Is it possible to prove that the sum of two independent r.v.'s $X$ and $Y$ with convex support cannot be uniformly distributed on an interval $[a,b]$, with $a < b$?

(Let us rule out the trivial case where $X$ is degenerate and $Y$ is uniform.) Note that $X$ and $Y$ need not be identically distributed.

I have a strong intuition supporting a negative answer because commutativity gets $X+Y$ to "pile up" away from the tails, but I cannot find a way to pin this down to an impossibility proof.

Related questions on Math.SE:

1) Can the sum of two independent identical random variables be uniform? and Can the sum of two i.i.d. random variables be uniformly distributed? answer negatively under the additional assumption that $X$ and $Y$ are identically distributed.

2) in Can sum of two random variables be uniformly distributed @GEdgar answers positively under the assumption that the two independent r.v.'s $X$ and $Y$ have non-convex support.

  • $\begingroup$ Are you asking for a general proof or a specific example? $\endgroup$ – copper.hat Mar 24 '17 at 23:56
  • $\begingroup$ An example would be nice, and a general proof would be even better. $\endgroup$ – mlc Mar 25 '17 at 6:11

Your intuition is correct : the answer is negative. Here is a proof.

Suppose, by contradiction, that $X$ and $Y$ are independent, non-degenerate variables with convex support, say $[c,d]$ and $[e,f]$, and $X+Y$ is uniform on $[c+e,d+f]$. Replacing $(X,Y)$ with $(X-c,Y-e)$, we may assume without loss that $c=e=0$. Interchanging $X$ and $Y$ if necessary, we may further assume that $d\leq f$. If $d<f$ we can also replace $(X,Y)$ with $((2f)X+(3f+d)Y,(4f)X+(3f-d)Y)$ (remember that a linear bijective transformation preserves rv independence), so we may further assume that $d= f$. Finally, replacing $(X,Y)$ with $(\frac{X}{d},\frac{Y}{d})$ we may assume that $d=1$.

At this point, we then have $X,Y$ independent with support exactly equal to $[0,1]$, and $X+Y$ distributed uniformly on $[0,2]$. Now, define the following events :

$$ \begin{array}{cccl} X_1= \lbrace X\in[0,\frac{1}{3}] \rbrace, & X_2= \lbrace X\in(\frac{1}{3},\frac{2}{3}] \rbrace, & X_3= \lbrace X\in(\frac{2}{3},1] \rbrace, \\ Y_1= \lbrace Y\in[0,\frac{1}{3}] \rbrace, & Y_2= \lbrace Y\in(\frac{1}{3},\frac{2}{3}] \rbrace, & Y_3= \lbrace Y\in(\frac{2}{3},1] \rbrace, & \end{array} $$

$$ P_k=\cup_{i+j=k}X_i\times Y_j, Q_l=\cup_{k=2}^l P_k, L(x)=\lbrace X+Y \leq x \rbrace $$

We then have the inclusions :

$$ L\big(\frac{l-1}{3}\big) \subseteq Q_l \subseteq L\big(\frac{l}{3}\big) $$

Putting $q_l=P(Q_l)$, we deduce

$$ \frac{l-1}{6}\leq q_l \leq \frac{l}{6} \tag{1} $$

Next, put $x_k=P(X_k), y_k=P(Y_k), p_k=P(P_k)$. It follows from (1) that the following number is nonnegative :

$$ s=(1-x_1)\big(q_2-\frac{1}{6}\big)+x_1\big(\frac{2}{3}-q_4\big)+x_1x_2y_2 \tag{2} $$

Clearly $q_2=x_1y_1$. And using $x_3=1-(x_1+x_2)$ and $y_3=1-(y_1+y_2)$, we find that

$$ q_4 =x_1y_1+x_2y_1+x_1y_2+x_1y_3+x_2y_2+x_3y_1=x_1+y_1+x_2y_2-x_1y_1 \tag{3} $$

Reinjecting this back into (2), we obtain

$$ s=\bigg(x_1-\frac{1}{2}\bigg)\bigg(\frac{1}{3}-x_1\bigg) \tag{4} $$

It follows that $x_1\in[\frac{1}{3},\frac{1}{2}]$. Using natural symmetries of the problem, it follows that $x_3,y_1$ and $y_3$ are also in $[\frac{1}{3},\frac{1}{2}]$.

