# Which "difference" functionals only depend on the margins?

Let the real-valued random variables $$(X_1,X_2)$$ have joint distribution $$P_1\times P_2$$. Let the distribution of $$X_1-X_2$$ be $$P_{\mathrm{diff}}$$.

Let $$\psi(P_{\mathrm{diff}})$$ denote a functional of distribution of the difference. The functional is identified by the margins if $$\psi(P_{\mathrm{diff}})=\psi_m(P_1, P_2)$$ for some functional $$\psi_m$$ that only depends on the marginal distributions $$P_1$$ and $$P_2$$.

My question: What is the class of all functionals of the distribution of the difference which are identified by the margins?

Clearly by the linearity of expectation, the expectation-of-difference functional $$\psi(P_\mathrm{diff}) = \int_{\mathbb{R}^2} x_1 - x_2 d(P_1\times P_2)(x_1,x_2)$$ is identified by the margins. I cannot find other examples or prove a general theorem.

• I think that $\psi:\mathcal U\to\mathbb R$ would be better, with $\mathcal U$ the set of joint distribution on $X_1$ and $X_2$ or something like that. Indeed $P_1\times P_2$ is not a set. Also is $\psi$ linear ? (some people say that functional is linear and some don't so I like specifying). Commented Feb 1, 2023 at 19:08
• @P.Quinton Thanks for the correction about the domain. I'm interested in both the linear and nonlinear case (so no restriction, but an answer concerning linear functionals would still be interesting).
– Ben
Commented Feb 1, 2023 at 19:13
• There are lots of other examples. For example, take the joint distribution, compute the marginal $P_1$, then output any functional of $P_1$ alone, such as $P_1^3[A]$ for some fixed Borel set $A \subseteq \mathbb{R}$, or such as $E[\cos(X)]$ where $X$ has distribution $P_1$. Or take the output to be $E[h(Y_1)]E[g(Y_2)]$ for some measurable functions $h, g$ and where $Y_i$ has distribution $P_i$. Commented Feb 1, 2023 at 21:50
• @Michael Thanks for the examples. You are correct - the issue is that I incorrectly formulated the question. I've updated the question.
– Ben
Commented Feb 1, 2023 at 22:23
• +1, interesting question! (Of course all functions of the expectation of the difference also work.) Commented Feb 2, 2023 at 5:37

If you mean $$P_1 \times P_2$$ to be the product measure of $$P_1$$ and $$P_2$$, you can define $$\bar P_2 : A \mapsto P_2(-A)$$, and observe that $$P_\mathrm{diff} = P_1 * \bar P_2$$ where $$*$$ denotes the convolution of measures. Then writing $$\phi : (P_1, P_2) \mapsto P_1 * \bar P_2$$, you get that $$\psi_m = \psi \circ \phi =: \phi_m$$ satisfies the condition. This means every functional on the set of all measures that can be written as $$P_\mathrm{diff}$$ is identified by the margins.

If on the other hand you mean that $$P_\mathrm{diff}$$ is a measure that is the distribution of the difference of two not necessarily independent random variables with respective distributions $$P_1$$ and $$P_2$$, the answer is less obvious and might depend on the properties of the measured group that the $$P_1, P_2$$ are defined on – and you need to clarify your post.

Since the second case implies the first, I will treat the second case. The Theorem below provides a full characterization when $$\psi$$ is continuous and restricted to measures with a second moment condition. But first, let us give an example of a family of functionals that are identified by their margins.

Lemma: Write $$\mathcal M_1$$ for the set of all measures $$\mu$$ on $$\mathbb R^2$$ such that $$\int |x_2 - x_1|d\mu(x_1, x_2) < \infty$$. For every function $$f:\mathbb R\to\mathbb R$$, the functional $$\psi_f : \mathcal M_1 \to \mathbb R \ , \ \mu \mapsto f\left(\int (x_2 - x_1)d\mu(x_1, x_2) \right) \qquad (*)$$ is identified by the margins.

This Lemma has a reciprocal under slightly stronger condition. Write $$\mathcal M_2$$ for the set of all distributions of $$(X,Y)$$ with $$X-Y \in L^2$$, equipped with the topology of weak convergence $$\mu_n \Rightarrow \mu \iff \forall f\in C_b(\mathbb R^2), \int fd\mu_n \to \int fd\mu$$:

Theorem: the set of continuous functionals on $$\mathcal M_2$$ that are identified by the margins is exactly the set of all $$\psi_f$$ with $$f$$ continuous.

The continuity of $$f$$ comes from considering the measures $$\delta_{(x,-x)}$$, with image $$f(2x)$$: this should be continuous in $$x$$. The condition that the functional be defined on $$\mathcal M_2$$ has two opposing effects. First, since every functional on the set of probability measures induces a functional on $$\mathcal{M}_2$$, every functional considered by OP falls under the scope of this theorem (provided it is continuous on $$\mathcal M_2$$). On the other hand, while $$(*)$$ must hold on $$\mathcal{M}_2$$, it might fail to generalize to the values taken by $$\psi(\mu)$$ when $$\mu \notin \mathcal M_2$$. In addition, $$\mathcal M_2$$ can likely be replaced by a larger set of probability measures without changing the validity of the theorem.

