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Let $G=\mathbf{Z}/p \mathbf{Z}$ where $p$ is prime, $X\in G$ be a uniform random variable and $Y\in G^{*}$ be any random variable.

Is it possible to have $Z=X+Y \in G$ with a uniform distribution?

If so, is there any condition on the variable $Y$ that guarantees that $Z$ is uniform?

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closed as unclear what you're asking by Did, JKnecht, drhab, Semiclassical, hardmath Apr 22 '16 at 22:30

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    $\begingroup$ Uniform on what? $\endgroup$ – Did Apr 22 '16 at 14:27
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    $\begingroup$ It's certainly possible. Say $X$ is uniform on $[0,1]$, $X$ and $Y$ are independent, and $P(Y=0)=P(Y=1)=1/2$. $\endgroup$ – David C. Ullrich Apr 22 '16 at 14:30
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    $\begingroup$ @NumBee Let us know when you decide what the question actually is - there's nothing about $\Bbb Z/ p\Bbb Z$ in the edited version... $\endgroup$ – David C. Ullrich Apr 22 '16 at 14:37
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    $\begingroup$ @NumBee If, as suggested in your comment, you actually mean to be talking about random variables with values in $G=\Bbb Z/p\Bbb Z$, and so the sum is also supposed to be the sum in $G$, then (assuming $X$ and $Y$ are independent) if $X$ is uniform on $G$ then $X+Y$ is also uniform on $G$ for any $Y$. $\endgroup$ – David C. Ullrich Apr 22 '16 at 14:40
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    $\begingroup$ If there's no assumption of independence then you can't expect to say anything about $X+Y$. This is sort of so clear that the other commenters have been assuming you're talking about independent variables, without, er, comment. $\endgroup$ – David C. Ullrich Apr 22 '16 at 14:53
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Guessing that $G^*$ denotes the set of non-zero elements of $G$, and hoping that the question has finally stabilized: If $X$ and $Y$ are independent then $X+Y$ is uniform on $G$. Without assuming independence I don't see how you can expect to say anything about $X+Y$.

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  • $\begingroup$ Yes, $G*:=G\setminus\{0\}$. So if independence is needed, is this a standard result for discrete variables? $\endgroup$ – NumBee Apr 22 '16 at 15:08
  • $\begingroup$ I don't know about "standard", but it's clear for any finite group. Say there are $p$ elements. So $P(X=a)=1/p$ for every $a$. Given $b in G$, $X+b=a$ if and only if $X = a-b$, so $P(X+b=a)=P(X=a-b)=1/p$. That's the result for constant $Y$; the result for a random variable $Y$ follows. $\endgroup$ – David C. Ullrich Apr 22 '16 at 18:51
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Let $U$ be uniform continuous on $[0,1]$ and $V$ be uniform discrete on $\{0,1,\cdots,m\}$. Then $U+V$ is uniform continuous on $[0,m+1]$.

More generally, assume $U$ to be uniform continuous on $[0,1]$ and $V$ to be another random variable such that $U+V$ is uniform continuous on some other interval. By translation, we can assume that this interval is $[0,L]$ for some $L>0$.

Taking moment-generating functions, we see that: $$\frac{e^t-1}{t}\cdot M_V(t)=\frac{e^{Lt}-1}{Lt}\implies M_V(t)=\frac{e^{Lt}-1}{L(e^t-1)}$$

From here, I can't entirely finish - I know that integer $L$ gives discrete uniform $V$, but I'm not sure for which other values of $L$ it gives a proper MGF.

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  • $\begingroup$ Good answer to what is alas no longer the question. People shouldn't do that... $\endgroup$ – David C. Ullrich Apr 22 '16 at 14:57
  • $\begingroup$ Sorry for the delay in editing my question. But I appreciate the answer! $\endgroup$ – NumBee Apr 22 '16 at 15:10

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