# The sum of two i.i.d. random variables cannot be uniformly distributed

Given $$X$$ and $$Y$$ iid (independent and identically distributed), show that $$Z=X+Y$$ cannot be uniformly distributed. The uniform distribution might be on any set, not necessarily an interval.

Notice that the density might not exist, so convolution might not work. For the trivial cases of where the pdf (probability density function) exists, we just use the continuity of convolution and it's done. This also shows that in this case any uniform distribution cannot work, whether it's an interval or not. So, I'm guessing for the more general case.

The original question is like this. *I'm not asking about this, but (1) might help as a hint, and the original condition might tell you whether my abstraction is correct.

$$X,Y$$ are iid on $$[0,\frac12]$$, cdf (cumulative distribution function) is $$F(x)$$, $$Z=X+Y$$,

(1) Show that $$\forall \epsilon\in(0,\frac14)$$, $$P(Z\leq\epsilon)\leq F(\epsilon)^2$$ $$P(\frac12-\epsilon\leq Z\leq \frac12+\epsilon)\geq2F(\epsilon)(1-F(\frac12-\epsilon))$$

(2) Prove that $$Z$$ is not uniformly distributed.

• Please add your efforts and thoughts into the question. There is no indication of what you have tried, or what parts of the problem you are having difficulty with. Commented Apr 8 at 14:54
• The original question is not important. I only ask for the general version of (2). I have tried convolution, which certainly doesn't work in this problem where the density might not exist. Commented Apr 8 at 14:57
• Thanks, the edited question is much more clear now, upvoting. Commented Apr 8 at 16:46
• Trivial exception: if $X$ and $Y$ are both constant, then $X+Y$ is uniform on a singleton set. Commented Apr 8 at 16:49
• The concept of uniform distribution is defined only when you fix a reference measure $\lambda$. We usually talk about the cases when $\lambda$ is either the Lebesgue measure or the counting measure, but this doesn't cover the cases like Cantor distribution. Commented Apr 9 at 8:56

For simplicity consider a random variable $$Z$$ uniform on $$(-1,1).$$ Its characteristic function is $$E(e^{itZ})=\frac {\sin t}{t}=\prod_{n=1}\left(1-\frac{t^2}{n^2\pi^2}\right).$$ If $$Z\sim X+Y$$ with $$X$$ and $$Y$$ iid, with characteristic fonction $$f(t)$$ then $$|X|<1/2$$. This implies that $$f$$ is the restriction to $$R$$ of an entire function. However the zeros of $$f^2=\frac {\sin t}{t}$$ are both of multiplicity even and 1: a contradiction.

Edit: Everybody seems to ignore the magnificent paper by T. Lewis 'The factorisation of the rectangular distribution' Journal of Applied Probability Vol 4 Issue 3 November 1967 pages 529-542. A related work is due to Roland Berthuet 'Sur la loi exponentielle tronquee' Z. Wahrsheinlichkeitstheorie verv Geb. vol 21 page 300-328 (1972).

• Is there some way to prove that any uniform distribution (like with multiple intervals, or even with any measurable set) is not possible, like the case with density, where the pdf has to be continuous and eliminating all of them? Commented Apr 10 at 16:37

When $$X,Y$$ are iid, the distribution of $$Z=X+Y$$ can be one of discrete or continuous uniform distributions, studied in the following.

For the continuous case, two cases are considered: uniformity on a single interval and on a number of disjoint intervals.

## Discrete Uniform Distribution

Let the support $$S_X$$ of $$X$$ be a finite set including $$a_1<\cdots with corresponding probabilities $$p_1,\cdots,p_n>0$$. For $$Z$$ to have a discrete uniform distribution, the following system:

$$\mathbb P(Z=2a_1)=p^2_1=\mathbb P(Z=2a_n)=p^2_n=\mathbb P(Z=a_1+a_n)=2p_1p_n+\beta, \, \beta\ge 0,$$

needs to have a non-zero solution $$p_1$$ and $$p_n$$, which is impossible (as $$x^2=2x^2+\beta$$ cannot have such a solution) unless $$n=1$$, i.e. $$X$$ is a constant.