Note: I am not assuming that $$\psi$$ is linear.

## Let’s play a game.

The game has one player, with two available actions. Starting from a distribution $$\mu$$ on $$\mathbb R^2$$, or equivalently a pair of random variables $$(X,Y)$$, they can change the distribution of $$(X,Y)$$ while keeping constant at least one of

• the marginals, i.e. the distribution of $$X$$ and that of $$Y$$,
• the distribution of $$X-Y$$.

In other words, either the "projection" on either axis must be conserved, or the "projection" on the anti-diagonal must be conserved. It is clear that if $$\psi$$ is identified by the margins, then those two operations leave $$\psi(\mu)$$ invariant.

The question is: what are the classes of the set of all probability measures on $$\mathbb R^2$$, when two measures are invariant when we can go from one to the other (not necessarily in both directions, one is enough!) by a finite number of steps in the above game? The functional $$\psi$$ will be constant on them.

Question 1: is the set of all $$\psi$$ that are identified by the margins, exactly the set of $$\psi$$ that are constant on the classes above?

## Some observations.

• if $$X-Y$$ has the same distribution as $$Y-X$$ (its distribution is symmetric), we can arrive to the distribution $$\delta_0$$ in three steps. Indeed, writing $$Z = (X-Y)/2$$, $$(Z, -Z)$$ has the same distribution of the difference as $$(X,Y)$$, then $$(Z,Z)$$ has the same distribution of the marginals, and finally, $$(0,0)$$ has the same distribution of the difference.

• The quantity $$\mathbb E[X-Y]$$ (if well-defined) is left unchanged by the operations above, and for every $$z\in\mathbb R$$ the "random" vector $$(z/2, -z/2)$$ is such that $$z/2+z/2 = z$$, so we have at least a family indexed by $$\mathbb R$$ of classes.

Question 2: starting from $$(X,Y) = (z, 0)$$ with probability one, can we reach any distribution of $$(X,Y)$$ such that $$\mathbb E[X-Y] = z$$?

If the answer is yes to questions 1 and 2, then the set of $$\psi$$ that are identified by the margins, restricted to the set of measures where $$X-Y$$ is integrable, can be identified as the set of functions $$f:\mathbb R \to \mathbb R$$ by the formula $$\psi(\mu) = f(\mathbb E[X-Y])$$ where $$(X,Y)$$ has distribution $$\mu$$. This would show that the Lemma is in fact not only sufficient, but necessary, i.e. that every functional on the set of measures with $$X-Y$$ integrable that is identified by the margin is of the form $$\psi_f$$.

## Some related questions

The following processes are legal sequences in the game.

Q1: fix $$(X_0, Y_0)$$, and define by induction $$(X_{n+1},Y_{n+1})$$ independent such that $$2X_{n+1}$$ and $$-2Y_{n+1}$$ both have the same distribution as $$X_n-Y_n$$. How does the sequence $$(X_n, Y_n)$$ behave as $$n\to\infty$$?

We can check that \begin{align} \mathbb E[(X_{n+1} - Y_{n+1})^2] &= \mathbb E [X_{n+1}^2] + \mathbb E [Y_{n+1}^2] + 2 \mathbb E [X_{n+1}]^2 \\ &= \frac12 \mathbb E[(X_n-Y_n)^2] + \frac12\mathbb E[X_n-Y_n]^2 \\ &= \frac12 \mathbb V[(X_n-Y_n)^2] + \mathbb E[X_{n+1}-Y_{n+1}]^2 \end{align} so that the variance is divided by half at each step (this is true, more generally, for every cumulant that is well-defined). This means that $$X_n - Y_n \to \mathbb E[X_0 - Y_0]$$ in probability (and even almost surely) as $$n\to\infty$$. Under a condition of continuity on $$\psi$$ this gives us a proof of question 2 and of the Theorem. The method of cumulants would even allow a very precise central limit theorem: in distribution, $$2^{n/2} \frac{X_n-Y_n - \mathbb E[X_0-Y_0]}{\sqrt{\mathbb V(X_0 - Y_0)}} \underset{n\to\infty}\longrightarrow \mathcal N(0,1) .$$

Q2: Same question, but with $$X_{n+1} = (X_n-Y_n) + \epsilon_n$$ and $$Y_{n+1}$$ an independent copy of $$\epsilon_n$$, where $$(\epsilon_n)_n$$ is a "nice" sequence of centered random variables (independent, possibly i.i.d.).

The central limit theorem applies, meaning that we can reach distributions $$(X_n, Y_n)$$ with arbitrarily large $$\| X_n - Y_n \|_1$$ from arbitrarily small $$\| X_0 - Y_0 \|_1$$.