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I've been stuck on this for a couple days and I feel like there's some obvious reason that what I want can't be done:

I want to create 3 random variables $p_1$, $p_2$, and $p_3$ so that every pairwise difference between them is uniformly distributed between $(-1, 1)$.

With 2 variables $p_1$ and $p_2$ this can easily be done by setting $p_1 = 0$ and having $p_2$ be drawn from the uniform distribution $U(-1, 1)$.

With 3 variables, if I ignored the pairwise differences $p_1 - p_3$ and $p_3 - p_1$ then I could construct $p_1$ and $p_2$ as before and set $p_3 = p_2 + x$ where $x$ is drawn from $U(-1, 1)$, but obviously in this construction we have that $p_3 - p_1$ will be the sum of two independently drawn uniformly distributed variables, which is not itself a uniform distribution.

I'll also throw out that it can't be done in the case where the $p_i$s are both independent and identical thanks to this question.

I'd welcome any thoughts on this.

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  • $\begingroup$ I'll add that another take-away from the previous question I posted is that the random variables have to have some dependence on each other: if we had some setting $S$ achieving the above constraint, then we could first perform a random permutation over the $p_i$s and then take the setting $S$. By doing the random permutation first, we have that each $p_i$ is distributed identically. Since we know that no valid algorithm for this problem can set each $p_i$ identically and independently, the $p_i$s must be dependent on each other. $\endgroup$ Commented Nov 26, 2021 at 16:23

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I think I figured it out.

Set $p_1 = 0$, let $p_2$ be sampled from $U(-1, 1)$. We then let $$f(x) = \begin{cases} -(x+1) & \text{if $x<0$}\\ 1-x & \text{otherwise} \end{cases}$$ and set $p_3 = f(p_2)$. Then we get: $$p_2 - p_3 = \begin{cases} 2p_2+1 &\text{if $p_2<0$}\\ 2p_2-1 &\text{otherwise,} \end{cases}$$ and $$p_3 - p_1 = p_3 = f(p_2).$$ One can verify using standard methods that both $p_2 - p_3$ and $p_3 - p_1$ follow the uniform distribution $U(-1, 1)$, which immediately also implies that $p_3 - p_2$ and $p_1 - p_3$ do too.

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