# Prove the relation between two distribution functions

I have been given a homework in a subject called "Non-Parametric Statistics" and I'm a bit stuck with it. I would be very thankful if you could give me any advice or help, which would lead to a solution!

Show that if $$F_x$$ is symmetrical and the relation between distribution functions $$F_x(x)$$ and $$G_y(y)$$ is $$F_x(x) = G_y(x - \Delta)$$, where $$\Delta$$ > 0, then
$$P(X > Y, X' > Y) = P(X > Y, X > Y')$$, whereas $$X'$$ is from the same distribution as $$X$$ and $$Y'$$ is from the same distribution as $$Y$$. A hint that has been given: Without loss of generality we may assume, that $$F_x$$ is symmetrical with respect to 0. In this case $$X$$ and $$-X$$ are from the same distribution.

We have this homework given in the Wilcoxon Rank Sum Test part, but in general case the solution idea stays the same.

My first thought was to use the independence of variables, so I could give $$P(X > Y, X' > Y)$$ like this $$P(X > Y)$$*$$P(X' > Y)$$ as a product, but I wasn't sure about it, whether they are independent or not ..

• If you write it down as: $$P(X > Y, X' > Y) = P(Y < X, Y' < X)$$ then the symmetry might become more clear (the probability that two draws from X are larger than one draw from Y versus two draws from Y smaller than one draw from X). Then flip the left side around the axis and shift by $\Delta$ $$Y < X \rightarrow -Y > -X \rightarrow \Delta-Y > \Delta-X$$ – Sextus Empiricus Nov 8 '18 at 17:02
• Basically you flip the equation around the axis at $\Delta/2$, you can visualize this by plotting an example Y and X. – Sextus Empiricus Nov 8 '18 at 17:31
• Okay, I generally understand your thoughts. But the $P(X > Y, X' > Y) = P(Y < X, Y' < X)$ part is the one that I have to prove. In short, I don't really understand at the moment, how to use your help efficiently to solve this task. Like .. I can't draw a line atm what I can use as an assumption and what I have to prove (show) – Martin Smith Nov 8 '18 at 20:05
• Take, just by example, a normal distribution X~N(-0.5,1) and a shifted one Y~N(0.5,1). You can do this without loss of generality (you shift the variables such that the mirror points of X and Y are at $\pm \Delta/2$). Plot them! Do you notice that the symmetry, how X becomes Y and Y becomes X when you mirror around 0? (which is this special case, and in general you mirror around $a_X$, the mirror point for $X$, plus $\Delta/2$). Now you only have to write this down algebraically, ie use rules like -X is similarly distributed as Y (for special case) or $-(X-a_X-\Delta/2)$ as $Y-a_X-\Delta/2$. – Sextus Empiricus Nov 8 '18 at 22:10
• So if you got the image correctly shifted such that (shifted) -X is the same distribution as (shifted) Y. Then P(X>Y,X' >Y) becomes P(-X<-Y,-X'<-Y) becomes P(Y<X,Y'<X) and P(X>Y,X>Y'). So it may be easier to let X be symmetrical with respect to $-\Delta/2$ (without loss of generality), such that -X and Y follow the same distribution (as well as -Y and X). – Sextus Empiricus Nov 8 '18 at 22:16