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Say $X$ and $Y$ are two random variables where $X\in [-\alpha,\alpha]$, $Y\in [-\alpha,\alpha]$ and $Z=X+Y$. Is it possible to find two independent random variables with certain pdf (not necessarily identically distributed) that force $Z$ to be uniformly distributed (i.e. $Z\sim \mathcal{U}[-2\alpha,2\alpha]$)?

As the sum of $N$ random variables with zero mean resembles Gaussian distribution with zero mean, I suspect it is not possible to find two such random variables. Do you know any counterexample?

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    $\begingroup$ Very rough sketch of argument: We can conclude that $P(|Z|>2\alpha-\epsilon) >0$ for every $2\alpha>\epsilon>0$, from which it follows that both $X$ and $Y$ have positive probability densities near the endpoints, but then $Z$ will more likely be $0$ than $\pm 2\alpha$... $\endgroup$
    – uvs
    May 8, 2014 at 13:18
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    $\begingroup$ No, the sum $Z$ of two independent random variables $X$ and $Y$ cannot be uniformly distributed on an interval. The density function of a uniformly distributed random variable is discontinuous at the end points of the interval. However, the density of $Z$ is the convolution of the densities of $X$ and $Y$, $$f_Z(z) = \int_{-\infty}^\infty f_X(x)f_Y(z-x)\,\mathrm dx$$ and is a continuous function of $z$ (though not necessarily a differentiable function of $z$) for all $z$. $\endgroup$
    – Dilip Sarwate
    May 8, 2014 at 14:14
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    $\begingroup$ The arguments in the previous comments apply when X and Y both have densities. Are you requiring this? $\endgroup$
    – Did
    May 8, 2014 at 15:32
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    $\begingroup$ @Did I was assuming that $X$ and $Y$ are continuous random variables with density functions and not taking the OP's statement "$X \in \{-\alpha, \alpha\}$" literally but rather as typographical errors; assuming that the OP meant to write "$X\in [-\alpha,\alpha]$" $\endgroup$
    – Dilip Sarwate
    May 8, 2014 at 16:54
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    $\begingroup$ @DilipSarwate I got that, and I agree that $X,Y\in\{-\alpha,\alpha\}$ makes little sense and should read $X,Y\in[-\alpha,\alpha]$. But my question to the OP is different--hence let us wait a reaction. $\endgroup$
    – Did
    May 8, 2014 at 21:25

2 Answers 2

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Here is another one.

Recall if $U_n$ are IID with Bernoulli distribution $$ \mathbb P[U_n=0]=\frac{1}{2},\qquad \mathbb P[U_n=1]=\frac{1}{2}, $$ then $$Z = \sum_{n=1}^\infty U_n 2^{-n} $$ is uniformly distributed on $[0,1]$. So let $$ X = \sum_{n\text{ even}} U_n 2^{-n},\qquad Y = \sum_{n\text{ odd}} U_n 2^{-n} $$ to get independent $X,Y$ with $X+Y=Z$.

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Let $ X \sim {\mathcal U}[-\alpha, \alpha] $ and let $ Y $ be a discrete r.v. taking values $ \pm\alpha $ with equal probabilities. Then $ X+Y \sim {\mathcal U}[-2\alpha, 2\alpha] $.

I believe that the sum of $ n \geq 3 $ independent r.v.'s distributed on $ [-\alpha, \alpha] $ cannot be uniform on $ [-n\alpha, n\alpha] $, but I have no proof of this.

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