Hello people I have a small confusion regarding a given problem.
Lets see it a bit.
We have given that:
$f(x,y) =\begin{cases} x y \text{ if } 0 \leq x \leq 2 \text{ and } 0\leq y\leq 1 \text{ , }\\ 0 \text{ else. }\end{cases}$
So, I have to find this probability $P(X/2 < Y < X)$
To begin with, I found $f$, $y$.
$f_x(x)=x/2$ and $f_y(y)= 2y$
So as you can see $f_x$ and $f_y$ are independent as $f_x \cdot f_y = f(x,y)$.
Now, my main problem is that I am not sure that integral limits should I take!
I have done quite a paper work already, and my "best bet" is this:
If $x$ belongs to $(0,1)$ then I take integral from $\int_{x/2}^x 2y dy= \frac 3 4 x^2.$ So finally, to calculate this interval, we take integral $\int_0^1 \frac 3 4 x^2\cdot \frac x 2 dx=3/32. \tag{Eq 1}$
If $x$ belongs to $ (1,2)$ then, $ \int_0^1 2y dy= 1.$ Basically I should take also another integral from $1$ to $y$, but that one equals to $0$. So finally, to calculate this interval, we take $\int_1^2 x/2 dx = 3/4 \tag{Eq 2}$
To sum up just add Eq1 + Eq2.
As you can see, I cant understand if my second bullet is correct, and why? (Maybe 1st bullet is wrong as well, but at least it makes more sense.
Thanks for your time Gents