# What is the largest possible expectation of difference between two i.i.d. random vectors in the separable Hilbert space?

Suppose $$M$$ is a compact subset of the separable Hilbert space $$l_2$$. Suppose, $$X$$ and $$Y$$ are i.i.d. random vectors with support in $$M$$. What is the largest possible $$E \| X - Y \|$$?

Suppose $$M$$ is a compact subset of the separable Hilbert space $$l_2$$. Then $$M$$ is closed and bounded, and there exist such $$x$$ and $$y$$ in $$M$$, such, that $$\| x - y \| = \operatorname{diam}(M)$$. Now, suppose $$X$$ and $$Y$$ are i.i.d. random vectors, such that $$P(X = x) = P(X = y) = \frac{1}{2}$$. One can see, that

$$E\| X - Y \| = \frac{1}{4}\| x - x \| + \frac{1}{4}\| x - y \| + \frac{1}{4}\| y - x\| + \frac{1}{4}\| y - y \| = \frac{1}{2}\operatorname{diam}(M)$$

Now let’s prove that it is the maximal possible expected distance, or to be more exact, that if $$X = (X_n)_{i = 1}^{\infty}$$ and $$Y = (Y_n)_{i = 1}^{\infty}$$ are i.i.d. random vectors with support in $$M$$, then $$E\| X - Y \| \leq \frac{1}{2}\operatorname{diam}(M)$$. By Hölder inequality:

$$E\| X - Y \| \leq \left(E(\| X - Y \|^2)\right)^{\frac{1}{2}}.$$

And one can see, that

\begin{align*} E(\| X - Y \|^2) &= E\langle X - Y, X - Y \rangle = 2E\langle X, X \rangle - 2E\langle X, Y \rangle \\ &= 2\left(\sum_{n = 1}^\infty E(X_n)^2 - \sum_{n = 1}^\infty EX_n Y_n \right) \\ &= 2\left(\sum_{n = 1}^\infty E(X_n)^2 - \sum_{n = 1}^\infty EX_n EX_n\right ) \\ &\leq 2\left(\sum_{n = 1}^\infty E(X_n)^2 \right) \\ &= 2E\langle X, X \rangle = 2E(\| X \|^2) \end{align*}

Now, suppose, that the least closed ball containing $$M$$ is the closed ball with radius $$\frac{1}{2}$$ and center $$0$$. That will result in $$\| X \|$$ being a random variable on $$[-1, 1]$$. So, its second moment does not exceed $$\frac{1}{4}$$ (There are several proofs of this fact here: What is the largest possible variance of a random variable on $[0; 1]$?), and we get $$E\| X - Y \| \leq \frac{1}{\sqrt{2}}$$.

And now let’s return to the general case. Suppose $$z$$ is the center of the least closed ball containing $$M$$. Then $$\frac{M - z}{\operatorname{diam}(M)}$$ is such a subset, that the least closed ball containing it is the closed ball with radius $$\frac{1}{2}$$ and center $$0$$. So

$$E\| X - Y \| = \operatorname{diam}(M)E \left\| \frac{X - z}{\operatorname{diam}(M)} - \frac{Y - z}{\operatorname{diam}(M)}\right\| \leq \frac{1}{\sqrt{2}}\operatorname{diam}(M)$$

So we know, that the largest possible $$E\| X - Y \|$$ we search is certainly $$\geq \frac{1}{2}\operatorname{diam}(M)$$ and certainly $$\leq \frac{1}{\sqrt{2}}\operatorname{diam}(M)$$. However, I do not know, how to find its exact value.

This question is partially inspired by the following question: Probability distribution to maximize the expected distance between two points

• I made some edit on your question so that formulas are better to read, I hope. – Sangchul Lee Feb 18 at 4:29
• It definitely depends on the choice of $M$. I can set up variational argument to obtain a characterization of the maximizer, and in the case where the convex hull of $M$ is a finite-dimensional polytope, then I can give a formula computing the optimal constant. For a very specific case where the convex hull of $M$ is a regular $n$-gon (so that its $n$ vertices $v_1, \cdots, v_n$ also lie in the original set $M$), the expectation is maximized by $X, Y$'s uniformly distributed on $\{v_1, \cdots, v_n\}$. – Sangchul Lee Feb 18 at 8:30

## 1 Answer

Here is a note on some of my observations, but this answer is far from being complete, as I did not explain many steps.

