Probability distribution optimization problem of distances between points in $[0,1]$

Question

Let $\mathcal{D}$ be a probability distribution with support $[0,1]$. Let $X, Y, Z$ three i.i.d. random variables with distribution $\mathcal{D}$, and $T$ a random variable uniformly distributed in $[0,1]$ independent from $X$, $Y$ and $Z$. We define $$\Delta=\mathbb{E}\left(1-|x-y|~\big|~x,y<t<z\right)$$ and $$\Delta'=\mathbb{E}\left(1-\min\left(|x-y|,|z-y|\right)~\big|~x,y<t<z\right)~.$$

---

Question: What is the minimum value of the ratio $\rho=\frac{\Delta}{\Delta'}$ over all probability distributions $\mathcal{D}$? (If $\mathcal{D}$ is uniform, then $\rho=\frac{16}{17}$. Is there a distribution $\mathcal{D}$ such that $\rho<\frac{16}{17}$?)

Answer 1

Sorry, my computation in the comments was wrong. I think it leads to something with $\rho < \frac{16}{17}$.

Namely, let $\mathcal{D}$ be the distribution with $\mathrm{Pr}(\mathcal{D}=0)=\mathrm{Pr}(\mathcal{D}=3/4)=1/N$, and $\mathrm{Pr}(\mathcal{D}=1)=(N-2)/N$, where $N$ is large.

Then the possibilities for $(x,y,z)$ which fit your conditional probability are:

• $0 < t < \frac{3}{4}$: $(0,0,\frac{3}{4})$, $(0,0,1)$
• $\frac{3}{4} < t < 1$: $(0,0,1)$, $(0,0,1)$, $(0,\frac{3}{4},1)$, $(\frac{3}{4},0,1)$, $(\frac{3}{4},\frac{3}{4},1)$

Only one of these has $z\neq 1$; if $N$ is very large, then that case will occur much less frequently and we can "ignore" it (so we're really doing the limit $N\to \infty$ computation, for convenience).

Let $\delta=1-|x-y|$ and $\delta'=1-\min(|x-y|,|z-y|)$. Then the events to consider, and their probabilities and values, are

• $0 < t < \frac{3}{4}$: $(0,0,1)$ - relative prob. $\frac{3}{7}$, $\delta=\delta'=1$
• $\frac{3}{4} < t < 1$: $(0,0,1)$ - relative prob. $\frac{1}{7}$, $\delta=\delta'=1$
• $\frac{3}{4} < t < 1$: $(0,\frac{3}{4},1)$ - relative prob. $\frac{1}{7}$, $\delta=\frac{1}{4}$, $\delta'=\frac{3}{4}$
• $\frac{3}{4} < t < 1$: $(\frac{3}{4},0,1)$ - relative prob. $\frac{1}{7}$, $\delta=\delta'=\frac{1}{4}$
• $\frac{3}{4} < t < 1$: $(\frac{3}{4},\frac{3}{4},1)$ - relative prob. $\frac{1}{7}$, $\delta=\delta'=1$

So we can compute $$\Delta=\frac{3}{7}+\frac{1}{7}+\frac{1}{7}(\frac{1}{4})+\frac{1}{7}(\frac{1}{4})+\frac{1}{7}=\frac{11}{14}$$ $$\Delta'=\frac{3}{7}+\frac{1}{7}+\frac{1}{7}(\frac{3}{4})+\frac{1}{7}(\frac{1}{4})+\frac{1}{7}=\frac{12}{14}$$ $$\rho=\frac{\Delta}{\Delta'}=\frac{11}{12}< \frac{16}{17}$$

As mentioned, really we took the limit $N\to \infty$; but since we got $\rho< \frac{16}{17}$, that means there should be some finite $N$ we can take with $\rho< \frac{16}{17}$, just the computation will be more annoying.