Does the Distribution of X + Y mod a Remain Uniform?

[areslagae]

If X is uniformly distributed over [0,a), and Y is independent, then X + Y (mod a) is uniformly distributed over [0,a), independent of the distribution of Y.

Can anyone point me to a statistics text that shows this?

Thanks,

 

[CRGreathouse]

If X is uniformly distributed over [0,a), and Y is independent, then X + Y (mod a) is uniformly distributed over [0,a), independent of the distribution of Y.

Can anyone point me to a statistics text that shows this?

Thanks,

Consider any possible value from Y. Since X and Y are independent, X + that value is uniformly distributed mod a. Now since this is true for any value, it is true for any combination of values.

 

[mathman]

I don't know of a text, but the proof is simple enough. Let Y=y, then X+y (mod a) is uniformly distributed, since X is independent of Y. Therefore the theorem holds irrespective of the distribution of Y.

 

[areslagae]

Thanks for both your replies!

At first, me (and my collegues) found this result somewhat counter-intuitive. It seems that you do not, but you most likely you have a deeper intuition.

Meanwhile, I also found the following paper which is interesting in this context:
The Distribution Functions of Random Variables in Arithmetic Domains Modulo a
P. Scheinok
http://www.jstor.org/stable/2310973

It seems that the theorem from my first post follows from 3.3, with g_Y()=1/a.

Thanks,

 

[areslagae]

Hi all,

I have a quick additional question.

A colleague pointed out to me that the cited paper only proves the theorem from my first post in the case that Y is defined over [0,a].

However, the random variables (X + Y) mod a and (X + (Y mod a)) mod a have the same distribution, so the original theorem should hold, no?

Thanks,