This problem allows variational formulation. Let $$\mathcal{P}$$ denotes the set of all probability measures on the compact set $$M \subset \mathbb{R}^d$$. Then, on $$\mathcal{P}$$, we define

$$\langle \mu, \nu \rangle = \int_{M^2} \| x - y \| \, \mu(\mathrm{d}x)\nu(\mathrm{d}y), \qquad Q(\mu) = \langle \mu, \mu \rangle.$$

This is related to our problem by noting that, if $$X$$ and $$Y$$ are i.i.d. random variables on $$M$$, its distribution $$\mu$$ is an element of $$\mathcal{P}$$ and $$\mathbb{E}[\| X - Y \|] = Q(\mu)$$ holds. In light of this, the question boils down to finding the maximum of $$Q$$ over $$M$$. If we equip $$\mathcal{P}$$ the topology of weak-* convergence, then $$\langle \cdot, \cdot \rangle$$ is continuous, and so, $$Q$$ attains maximum on $$\mathcal{P}$$. Then the following is a variational formulation of characterizing these maximums:

Claim. For $$\mu \in \mathcal{P}$$, the followings are equivalent:

1. $$\mu$$ maximizes $$Q$$.
2. $$\langle \mu, \delta_z \rangle \leq \langle \mu, \delta_x \rangle$$ for any $$z \in M$$ and $$x \in \operatorname{supp}(\mu)$$.

The argument is quite similar to one of my previous answer to a similar question, so let me skip the proof at this moment and rejoice its consequence. Let $$\mu$$ be a maximizer of $$Q$$ over $$\mathcal{P}$$. Then the following transform

$$L\mu(z) := \langle \mu, \delta_z \rangle = \int_M \|x - z\| \, \mu(\mathrm{d}x)$$

is a continuous convex function on $$\mathbb{R}^d$$. Since every point in the support of $$\mu$$ is a maximum point of $$L\mu$$ over $$M$$, by writing $$m = \max_M L\mu$$, we easily read out that

$$\operatorname{supp}(\mu) \subseteq (L\mu)^{-1}(\{m\}) \qquad \text{and} \qquad M \subseteq (L\mu)^{-1}((-\infty, m]).$$

To make further progress, we separate exceptional case from the general argument:

1. If it happens that $$\operatorname{supp}(\mu)$$ lies in a line, then it reduces to 1-d problem. In such case, it is not hard to check that $$\mu$$ must be of the form $$\mu = \frac{1}{2}(\delta_{x_0} + \delta_{x_1})$$ for some distinct points $$x_1$$ and $$x_2$$. In such case $$M$$ must be a compact subset of the line segment $$\overline{x_0x_1}$$.

2. Otherwise, $$(L\mu)^{-1}((-\infty, m])$$ is a strict convex set and $$(L\mu)^{-1}(\{m\}) = \partial (L\mu)^{-1}((-\infty, m])$$. This may be used to further restrict the possible form that $$\operatorname{supp}(\mu)$$ can take.

3. As a special case, consider the situation where the convex hull of $$M$$ is a convex polytope. Then $$\operatorname{supp}(\mu)$$ is supported on the vertex-set of that polytope. If $$x_1, \cdots, x_n$$ denotes these vertices, then the problem reduces to solving the system of linear equations

$$L \mu = m\mathbf{1}, \qquad \mathbf{1}^{\mathsf{T}}\mu = \mathbf{1}$$

Here, we identify $$L$$ as the symmetric matrix $$L = (\| x_j - x_k \|)_{j,k=1}^{n}$$ and the measure $$\mu$$ as the column vector $$(\mu(\{x_j\}))_{j=1}^{n}$$. Also, $$\mathbf{1}$$ is the column vector of length $$n$$ consisting of only $$1$$'s. The conclusion is that $$m = 1/(\mathbf{1}^\mathsf{T}L^{-1}\mathbf{1})$$, or equivalently, $$m$$ is the inverse of the sum of all entries of $$L^{-1}$$.

4. In some cases, we may use the characterization of the maximizer directly. For instance, if the convex hull of $$M$$ is a ball, then $$\mu$$ must be the uniform distribution over the boundary of the ball, again allowing an explicit computation of the optimal constant.