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If I have two processees $x_{1}$ and $x_{2}$, what distribution(s) would they need to have in order for $x_{1} + x_{2}$ to be uniformly distributed?

Arnfinn
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  • Are the variables independent? Do you want the same distribution for both (in which case the answer is "this is not possible") or can they differ? (in the second case there are trivial answers, such as adding a discrete uniform variate to continuous uniform variate whose range matches the gap in the discrete one) – Glen_b Feb 21 '17 at 22:20

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The answer is yes.

Dependent $X_1$ and $X_2$ example:

If $X_1$ and $X_2$ need not be independent, then it's trivial.

Let $X_1$ be a random variable following any distribution. Let $Z$ be a random variable following the uniform distribution (eg. from 0 to 1). Compute the distribution of $X_2 = Z - X_1$. Then $X_1 + X_2$ will be uniform.

Independent $X_1$ and $X_2$ example:

As the answer by @grand_chat points out, this answer on math.stackexchange provides a clever example.

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Matthew Gunn
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    It would be kind of you to make it clear that this construction only works because $X_1$ and $X_2$ are not independent. – whuber Feb 21 '17 at 19:48
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If you are looking for two random variables whose sum is uniform over an interval, here are some pointers in the independent case.

If $X$ and $Y$ are independent random variables, and ...

  • ... $X$ and $Y$ both have densities, then the answer is no. This can be seen from the density of $X+Y$, which will be the convolution of the densities of $X$ and $Y$. The convolution $f_{X+Y}(x)$ will peak when the densities of $X$ and $Y$ are 'aligned', and will decrease from that peak.

  • ... $X$ and $Y$ have the same distribution, then the answer is no. One proof involves characteristic functions and showing that the CF of $X$ is impossible.

If $X$ and $Y$ are independent but not identically distributed, the answer is yes, it is possible. By the above, at least one of $X$ and $Y$ cannot have a density. Examples:

  • If $Y$ is constant, there's an example in Zahava Kor's answer: Let $X$ have uniform distribution.

  • A more bizarre example is seen here.

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grand_chat
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  • Could you explain the sense in which you mean the PDF of a convolution of continuous random variables must be continuous? There seem to be plenty of counterexamples: for instance, a sum of independent $\chi^2(1)$ distributions will be a $\chi^2(2)$ distribution whose PDF is not continuous at $0$. Incidentally, your second bullet is demonstrated here at http://stats.stackexchange.com/questions/125360. – whuber Feb 21 '17 at 20:01
  • @whuber Right, bad explanation. I've made an edit to fix that. – grand_chat Feb 21 '17 at 20:18
  • Thank you. Are you aware of a rigorous demonstration of your modified claim? After all, the densities of $X$ and $Y$ could have infinitely many peaks, so the "alignment" argument--although suggestive--is unconvincing. – whuber Feb 21 '17 at 20:22
  • Your "bizarre example" turned out to work really well for my application. Thanks for bringing it to my attention! How did you find it? – Arnfinn Feb 22 '17 at 00:18
  • @Arnfinn I used the search box on math.stackexchange. The query "sum uniformly distributed possible" gets you right to the bizarre example. – grand_chat Feb 22 '17 at 15:42
  • @grand_chat Great! I have spent days searching for something like that... – Arnfinn Feb 22 '17 at 22:57
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First - X1 and X2 are not processes but random variables. Second - you did not specify whether they have to be independent ot not. If Independent - X+c will work where X is Uniform and C is a constant. If not necessarily independent - then X1=X2=X/2 where X is Uniform will work, because X/2 + X/2 = X and X is Uniform.

Zahava Kor
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