Getting the marginal distribution from the joint pdf 
To bein with, I did the following calculations:
$$
Y\sim Uniform(0,x)\\
f_x(x)=\{\frac{1}{x^2},x\ge1\}\\
f_{y|x}(y)=\{\frac{1}{x},0\le y \le x\}\\
f(x,y)=f_x(x)f_{y|x}(y)=\frac{1}{x^3},x\ge 1,0\le y\le x\\
$$
To find $f_y(y)$, we must integrate $f(x,y)$ in the $dx$ axis. However I don't understand why the lower boundary of the integral is not $1+y$ instead of $y$ because if we draw $x\ge1, 0\le y\le x$ in a plane $x$ starts at $y+1$. Could someone explain why the lower boundary must start at $y$? Thanks!
$$
f_y(y)=\int_{y}^{\infty}f(x,y)dx
$$
 A: In the general case, neither of these is right. You have two inequalities for $x$ in $f(x,y)$, namely $x\ge1$ and $x\ge y$, so the lower limit of the integral is $\max(1,y)$. You appear to be implying that the lower limit $y$ was specified in some book or lecture or the like. If so, I suspect that this was done because you need the value for $Y=\frac32$, and $\max\left(1,\frac32\right)=\frac32$. I can't say much about why you thought that a drawing suggests that $x$ starts at $y+1$, since you didn't provide the drawing.
A: It is because $0\leq y\leq x$ is the support if $Y$.  When you factor in the support of $X$, the support of the joint distribution is , $0\leq y, \max(1,y)\leq x$
$$\begin{align}f_{Y}(y) =&~ \int_{y}^\infty f_{X,Y}(x,y)\operatorname d x~\big[0\leq y\big]\\[2ex] =&~ \int_{\max(1,y)}^\infty \tfrac 1{x^3}\operatorname d x~\big[0\leq y\big]\\[2ex]=&~\begin{cases}\int_1^\infty x^{-3}\operatorname d x & : 0\leq y< 1\\[1ex]\int_y^\infty x^{-3}\operatorname dx & : 1\leq y \end{cases}\end {align}$$ 


 PS: You are seeking $\mathsf E(X\mid Y=3/2)$ which is $\frac{\int_{3/2}^\infty x~f_{X, Y}(x, 3/2)\operatorname d x}{\int_{3/2}^\infty f_{X, Y}(x, 3/2)\operatorname d x}=\frac{\int_{3/2}^\infty x^{-2}\operatorname d x}{\int_{3/2}^\infty x^{-3}\operatorname d x}$, since $3/2>1$

