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Let $X,Y,Z$ be random variables. $X \sim \text{Unif}(0,1)$. Let $0<a<b<1$

$$Y = \begin{cases} 0, & \text{if } 0 \leq X\leq a, \\ 1, & \text{otherwise} \end{cases}$$

$$Z = \begin{cases} 0, & \text{if } b \leq X\leq 1, \\ 1, & \text{otherwise} \end{cases}$$




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$Y$ and $Z$ are not independent. This is because $P(Y,Z) != P(Y) ^{*} P(Z)$

Find $E_{Y}(Y | Z=z)$ for each value of $z$.

$Z=0$

$E(Y|Z=0)= 0P(Y=0|Z=0)+1P(Y=1|Z=0)$

$= 0+1^{*}\dfrac{a}{b}=\dfrac{a}{b}$

$E(Y|Z=1)= 0P(Y=0|Z=0)+1P(Y=1|Z=0) = 0$

by Wooden (693 points)