## Continuous Uniform Distribution on a Single Interval

Here, I provide an elementary proof (not using complex analysis) based on the inequalities given in the OP for iid $$X,Y \in [0,\frac{1}{2}]$$ and $$\epsilon \in \left (0,\frac{1}{4} \right)$$:

$$\color{blue}{P(Z\leq\epsilon)\leq F(\epsilon)^2} \tag{1}$$

$$\color{blue}{\mathbb P \left (\frac12-\epsilon\leq Z\leq \frac12+\epsilon\right )\geq2F(\epsilon)\left (1-F\left (\left [\frac12-\epsilon \right ]^- \right) \right)} \, \tag{2}.$$

To indirectly prove the statement, let us assume $$Z\sim \mathcal U(0,1)$$. Then, it can be shown that the mfgs of both $$\frac12-X$$ and $$X$$ are the same, so we have $$\frac12-X\sim X,\frac12-Y\sim Y \tag{3},$$ (an alternative way to prove (3) is to show that $$X-\frac{1}{4},Y-\frac{1}{4}$$ have symmetric distributions by the fact that the characteristic function of any symmetric distribution is real-valued.)

Considering (3), we have

$$F(\epsilon)=\mathbb P(X \le \epsilon)=\mathbb P \left (\frac12-X \ge \frac12 -\epsilon \right )=\mathbb P \left (X \ge \frac12 -\epsilon \right)=1-F\left (\left [\frac12-\epsilon \right ]^- \right).$$

Hence, by (3) from (1) and (2) for $$Z\sim \mathcal U(0,1)$$, for $$\epsilon \in \left (0,\frac{1}{4} \right)$$ we get $$\epsilon\leq F(\epsilon)^2$$ and $$2\epsilon\geq 2 F(\epsilon)^2$$, so

$$F(\epsilon)=\begin{cases} \sqrt{\epsilon}, & 0\le \epsilon \le \frac14\\ 1-\sqrt{\frac12 - \epsilon} & \frac14 <\epsilon \le \frac12 \end{cases},$$

for which $$Z \nsim \mathcal U(0,1)$$, a contradiction that proves the statement.

Proofs for (1) and (2)

As $$X,Y\ge 0$$, we have the following implication for $$\epsilon \in \left (0,\frac{1}{2} \right)$$:

$$Z\leq\epsilon \Rightarrow X\leq\epsilon, Y\leq\epsilon,$$

which shows the left event is a subset of the right event, so we have (1).

Moreover, considering $$0\le X,Y \le \frac{1}{2}$$, we have the following implication for $$\epsilon \in \left (0,\frac{1}{4} \right)$$:

$$X\leq\epsilon, Y\ge \frac12-\epsilon \, \text{or}\, Y\leq\epsilon, X\ge \frac12-\epsilon\Rightarrow \frac12-\epsilon\leq Z\leq \frac12+\epsilon,$$

by noting that $$Z=Y+X\leq\epsilon+Y\le \frac12+\epsilon, Z=X+Y\ge \frac12-\epsilon +Y\ge \frac12-\epsilon$$. Since $$\epsilon\in(0,\frac14)$$, the two events on the LHS of the above implication are disjoint, and we obtain (2).

## Continuous Uniform Distribution on Disjoint Intervals

By contradiction, let $$Z$$ have a uniform distribution over the union of two disjoint intervals $$[0,a] \cup [1-b,1]$$ with $$0. Then, the cdf $$F$$ satisfies the following equation for $$0:

$$\frac{z}{a} \frac{a}{a+b} = \int_{0}^x F(z-t)\text{d}F(t).$$

Defining $$\bar{F}=\sqrt{\frac{a+b}{a}}F$$ for $$0, this can be written as

$$\frac{z}{a}= \int_{0}^x \bar{F}(z-t)\text{d}\bar{F}(t),$$

which means that the sum of iid $$X,Y$$ with the cdf $$\bar{F}$$ has a uniform distribution on $$(0,a)$$, already shown not possible in the previous section, and we get a contradiction.

