Calculate PDF for 2 random variables?

Question

Given $(X,Y)$ uniformly distubuted in $([1,2]\times[1,4]) \cup ([2,3]\times[2,3])$.

I found that $f_{X,Y}(x,y) = 1/4$ if $(x,y)$ in $([1,2]\times[1,4]) \cup ([2,3]\times[2,3])$, $0$ otherwise.

I'm stuck calculating $f_Y(y)$:

$f_Y(y)$ = $\int_{-\infty}^{\infty} f_{X,Y}(x,y) dx=\int 1/4 dx = x/4$

I don't know what limits to use here... as we have 2 cases for $x$. any help?

Answer 1

The limits depend on the value of $y$. If $y$ is between 1 and 2 or between 3 and 4 you have one region, from which you get $$\int_1^2 \frac{1}{4} dx=1/4$$ If y is between 2 and 3 you get both regions, or $$\int_1^2 \frac{1}{4} dx + \int_2^3 \frac{1}{4} dx=1/2$$

Answer 2

$(X, Y)$ is uniformly distributed in $([1,2]\times[1,4]) \cup ([2,3]\times[2,3])$.

As you mentioned, $ \displaystyle f_{XY}(x, y) = \frac 14$

$(i)$ For $~y \in [1, 2] \cup [3, 4]$, $x \in [1,2]$

$ \displaystyle f_Y(y) = \int_1^2 \frac 14 ~dx = \frac 14$

$(ii)$ For $~y \in [2, 3], x \in [1,3]$

$ \displaystyle f_Y(y) = \int_1^3 \frac 14 ~dx = \frac 12$

So,

$f_Y(y) = \left \{\begin{array} {l}\frac 14 & 1 \lt y \lt 2 \\ \frac 12 & 2 \lt y \lt 3 \\ \frac 14 & 3 \lt y \lt 4 \\ 0 & \text {otherwise} \\ \end{array} \right.$