup$ MathOverflow is primarily for asking questions on mathematics research. $\endgroup$
    – Christophe Leuridan
    Commented Jan 19, 2023 at 15:39

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It is enough to show $$ Var[X] - Var[E[X|Y]] \ge 0. $$ By the fact $Var[X] = E[X^2] - E[X]^2$, $$ Var[X] - Var[E[X|Y]] = E[X^2] - E[X]^2 - E\big[ E[X|Y]^2 \big] + E \big[E[X|Y]\big]^2 \\ = E\big[ X^2 - E[X | Y]^2 \big] \\ = E \big[ E\big[X^2 - E[X|Y]^2 \big| Y\big] \big] = E[Var[X|Y]\big] \ge 0, $$ where the second and the third equality is due to the law of total expectation.

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  • $\begingroup$ thank you, @SeungHyeonYu! $\endgroup$
    – Adrian Leverkuhn
    Commented Jan 18, 2023 at 23:09

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