$\mathbb E[X]=0$ implies existence of two random variables $Y\stackrel{\mathrm d}=Z$, equal in distribution, such that $X\stackrel{\mathrm d}=Y-Z$

Question

Let $X$ be a bounded random variable on a non-atomic probability space, i.e. $X\in L^\infty(\mathbb P)$, such that $\mathbb E[X]=0$. In a lecture, I heard someone claim that this implies the existence of two other random variables $Y,Z\in L^\infty(\mathbb P)$ such that $Y\stackrel{\mathrm d}=Z$ and $X\stackrel{\mathrm d}=Y-Z$, although they claimed the proof was difficult and tedious.

Is this result even true? Even in simple examples I don't really see what this decomposition should look like. What is the intuition behind such a result? Is there perhaps a proof in the literature or in a textbook?

It's certainly not true if you add the restriction that $Y,Z$ are independent. — Dark Malthorp, Feb 02 '24 at 14:07
This is supposed to be true in any probability space? Or in a specific probability space? — , Feb 02 '24 at 14:10
@jwhite Thanks! I forgot to mention that we work with non-atomic r.v.s — Václav Mordvinov, Feb 02 '24 at 14:12
$A \stackrel{\mathrm d}= B$ implies $|| A-B ||_1 = 0$ ie $L_1$ equivalent? If so I'm really not seeing how this could be true. — user24142, Feb 04 '24 at 16:46
@user24142 you're claim is not correct, for $A,B$ are allowed to be dependent. — Václav Mordvinov, Feb 04 '24 at 17:21
e.g. take $A$ standard normal, $B=-A$, $A-B=2A$, which of course has zero expectation but $|A|_{L^1}=\mathbb E|A|>0$ — Václav Mordvinov, Feb 04 '24 at 17:26
Well, clearly if X is symmetrically distributed, we can take Z=-X/2 and Y=X/2. So the difficulty is for X no symmetric. — Enredanrestos, Feb 04 '24 at 19:14
@user24142 $\stackrel{\mathrm d}=$ denotes equality in distribution — Václav Mordvinov, Feb 04 '24 at 19:36

score 8 · Accepted Answer · edited Feb 06 '24 at 15:12

Here's an attempt at a proof. I don't have a reference, or much intuition beyond this construction.

Consider first the case of a variable $X$ taking the value $a<0$ with probability $b/(|a|+b)$ and taking the value $b>0$ with probability $|a|/(|a|+b).$ Pick random variables $Y,Z$ uniformly distributed on $[a,b],$ with $Y=Z+b$ when $Z<0$ and $Y=Z+a$ when $Z\geq 0.$ Then $Y-Z\stackrel{\mathrm d}= X$ as required.

For the general case we can take a mixture, by pairing up the events $X<0$ and $X>0$ in a suitable way.

Remark. We just need to write the law of $X$ as a weighted average of zero-mean probability distributions supported on two points. A slick way to do this is to use Choquet's theorem and a characterization of extreme points of the set of zero-mean variables, e.g. Winkler's "Extreme Points of Moment Sets". The remainder of this answer is just a more explicit version of this.

Let $\mu$ be the law of $X.$ Let $M=\tfrac 1 2 \mathbb E|X|.$ There exists a Borel function $f^+:[0,M]\to\mathbb R_{>0}$ such that the pushforward of Lebesgue measure $f^+_*\lambda$ is the measure $x d\mu(x).$ For example, the generalized inverse of the function $t\mapsto \int_0^t x d\mu(x).$ Similarly there exists a Borel function $f^-:[0,M]\to\mathbb R_{<0}$ such that $f^-_*\lambda$ is the measure $|x| d\mu(x).$ The pushforward of $(1/f^+(x))d\lambda(x)$ under $f^+$ is $\mu$ restricted to $\mathbb R_{>0}.$ And similarly the pushforward of $(1/|f^-(x)|)d\lambda(x)$ under $f^-$ is $\mu$ restricted to $\mathbb R_{<0}.$

Pick independent random variables $R,S,T$ such that

$R$ is a Bernoulli random variable with $\mathbb P[R=0]=\mathbb P[X=0],$
$S$ is a random variable on $[0,M]$ distributed according to the measure $((1/f^+(x))+(1/|f^-(x)|))d\lambda(x),$
$T$ is uniformly distributed on $[0,1].$

Then define $Y$ and $Z$ by:

wherever $R=0,$ we take $Y=Z=0,$ and skip the next steps.
$Z$ is $T$ rescaled from $[0,1]$ to the interval $[f^-(S),f^+(S)].$
$Y$ is $Z+f^+(S)$ when $Z<0$
$Y$ is $Z+f^-(S)$ when $Z\geq 0.$

$Z$ has the same distribution as $Y.$ The random variable $Y-Z$ has the same probability of being zero as $X.$ For any Borel $U\subseteq\mathbb R_{>0},$ the probability that $f^+(S)\in U$ and $Y-Z=f^+(S)$ is $\int (1/f^+(x))1_{f^+(x)\in U}d\lambda(x)=\mu(U).$ So the laws of $Y-Z$ and $X$ are the same on $\mathbb R_{>0}$. And similarly for $\mathbb R_{<0}.$ So this construction ensures $Y-Z\stackrel{\mathrm d}= X.$

This looks fine to me, and I bet this construction is way simpler than the `tedious' proof talked about in the lecture. Thanks for your inventive answer, +1, and I leave the bounty open to allow for other answers as well. — Václav Mordvinov, Feb 06 '24 at 15:06

$\mathbb E[X]=0$ implies existence of two random variables $Y\stackrel{\mathrm d}=Z$, equal in distribution, such that $X\stackrel{\mathrm d}=Y-Z$

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