I'm currently studying for an upcoming statistics exam and I happened to get stuck on the following question:
Let X and Y be continuous random variables with joint p.d.f. $$f_{x,y}(x, y) = \begin{cases} \ 1/2 & x\geq 0, y\geq 0, x+y \leq 2,\\ 0 & \text{otherwise}. \end{cases} $$ Find P(X > 1|Y = 0.5).
I feel like to get to the answer I have to calculate $\int_1^{2-y}f_{x|y}(x|y=0.5)dx$ but to get to that answer i first need to calculate $f_y(y) = \int_0^{2-y}\frac12dx = \frac12(2-y) \implies f_{x|y}(x|y) = \frac{\frac12}{\frac12(2-y)} = \frac{1}{2-y}$
However this implies that P(X > 1 | Y = 0.5) = $\int_0^{2-y}\frac23 dx$ which doesn't give a number. The only problem I can think of is that my supports are wrong for one of my integrals but I don't see any alternatives. Help would be much appreciated.
It has been a long time since I did these kinds of computations, but if I am not wrong it should go like this:
You did everything correctly until: $f_{x|y} (x|y) = \frac{1}{2-y}$. After that however we compute: $$ \mathbb{P} (X > 1 | Y = y) = \int_{1}^{2-y} f_{x|y} (x|y) dx = \int_{1}^{2-y} \frac{1}{2-y} dx = 1 - \frac{1}{2-y}$$
This correctly depends on $y$. Remember that the conditional probabilities are random variables so they depend on the outcome of the conditioning part.
Now set $y= \frac{1}{2}$ and you will obtain:
$$ \mathbb{P} (X > 1 | Y = \frac{1}{2}) = 1 - \frac{2}{3} = \frac{1}{3}$$.
Which I think is the expected answer (you can try to draw it to see what it should be the expected answer, as I did).