Intuitive probability inequality $X$ and $Y$ are two continuous i.i.d random variables. They are both symmetric about zero.
How do I prove the intuitive inequality:
$P(\ |X+Y| < 2|X|\ \big | \ XY<0 \ ) > 0.5$
I know that $P(XY < 0) = 0.5$, I tried finding $P(\ |X+Y| < 2|X|\ and \ XY<0 \ )$ by integrating the probability density functions but I didn't get very far. I feel like I am missing something obvious.
 A: As mentioned in comments, there are two main cases depending on the signs of $X$ and $Y.$
Try just one case to start with, for example $X > 0, Y < 0.$
That is, consider values of $X$ and $Y$ only in the fourth quadrant.
In the fourth quadrant, plot the region of points $(x,y)$ for which
$\lvert x + y \rvert < 2\lvert x\rvert.$
Since $x > 0$ in this quadrant, the formula to plot can be simplified to
$\lvert x + y \rvert < 2x.$
You can deal with the remaining absolute value by subdividing into additional cases, $x + y < 0$ and $x + y > 0.$
The region in the other quadrant should then be relatively easy to plot.
In fact you can plot it immediately by symmetry.
Now you have a region in the plane over which to integrate the joint distribution $f(x,y) = f(x)f(y).$
When you look at a plot of this region it should be relatively obvious
how to subdivide it into one region of total probability $\frac14$
and another region whose probability you can show is positive.
(Note: I say "obvious" only because I have looked at the plot;
it wasn't obvious to me at all from the problem statement itself.)