Next, consider

$$ \begin{array}{clc} t_1 &=& (x_1x_3-x_2^2)\big(\frac{5}{6}-q_5\big)+x_3x_2\big(\frac{2}{3}-q_4\big)+x_3^2\big(q_3-\frac{1}{3}\big) \\ t_2 &=& (x_1x_3-x_2^2)\big(q_2-\frac{1}{6}\big)+x_1x_2\big(q_3-\frac{1}{3}\big)+ x_1^2\big(\frac{2}{3}-q_4\big) \\ t &=& t_1+t_2 \end{array} $$

Note that since $x_1x_3 \geq \frac{1}{9} \geq x_2^2$, the numbers $t_1,t_2$ and $t$ are all nonnegative. Simplifying $t$ as we did for $s$, we obtain :

$$ t=-\frac{x_2}{4}(1+x_2)(\frac{1}{3}-x_2)-\frac{3(x_3-x_1)^2}{4}\bigg(\frac{4}{9}-x_2\bigg) \tag{5} $$

In (5) above we have a nonnegative LHS and a nonpositive RHS. Note also that $x_2>0$ because the support of $X$ is $[0,1]$. This forces $x_2=\frac{1}{3}$ and $(x_3-x_1)^2=0$, so that $x_1=x_2=x_3=\frac{1}{3}$, and similarly $y_1=y_2=y_3=\frac{1}{3}$. But then $q_2=\frac{1}{9}$ contradicts (1). This finishes the proof.

UPDATE: Since (5) was questioned in the comments, I append a computer verification of it.

                  GP/PARI CALCULATOR Version 2.7.4 (released)
          i386 running darwin (x86-64/GMP-6.0.0 kernel) 64-bit version
compiled: Sep 27 2015, Apple LLVM version 6.0 (clang-600.0.57) (based on LLVM 3.5svn)
                            threading engine: single
                 (readline v6.3 enabled, extended help enabled)

                     Copyright (C) 2000-2015 The PARI Group

PARI/GP is free software, covered by the GNU General Public License, and comes 

Type ? for help, \q to quit.
Type ?12 for how to get moral (and possibly technical) support.

parisize = 8000000, primelimit = 500000
? x3=1-(x1+x2)
%1 = -x1 + (-x2 + 1)
? y3=1-(y1+y2)
%2 = -y1 + (-y2 + 1)
? x(k)=if((1<=k)&&(k<=3),eval(Str("x",k)),0)
%3 = (k)->if((1<=k)&&(k<=3),eval(Str("x",k)),0)
? y(k)=if((1<=k)&&(k<=3),eval(Str("y",k)),0)
%4 = (k)->if((1<=k)&&(k<=3),eval(Str("y",k)),0)
? p(k)=sum(j=1,k,x(j)*y(k-j))
%5 = (k)->sum(j=1,k,x(j)*y(k-j))
? q(l)=sum(k=2,l,p(k))
%6 = (l)->sum(k=2,l,p(k))
? s_term=(1-x1)*(q(2)-(1/6))+x1*((2/3)-q(4))+x1*x2*y2
%7 = -x1^2 + 5/6*x1 - 1/6
? check1=s_term-(x1-(1/3))*((1/2)-x1)
%8 = 0
? t1_term=(x1*x3-(x2^2))*((5/6)-q(5))+x3*x2*((2/3)-q(4))+(x3^2)*(q(3)-(1/3))
%9 = x1^3 + (3*x2 - 13/6)*x1^2 + (3*x2^2 - 25/6*x2 + 3/2)*x1 + (x2^3 - 11/6*x2^2 + 4/3*x2 - 1/3)
? t2_term=(x1*x3-(x2^2))*(q(2)-(1/6))+x1*x2*(q(3)-(1/3))+(x1^2)*((2/3)-q(4))
%10 = -x1^3 + 5/6*x1^2 + (-1/6*x2 - 1/6)*x1 + 1/6*x2^2
? t_term=t1_term+t2_term
%11 = (3*x2 - 4/3)*x1^2 + (3*x2^2 - 13/3*x2 + 4/3)*x1 + (x2^3 - 5/3*x2^2 + 4/3*x2 - 1/3)
? easy_part=(x2/4)*(x2+1)*((1/3)-x2)
%12 = -1/4*x2^3 - 1/6*x2^2 + 1/12*x2
? hard_part=(3/4)*((x3-x1)^2)*((4/9)-x2)
%13 = (-3*x2 + 4/3)*x1^2 + (-3*x2^2 + 13/3*x2 - 4/3)*x1 + (-3/4*x2^3 + 11/6*x2^2 - 17/12*x2 + 1/3)
? check2=(t_term+(easy_part+hard_part))
%14 = 0
  • $\begingroup$ I have been unable to verify (5). I have been doing it by hand, and got a slightly different result. Everything else works out. (I fixed three minor typos.) This approach is laborious but compelling. $\endgroup$ – mlc Apr 2 '17 at 20:24
  • $\begingroup$ @mlc Thank you for your feedback. I have added a verification of (5) by computer, and I invite you to double-check it, for example at tutorialspoint.com/execute_pari_online.php $\endgroup$ – Ewan Delanoy Apr 3 '17 at 7:01
  • $\begingroup$ I did not mean to question the computation, simply to plainly state that I could not (yet) vouch for the correctness of the proof. I have accepted your answer because it is most useful, but this is not equivalent to asserting that the proof is complete. And now it is :-) $\endgroup$ – mlc Apr 3 '17 at 8:45

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