What Jordan curve of length $1$ maximizes the expected length of a chord between two uniformly picked points on it?

Question

For a (sufficiently nice) Jordan curve $\sigma :[0,1] \to \mathbb{R}^2$ with unit length and natural (w.r.t. arc length) parametrization, we denote the expected value of the distance between two points that are picked from the path independently and uniformly by $c(\sigma)$. So

$$c(\sigma) = \mathbb{E}[ \space|\sigma(T_1)- \sigma(T_2)| \space],$$

where $T_1, T_2 \sim U([0,1])$ are independent.

What path maximizes $c$?

For example, for $\sigma =$ circle (of radius $\frac{1}{2\pi}$ to have unit perimeter), $$c(\sigma) = \frac{2}{\pi^2}$$

(It is known that for unit circle we get $\frac{4}{\pi}$ so scaling leads to the value $\frac{2}{\pi^2}$.)
Is circle the maximizing path?

PS. Here is a small application for testing different kinds of paths. The circle seemed to be the best among all paths I tried out.

Answer 1

I'm pretty certain that a straight line segment is the optimizing path.

For then the expected length appears to be $\frac{1}{3}$ (if my back-of-the-envelope computation is correct), which is much larger than $\frac{2}{\pi^2}$.

Answer 2

Here's a tentative partial answer to the revised question (where "arc" has been replaced by "curve"). Let's call the curve $u$, with $\|u'\| = 1$, i.e., it's parameterized by arclength.

I'm going to make one leap here, which is to replace the expectation of distance with the expectation of squared distance. This is similar to replacing "length" with "energy" when finding geodesics on a manifold, but without my copy of Morse Theory here at home, I cannot reproduce the argument...so much of what follows may well be incorrect.

Now the optimal curve can be translated so that its center of mass is at the origin, i.e., so that $$ \int_0^1 u(t) ~ dt = 0. $$

The thing we're optimizing is \begin{align} E(u) &= \int_0^1 \int_0 ^ 1 |u(t) - u(s)\|^2 ~ds ~dt \\ &= \int_0^1 \int_0 ^ 1 (u(t) - u(s)) \cdot (u(t) - u(s)) ~ds ~dt \\ &= \int_0^1 \int_0 ^ 1 u(t)\cdot u(t) - 2u(t) \cdot u(s)) + u(s) \cdot u(s) ~ds ~dt \\ &= \int_0^1 \int_0 ^ 1 u(t)\cdot u(t) ~ds ~dt - \int_0^1 \int_0 ^ 1 2u(t) \cdot u(s)) ~ds ~dt + \int_0^1 \int_0 ^ 1 u(s) \cdot u(s) ~ds ~dt \\ &= \int_0^1 u(t)\cdot u(t) ~dt - \int_0^1 \int_0 ^ 1 2u(t) \cdot u(s)) ~ds ~dt + \int_0 ^ 1 u(s) \cdot u(s) ~ds \end{align} The first and last terms here are the same, so they can be combined: \begin{align} E(u) &= \int_0^1 u(t)\cdot u(t) ~dt - \int_0^1 \int_0 ^ 1 2u(t) \cdot u(s)) ~ds ~dt + \int_0 ^ 1 u(s) \cdot u(s) ~ds \\ &= 2\int_0^1 u(t)\cdot u(t) ~dt - \int_0^1 \int_0 ^ 1 2u(t) \cdot u(s)) ~ds ~dt \\ &= 2\int_0^1 u(t)\cdot u(t) ~dt - 2\int_0^1 u(t) \cdot\int_0 ^ 1 u(s)) ~ds ~dt \\ &= 2\int_0^1 u(t)\cdot u(t) ~dt - 2\int_0^1 u(t) \cdot0 ~dt \\ &= 2\int_0^1 u(t)\cdot u(t) ~dt. \end{align}

So now the problem is reduced to this:

Among all unit-length loops whose average is the origin, which one maximizes $$ E(u) = \int_0^1 u(t)\cdot u(t) ~dt? $$

I suspect that's a standard Calc. of Variations problem with the circle being a well-known answer, but I can't promise that.

Anyhow, there you have at least a start at an attempt on the problem.