The following distribution is given for this problem:
$f(x, y)= \begin{cases}8 x y & \text { if } 0 \leq y \leq x \leq 1 \\ 0 & \text { otherwise }\end{cases}$
given is $x= \frac{1}{2}$.
The problem is
Calculate the expected value of $Y^3$ if $X=\frac{1}{2}$.
From the definition I would integrate $f(\frac{1}{2},y)Y^3$ from $0$ to $\frac{1}{2}$, since that is the upper $y$ value at $x=\frac{1}{2}$. However this is wrong. The manual gives the solution
$E\left(Y^3 \mid X=\frac{1}{2}\right)=\int_0^{\frac{1}{2}} y^3 f\left(y \mid \frac{1}{2}\right) d y=\frac{1}{20}$
My question is:
Why do I need to condition first on $x$, what is the intunition behind it?
If you were using discrete variables, say rolling a 6-sided and a 12-sided die, summing $p(3,y)$ for y from 1 to 12 would tell you the chance of getting a 3 on the 6-sided die. Summing $yp(3,y)$ would tell you expected value of "Y times the chance of getting a X=3 for that Y", which is not on the surface a terribly useful thing to know.
For continuous distributions, calculating $\int y f(0.5,y)dy$ is even less useful, since $f(x,y)$ doesn't give probabilities, it gives probability densities - so $\int f(0.5 , y)$ means something like "the expected value of "Y times the probability density of X, near X=0.5, for that Y". Again, not really useful.
That's why you have to work out $f(y|X=0.5)$, which is "the probability density function for Y given X=0.5" and then only work out the expectation value.