# Can the difference of random variables be uniform distributed? [duplicate]

Given two random variables X and Y with some distribution D, is it possible to choose a D such that Z = X - Y is uniform? Is there a standard distribution D that would satisfy this?

• @ReinstateMonica nice duplicate find! Though it's only a duplicate if the OP is looking for IID random variables (which I found unclear from their question, so I answered it both ways). – josliber Feb 23 at 3:32

Consider the following example:

$$X\sim\text{Unif}(0, 1) \\ Y = 1-X$$

$$X$$ and $$Y$$ are identically distributed as the standard uniform distribution, and $$X-Y = 2X-1$$, so $$X-Y\sim\text{Unif}(-1, 1)$$.

Note that this example relied on $$X$$ and $$Y$$ being dependent, identically distributed random variables. It is impossible for two independent, identically distributed random variables $$X$$ and $$Y$$ to have a difference that follows a uniform distribution. Clearly for $$X-Y$$ to be uniform we would need $$X$$ and $$Y$$ to be bounded, continuous random variables (assume bounds $$\underline{x}$$ and $$\overline{x}$$). However, since $$X$$ and $$Y$$ are continuous, their difference will have density 0 at its bounds $$\overline{x}-\underline{x}$$ and $$\underline{x}-\overline{x}$$, leading us to conclude that $$X-Y$$ does not follow a uniform distribution.

• What if $X \sim \operatorname{Unif}(0,1)$ and $Y$ is the constant 0? They are independent, and their difference is uniformly distributed. – Federico Poloni Feb 23 at 10:47
• @FedericoPoloni Excellent catch! I've removed the claims about independent, not identically distributed RVs. – josliber Feb 23 at 13:34

Independent and identically distributed. If you ask for IID $$X$$ and $$Y$$, as others have noted this is not possible. See the answers to this question.

If you are happy to drop either independence between $$X$$ and $$Y$$ or them having the same distribution, then there's hope.

Same distribution. If you allow for dependent but identically distributed variables, then you can build the joint distribution of $$X$$ and $$Y$$ so that the marginal distributions $$p_Y(y)=\int p_{X,Y}(x,y)\,dx$$ and $$p_X(x)=\int p_{X,Y}(x,y)\,dy$$ are the same and the integral over diagonal lines $$p_Z(z)=\int p_{X,Y}(y+z,y)\,dy$$ is constant in an interval. One such example is: $$p_{X,Y}(x,y)=\begin{cases}\frac12 & \text{if |x|+|y|<1}\\0 & \text{otherwise}\end{cases}.$$ The marginal distribution of $$X$$ is $$p_X(x) = 1 - |x|$$ for $$x\in[-1,1]$$, which is the same as that of $$Y$$, while $$Z \sim \mathcal{U}(-1,1)$$. A simple way to build such distributions for $$X$$ and $$Y$$ is to choose $$Z \sim \mathcal{U}(-a,a)$$ and $$W$$ distributed according to a distribution $$D'$$, then $$X=W+Z/2$$ and $$Y=W-Z/2$$ have the chosen properties (their difference is uniform and they are distributed according to a common distribution $$D$$ whose pdf is the convolution of the pdf of $$D'$$ and a rect). For example for Gaussian $$W$$, we could obtain something like the following joint and marginal densities:

Independent. If $$X$$ and $$Y$$ are independent, then $$Z$$'s pdf must be the convolution of that of $$X$$ and $$-Y$$. As noted by josliber, $$X$$ and $$Y$$ cannot be both continuous variables otherwise the convolution of their pdfs would be continuous and would need to approach $$0$$ at the boundaries of the support of $$Z$$. This limit can be overcome if the pdf of either variable is not a function but a distribution (in the mathematical sense). For example the convolution of a rect (pdf of a uniform distribution) and an appropriately chosen train of Dirac's deltas (pdf of a discrete variable) can be a rect. One such example is when $$Y \sim \mathcal{U}(-a,a)$$ and $$X = b + 2\,a\,N$$ with $$N \sim \mathcal{U}\{c,d\}$$ (discrete uniform distribution).

• +1 for the Independent section, which adds something new – Henry Feb 23 at 13:41