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Given $\mathbb E\left(X^2\right) \lt \infty $, why $$\mathbb E\Big(\big(X -\mathbb E(X \mid Y )\big)^2\Big) \leq 2\mathbb E\left(X^2\right) \lt \infty \;\text?$$

Mohsen Shahriari
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Songlin Wei
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    Not sure about the prefactor $2$, but we even have a better version: $$E(X^2)=E\left((X-E(X\mid Y))^2\right)+E\left((E(X\mid Y))^2\right)$$ – Sangchul Lee Aug 04 '20 at 03:48
  • Thanks for helping, i find this on a book, this inquality was referenced when trying to prove another theorem. But I have problem understanding your equality, too. – Songlin Wei Aug 04 '20 at 04:25
  • @SangchulLee Thank you! I found the theorem for the equality you wrote. this clearly implies the inequality i asked. – Songlin Wei Aug 04 '20 at 08:30

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