Range of Marginal Probability Distibution 
Set 
  $$f_{X,Y}(x,y) = \begin{cases}2,& \text{ if }x\in(0,1), y\in (0,x]\\0,& \text{ otherwise.}
\end{cases} $$
  (a) Let $W=X$ and $Z=X/Y$ and compute $f_{W,Z}$.
(b) Hence obtain $f_{W\mid Z}(w\mid Z)$ and use $f_{W\mid Z}(w\mid 1)$ to compute $\mathbb P(X>1/2\mid Z=1)$.

In order to find the conditional pdf w|z we need to find the marginal pdf of z but what are the limits we put on this integral? Is it the values that z can take, w, or a combination of both? When I used the limits of z, or w, I get a conditional pdf with only w in it so I know this can not be right. 
Thanks 
 A: It should be possible to see that the support for $W$ is $[0,1]$ and for $Z$ is $[1, \infty)$
So let's consider $\mathbb P(W \le w, Z \le z)=\mathbb P\left(X \le w, \frac XY \le z\right)=\mathbb P\left(X \le w, Y \ge \frac X z\right)$ in that support.  Either doing the integrations or drawing a square and some triangles and multiplying the areas by $2$ suggests that turns out to be $F_{W,Z}(w,z)=1-\frac{w^2}{z}$ in the support
Now taking the derivatives with respect to $w$ and $z$ suggests that the joint density  is $f_{W,Z}(w,z)= \frac{2w}{z^2}$ in the support
That then suggests that the marginal density is $f_{W \mid Z}(w \mid Z)=2w$ for $w \in [0,1]$ and thus $f_{W \mid Z}(w \mid Z=1)=2w$ and then $\mathbb P(W \gt 1/2 \mid Z=1) = \frac34$ i.e. $\mathbb P\left(X \gt 1/2 \mid \frac XY =1\right) = \frac34$ 
You have now managed to produce an example of the Borel–Kolmogorov paradox by conditioning on an event of probability $0$.  We know $W=X$ while $Z= \frac X Y=1 \implies X=Y \implies X-Y=0$.  If you had try to find $\mathbb P(X \gt 1/2 \mid X-Y=0)$ the same way, the result would have been $\frac12$ not $\frac34$ 
Here is a little simulation in R to demonstrate the point
cases <- 10^6
set.seed(1)
X0 <- runif(cases)
Y0 <- runif(cases) 
X <- X0[Y0 <= X0]  # remove cases where Y0 > X0
Y <- Y0[Y0 <= X0]
margin <- 0.01
mean(X[abs(X / Y - 1) < margin] > 1/2) # cond.prob X > 1/2 given X / Y about 1
# 0.7507466
mean(X[abs(X - Y - 0) < margin] > 1/2) # cond.prob X > 1/2 given X - Y about 0
# 0.5035035 

