Conditional expectation on $\Omega = [0, 1]$ with lebesgue measure

Question

Consider $\Omega=[0, 1]$ with $X(w) = \mathbb{1}_{[0, \frac{1}{3})}(w)+ 2\mathbb{1}_{[\frac{1}{3}, \frac{2}{3})}(w)$ and $Y(w) = 2w+1$. How do i calculate $E[X\mid Y]$ and $E[Y \mid X]$? I tried finding joint density of $(X, Y)$ and then using formula $E(X∣Y)=\int_{\mathbb{R}}{xf_{X \mid Y} (x, Y)}dx$ but I failed, and it seems like there is a easier way, but i dont even know where to start my approach.

Answer 1

The $\sigma$-algebra generated by $Y$ is the Borel $\sigma$-algebra hence $\mathbb E[X\mid Y]=Y$. An other way to see that is to express $X(\omega)$ as a function of $2\omega+1$. To do so, note that $0\leqslant\omega<1/3$ is equivalent to $1\leqslant Y(\omega)<2/3+1$ and do the same treatment for the other indicator.

The $\sigma$-algebra generated by $X$ is generated by the partition $\{[0,1/3),[1/3,2/3),[2/3,1]\}$ and there is a formula for the conditional expectation with respect to a $\sigma$-algebra generated by a partition.