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Problem and solution.

(the way I understood the solution) The condition is manipulated so as to represent it in a graphical way that allows one to find the area of the CDF.

My problem is, I find the method used here quite ad hoc; what if you're in 4 dimensions instead of 2? Basically what I'm asking is how could one find the formula for the area analytically, instead of graphically? I have spent some time thinking about it myself but with no success.

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$X$ and $Y$ are random locations. The distance in this one-dimensional problem is simply the random variable $|X-Y|$. In a more general problem, $X$ and $Y$ would be random vectors and the distance simply the distance (perhaps euclidean) between the vectors $d(X,Y)$. This distance is itself a one-dimensional random variable. The distribution of this random variable can be found using the distributions of $X$ and $Y$. See these notes on finding distributions of functions of two random variables.

Here is a paper about the distribution of the distance between two random points in a box

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  • $\begingroup$ So the feeling I'm getting is that there is no general way to get the CDF/PDF then? When I mentioned 4 dimensions I also was trying to imply a more complicated formula. Basically, what if we're working with 4, 10 or 100 random variables? In the notes you linked, he's still abusing the fact that it's 2d which allows him to display it graphically and get the bounds of the integral from it. For example the problem I posted. The tricky part for finding the PDF (analytically) would be getting the bounds for the integral right. I don't see how one could find those without a graphical representation. $\endgroup$ – Nimitz14 Dec 31 '15 at 15:18

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