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I'm looking for proof using CDFs or MGFs to show for $X$ and $Y$ independent random variables,

$$ X \sim \mathrm{Unif}(0,1) \text{ and } Y \sim \mathrm{Unif}(0,1) \implies (X+Y \bmod 1) \sim \mathrm{Unif}(0,1). $$

I've tried using MGFs for $X+Y$ but can't seem to figure out how to incorporate the $\bmod 1$ aspect. I've read this thread but I'm really looking for a proof using MGFs (or CDFs). Any hints are appreciated!

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    $\begingroup$ you cant do this via mgf $\endgroup$ – Lost1 Jan 23 '14 at 0:49
  • $\begingroup$ Why only using CDFs? To make things more complicated? $\endgroup$ – Did Jan 23 '14 at 7:03
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Not sure if it's the sort of thing you are looking for, but here is a hint.

The joint PDF of $X$ and $Y$ is $f(x,y)=1$ in the unit square $0\le x<1$, $0\le y<1$. Now for $0\le t<1$ we have $$X+Y\ {\rm mod}\ 1\le t\quad\hbox{if and only if}\quad X+Y\le t\ \hbox{or}\ 1\le X+Y\le1+t\ .$$ So, to find the CDF of $X+Y$ modulo $1$ you have to integrate $f(x,y)$ over two parts of the unit square: the triangle below the line $x+y=t$ and the strip between $x+y=1$ and $x+y=1+t$. If you draw a diagram and look carefully you should see that you don't really even need to integrate, you can evaluate this by a simple geometric argument. Good luck!

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You want $P(Z\leq z) = P(\{X+Y \leq z\} \cup \{1<X+Y\leq 1+z\})$, this is equal to

$$\int^z_{y=z-x}\int^z_{x=0}1\text{ d}x\text{d}y + \int_{y=1-z}^{1+z-x}\int_{x=z}^11\text{ d}x\text{d}y$$

You need to do the integral with respect to $y$ first, then $x$.

also note: MGF does not work. MGF only works with composition of distributions/sum of distributions. Taking mod is not particularly nice.

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