I have seen this problem solved for the discrete case, but unsure about continuous. Let $X$ and $Y$ be two continuous random variables scaled to $[0,a]$ where $a \in \textbf{R}$. The pdf of $X$ is uniform on the interval while $Y$ has an unknown distribution.
I would like to show that $Z = (X + Y)$ mod $a$ is a random variable with a uniform distribution on $[0,a]$. I've attached my work below, but I'm not a 100% sure on whether it makes sense.
Let the pdfs of $X$ and $Y$ be $f_{_X}(\cdot)$ and $g_{_Y}(\cdot)$ respectively and the pdf of $Z$, $h_{_Z}(\cdot)$, is defined via convolution as follows:
$h_{_Z}(u)$ = $\int_0^u f_{_X}(y)g_{_Y}(u-y)dy$ + $\int_u^a f_{_X}(y)g_{_Y}(a+u-y)dy \quad , 0 \leq u \leq a$
Since $X$ is uniform, its pdf is $\frac{1}{a}$, which allows me to make the following simplification:
$ah_{_Z}(u)$ = $\int_0^u g_{_Y}(u-y)dy + \int_u^a g_{_Y}(a+u-y)dy, \quad 0 \leq u \leq a$
If the cdf of $Y$ is given by $G_{_Y}(\cdot)$, then the above reduces to:
$ah_{_Z}(u)$ = $G_{_Y}(u) - G_{_Y}(0) + G_{_Y}(a) - G_{_Y}(u)$ = $G_{_Y}(a) - G_{_Y}(0)$
Since $G_{_Y}(\cdot)$ is a cdf on $[0,a]$, we have:
$ah_{_Z}(u) = 1$
$h_{_Z}(u) = \frac{1}{a}$
So the above result tells me that no matter what the distribution of $Y$ is, $Z$ will be uniformly distributed as long as one of them is uniformly distributed. I have seen this proved in the discrete case, but was curious if the above proof was correct for the continuous case. Random experiments on the computer seemed to match my results, but that's not "proof".
Also, if someone could give intuition (assuming the above is right), that'd be great. Why does the modulo operator care only about the distribution of one variable?