Which coupling of uniform random variables maximises the essential infimum of the sum?

Question

Recall that a coupling of probability measures $\mu_i$ is a set of random variables $X_i$ defined on the same probability space $\Omega$ such that $X_i \sim \mu_i$.

Question: Let $\mu_1, \dots, \mu_n$ all be uniform on $[0, 1]$. What coupling maximises the essential infimum

$$I_n := \text{essinf}_{\omega \in \Omega} (X_1 (\omega) + \dots + X_n (\omega))$$

and what is the achieved value?

In particular, if we let $I^*_n$ be the optimal value, I wonder if the sequence $\frac{1}{n}I^*_n$ exhibits any kind of predictable behaviour.

Answer 1

Let me complete Iosif Pinelis’ answer by presenting an example for $n=3$ (in fact, it seems that a similar construction itself works for all $n$).

If $X_1<1/2$, set $X_2=1-2X_1$ and $X_3=1/2+X_1$.

If $X_1\geq 1/2$, set $X_2=2-2X_1$ and $X_3=X_1-1/2$.

Answer 2

Note that $$I_n^*\le E(X_1+\cdots+X_n)=n/2.$$ This upper bound $n/2$ on $I_n^*$ is attained if $n=2m$ is even and $X_{2i}=1-X_{2i-1}$ for $i=1,\dots,m$.

So, $$I_n^*=n/2$$ for even $n$.

It also follows that $$(n-1)/2=I_{n-1}^*\le I_n^*\le n/2$$ for odd $n$.

So, $$I_n^*\sim n/2$$ for $\Bbb N\ni n\to\infty$.

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Conjecture: $I_n^*=n/2$ for all natural $n\ge2$.

Remark 1: Obviously, $I_1^*=0$. In view of the above, it is enough to show that $I_n^*=n/2$ for all odd $n\ge3$. Moreover, in view of the above cancellations, it is enough to show that $I_3^*=3/2$.