## Unsolved Special Case: An Infinite Sequence of Disjoint Intervals with Unbounded Below Union

The only case that I could not yet find an answer for it is the following case (hope readers can help to solve this case too). Suppose that the support of $$Z$$ is not bounded from below and includes disjoint intervals $$(l_i,l_i+\delta_i), i\in \mathbb Z$$ with $$\Delta=\sum_{i\in \mathbb Z}\delta_i<\infty$$. Note that if $$\inf_{i\in \mathbb Z} \{l_i \}>-\infty$$, we can handle it using the method used in the previous section.

• Thanks for your answer. This makes most of the problem clear. But I still have some problems: 1. you said that it's the "alternative way to prove (3)", but what is the first way? Can it be trivially deduced from $Z \sim U(0,1)$ that $X\sim 1/2-X$? Or am I missing something? 2. Is there some way to prove that any uniform distribution is not possible, like the case with density? Commented Apr 10 at 16:32
• I mean by any uniform distribution, is like with multiple intervals, or even with any measurable set. In the case where pdf exists, the continuity eliminates all of them. Commented Apr 10 at 16:43
• @ZhenRanZR You are welcome! 1. Note that $X,Y$ are not necessarily uniformly distributed and can have any distribution over $[0,0.5]$, and we only assume that $Z$ is uniform over $[0,1]$ (to prove by contradiction). 2. The first way is "to compare the mgfs of $X$ and $1/2-X$, which both exist as $Z\sim \mathcal U(0,1)$ has a finite mgf".
– Amir
Commented Apr 10 at 19:57
• wait... if $X,Y \in [a,b]\cup [c,d]$, shouldn't $Z\in [2a,2b]\cup[a+c,b+d]\cup[2c,2d]$? Commented Apr 11 at 3:16
• @ZhenRanZR Using a different approach, I addressed the case of uniformity on disjoint intervals. Please check the third part.
– Amir
Commented Apr 11 at 11:43

### Part (1) Analysis

Given $$X$$ and $$Y$$ are i.d.d. random variables uniformly distributed on $$[0,\frac{1}{2}]$$, we denote their common CDF by $$F(x)$$. For a random variable $$Z = X + Y$$, we analyze its distribution characteristics through the probabilities $$P(Z \leq \epsilon)$$ and $$P(\frac{1}{2} - \epsilon \leq Z \leq \frac{1}{2} + \epsilon)$$ for $$\epsilon \in (0, \frac{1}{4})$$.

#### For $$P(Z \leq \epsilon)$$:

Considering $$X$$ and $$Y$$ contribute to $$Z$$'s value being less than or equal to $$\epsilon$$, both must be within $$[0, \epsilon]$$. The joint probability, due to their independence, is the product of their individual probabilities, yielding:

$$P(Z \leq \epsilon) = P(X \leq \epsilon)P(Y \leq \epsilon) = F(\epsilon)^2$$

Thus, directly relating the probability $$P(Z \leq \epsilon)$$ with $$F(\epsilon)^2$$.

#### For $$P(\frac{1}{2} - \epsilon \leq Z \leq \frac{1}{2} + \epsilon)$$:

This probability involves the sum $$Z = X + Y$$ lying within an interval around $$\frac{1}{2}$$. Here, the contributions of $$X$$ and $$Y$$ are asymmetric around $$\frac{1}{2}$$, involving cases where one variable is small ($$\leq \epsilon$$) and the other compensates to bring the sum into the target interval. We express this through:

$$P(\frac{1}{2} - \epsilon \leq Z \leq \frac{1}{2} + \epsilon) \geq P(X \leq \epsilon)P(\frac{1}{2} - \epsilon < Y \leq \frac{1}{2}) + P(Y \leq \epsilon)P(\frac{1}{2} - \epsilon < X \leq \frac{1}{2})$$

Given the symmetry and i.d.d. nature of $$X$$ and $$Y$$, this simplifies to:

$$P(\frac{1}{2} - \epsilon \leq Z \leq \frac{1}{2} + \epsilon) \geq 2F(\epsilon)(1 - F(\frac{1}{2} - \epsilon))$$

### Part (2) Proving $$Z$$ Is Not Uniformly Distributed

For $$Z$$ to be uniformly distributed on any set, its distribution function must be linear (or constant in intervals) within the set's bounds. The analysis in Part (1) reveals two critical points:

• $$P(Z \leq \epsilon)$$ scales quadratically with $$F(\epsilon)$$, indicating a non-linear relationship in the distribution function near $$0$$.
• $$P(\frac{1}{2} - \epsilon \leq Z \leq \frac{1}{2} + \epsilon)$$ suggests a more complex behavior around $$\frac{1}{2}$$, incompatible with uniform distribution's requirement for a constant probability density.

