# Conditional Expectation Inequality for bounded moment

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?

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}