Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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

Some hints:

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)).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.