Suppose X is uniform in the interval [0, Y] and Y can either be 1 or 2 with equal probability.
How would I calculate P(X>1)?
This is what I'm trying to do:
$f_x(x) = \int_a^b \! f_{X,Y}(x,y) \, \mathrm{d} y $
where,
$f_{X,Y}(x,y) = \sum\limits_{y=1}^{2}f_{X\mid Y=y}(x\mid y) * p_Y(y) = 3/4$
Did I calculate the joint density correctly, because it feels a bit awkward adding pdf's like that.
Another trouble I'm having is figuring how to decide the limits of integration. I know a should be 1 but what should b be? Should it be 2 because that is technically the highest value X could take but can you only say that for the distribution $X\mid Y$ or can you say that for X also?