Hence, through these probabilities, we infer a deviation from the linear (or constant) behavior expected of a uniformly distributed random variable. Therefore, $$Z$$ cannot be uniformly distributed on any set, as its distribution characteristics inherently contradict the uniform distribution's fundamental properties.

############ EDIT PART 2 ################################

Uniform distributions, even on complex sets like the Cantor set, have a defining feature: for any two subsets of the distribution's support with equal measure, the probability assigned to these subsets is the same. This uniformity in probability distribution is central to our argument.

Given $$X, Y$$ are i.d.d. on $$[0, \frac{1}{2}]$$ with CDF $$F(x)$$, and $$Z = X + Y$$, we examine the distribution of $$Z$$ through the lens of these properties.

### Revisiting the Quadratic Upper Bound

From Part (1), we established that for $$\epsilon \in (0, \frac{1}{4})$$:

• $$P(Z \leq \epsilon) \leq F(\epsilon)^2$$ provides an upper bound on the probability that $$Z$$ falls within an interval near $$0$$.

This relation implicitly suggests that as $$\epsilon$$ approaches $$0$$, the growth rate of $$P(Z \leq \epsilon)$$ is no faster than quadratic in $$F(\epsilon)$$.

### Uniformity Analysis

Uniform distributions are characterized by a linear relationship between the measure of the set and the probability assigned to that set. For a distribution to be uniformly distributed over any arbitrary set (including fractals like the Cantor set), its cumulative distribution function (CDF) would need to increase linearly with the measure of the set.

• Non-linear Relationship: The quadratic nature of the upper bound on $$P(Z \leq \epsilon)$$ is inherently non-linear. This non-linearity contradicts the expected linear behavior of a CDF for a uniformly distributed variable, even in complex sets where the linear relationship might not be immediately apparent due to the set's fractal nature.
• Behavior at Extremes: The behavior of $$P(Z \leq \epsilon)$$ and $$P(\frac{1}{2} - \epsilon \leq Z \leq \frac{1}{2} + \epsilon)$$ as detailed previously indicates a distribution of $$Z$$ that does not uniformly assign probabilities to equal-measure subsets around $$0$$ and $$\frac{1}{2}$$.
The essence of uniform distribution is its constant density over its support. The analysis of $$Z=X+Y$$ shows that the probability density of $$Z$$ is not constant—evident from the quadratic relationship in small $$\epsilon$$ regions and the more complex behavior near $$\frac{1}{2}$$. Even if $$F(\epsilon)$$ does not behave linearly in a neighborhood of $$0$$, the fact that the growth of $$P(Z \leq \epsilon)$$ is bounded in a way that suggests non-linearity (quadratic) is key. For a uniformly distributed variable, any non-linearity in the assignment of probabilities to subsets of its support indicates a deviation from uniformity.
This deviation is fundamentally why $$Z$$ cannot be uniformly distributed, regardless of the specific set (even those with singular distributions like the Cantor set) on which the uniform distribution might be hypothesized. The argument leverages the inherent properties of uniform distributions—constant probability density over the support—and shows that $$Z$$’s distribution does not exhibit this property, thus cannot be uniform.
• For part (2), I would like to see this formalized more. When it comes to singular distributions like uniform distributon on Cantor set, it's difficult to see exactly where the contradiction lies, as $F(\epsilon)$ doesn't really behave linearly in a neighbourhood of $0$. It seems plausible that you could get quadratic growth rate with a suitable set, and the quadratic curve is just an upper bound rather than an estimate. Commented Apr 8 at 16:58
• In $P(Z \leq \epsilon) = P(X \leq \epsilon)P(Y \leq \epsilon)$ the $=$ should be $\leq$. We know that $Z\leq\epsilon \implies X \leq \epsilon \land Y \leq \epsilon$ but we do not know that $X \leq \epsilon \land Y \leq \epsilon \implies Z\leq\epsilon$. Commented Apr 9 at 17:24