Given a stick of length $1$. $A$ then pick the first point $X$ uniformly distributed over this stick. Then $B$ picks a point $Y$ uniformly distributed in the portion $1-X$ (so, $Y\leq 1-X$). Finally, $C$ just choose the entire portion $Z := 1-X-Y$. Find the expected value of $Z$.
My attempt:
We apply the formula to compute
$E(Z) = \int_{0}^{1} \int_{0}^{1-x} (1-x-y)f_{1-X-Y}(1-x-y) dydx = \int_{0}^{1} \int_{0}^{1-x} (1-x-y)f_{X,Y}(x,y)\ dydx$.
Note that $1-X$ and $Y$ in this case are independent, so $f_{X,Y}(x,y) = f_{X}(x)f_{1-X}(y) = 1\frac{1}{1-x}$ (since the density function of $Y$ is $U[0, 1-X]$, and once $X, Y$ are determined, $Z$ is always fixed.
Thus, $E(Z) = \int_{0}^{1} \int_{0}^{1-x} (1 - y/(1-x)) dydx = \int_{0}^{1} (1-x - (1-x)/2)\ dx = \int_{0}^{1} \frac{1-x}{2} dx = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$.
Question. Is the above solution correct? If not, where did it go wrong?
in a very very simple way, remembering that
$$\mathbb{E}[Y]=\mathbb{E}[\mathbb{E}[Y|X]]$$
and observing that $(Y|X)\sim U[0;X]$ you have
$$\mathbb{E}[Y]=\mathbb{E}[\frac{X}{2}]=\frac{1}{2}\mathbb{E}[X]=\frac{1}{4}$$