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I was going through a proof in Agnostic Estimation of Mean and Variance (Lemma 3.12, page 21) and encountered the following:

  • Let $X$ be a random variable with $\mathbf{E}[X] = \mu$ and $\mathbf{E}[(X-\mu)^2] = \sigma^2$
  • Let $A$ be an event that occurs with probability $1-\epsilon$ and let $A^c$ be the complement (i.e. $P(A^c)=\epsilon$)
  • Let $d \Omega$ be the probability measure

Using $E\big[(Y - E[Y])^4\big] \geq \big(E\big[(Y-E[Y])^2\big]\big)^2$ for a random variable $Y$, and $P(A^c)=\epsilon$, we have

$$ \frac{1}{\epsilon} \bigg( \int_{A^c} (X - \mu)^2 d\Omega \bigg)^2 \leq \int_{A^c} (X - \mu)^4 d\Omega $$ What I do not understand is where the $\frac{1}{\epsilon}$ is coming from when just using the proposed inequality with $Y = X \mid A^c$?

And how to show the inequality still holds?

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So this is actually trivial but the notation was confusing me.

Because $d\Omega$ is the probability measure for the entire space, $\int_{A^c}(X-\mu)^2 d\Omega=P(A^c)Var(X \mid A^c)$ so we'd have to normalize it.

$$ \begin{aligned} &\bigg( \int_{A^c} (X - \mu)^2 d \Omega \bigg)^2 \leq \int_{A^c} (X - \mu)^4 d\Omega \\ &\implies \bigg( \frac{1}{P(A^c)} \int_{A^c} (X-\mu)^2 d\Omega\bigg)^2 \leq \frac{1}{P(A^c)} \int_{A^c} (X - \mu)^4 d\Omega\\ &\implies \frac{1}{P(A^c)}\bigg( \int_{A^c} (X-\mu)^2 d\Omega\bigg)^2 \leq \int_{A^c} (X - \mu)^4 d\Omega \end{aligned} $$

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