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Consider a conditional random variable

\begin{equation} X = \begin{cases} Y & \quad\quad ,X \in A \\ Z & \quad\quad ,X \in A^\complement \end{cases} \end{equation}

$Y$ and $Z$ are bounded random variables, both strictly positive, continuous and $A$ is some event. I am interested in the upper bound for $\mathbb{P}(|X - \mathbb{E}[X]| > \epsilon)$. So I start as follows

\begin{equation} \begin{aligned} \mathbb{P}(|X - \mathbb{E}[X]| > \epsilon) &= \mathbb{P}(|Y\boldsymbol{1}_A - \mathbb{E}[Y\boldsymbol{1}_A] + Z\boldsymbol{1}_{A^\complement} - \mathbb{E}[Z\boldsymbol{1}_{A^\complement}]| > \epsilon) \\ &\leq \mathbb{P}(|Y\boldsymbol{1}_A - \mathbb{E}[Y\boldsymbol{1}_A]| + |Z\boldsymbol{1}_{A^\complement} - \mathbb{E}[Z\boldsymbol{1}_{A^\complement}]| > \epsilon) \\ &\leq \mathbb{P}(|Y - \mathbb{E}[Y\boldsymbol{1}_A]| + |Z - \mathbb{E}[Z\boldsymbol{1}_{A^\complement}]| > \epsilon) \\ &\leq \mathbb{P}(\max\left\{|Y - \mathbb{E}[Y\boldsymbol{1}_A]|, |Z - \mathbb{E}[Z\boldsymbol{1}_{A^\complement}]|\right\} > \frac{\epsilon}{2}) \\ &\leq \mathbb{P}(|Y - \mathbb{E}[Y\boldsymbol{1}_A]| > \frac{\epsilon}{2}) + \mathbb{P}(|Z - \mathbb{E}[Z\boldsymbol{1}_{A^\complement}]| > \frac{\epsilon}{2}) \end{aligned} \end{equation}

Now how do I get rid of the conditional expectations. Dropping the indicators will violate the upper bound in the last line. How do I get concentration bounds for the conditional variable $X$. Is there a conditional variant of Hoeffding's inequality that can be used in this case.

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I think you could just use Chebyshev's inequality. Since $Y$ and $Z$ are both bounded and positive, then $X$ is bounded and positive. This means the variance should be finite, and Chebyshev's should be applicable.

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  • $\begingroup$ Notice, I don't have means for random variables but scaled conditional expectations. To use Chebyshev, I need simple means. $\endgroup$ – Rohit Arora Aug 22 '18 at 23:18

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