# Distribution of the fractional part of a sum of two independent random variables

Let $X$ and $Y$ be independent random variables. $X$ is a uniform random variable on $[0,1]$. Let $Z=X+Y-\lfloor X+Y\rfloor$. What is the distribution of $Z$?

Let $X_1, X_2, \dots, X_n$ be IID with the same distribution of as $X$, all independent of $Y$. Let $Z_i=X_i+Y-\lfloor X_i+Y\rfloor$ I need to prove that $Z_1, \dots, Z_n$ are IID.

Can anyone point out how to go about solving this problem?

• Let $f(x,y) = x+y - \lfloor x+y\rfloor$, then the distribution of $Z$ is just $\mu_Z = f_*(\mu_X\otimes \mu_Y)$. Without specifying the distribution of $Y$ there is no more specific answer for $Z$ either.
– Ilya
Jun 3 '13 at 13:47
• Without specifying the distribution of Y there is no more specific answer for Z either. There is.
– Did
Jun 3 '13 at 19:16

For your first problem, note that $Z$ is just the fractional part of $X+Y$. Using this fact, it should be easy for you to argue that, for any fixed value of $Y$, the distribution of $Z$ is uniform on $[0, 1]$, and therefore overall $Z$ is uniform on $[0, 1]$.
For your second problem, using similar observations, you should be able to argue that all the $Z_i$ are independent uniform $[0, 1]$ random variables (one way to do this is to directly show that p(X_1 = x_1, X_2 = x_2, \dots, X_n = x_n) = p(X_1 = x_1)p(X_2 = x_2)\dots p(X_n = x_n)).