Probability distribution optimization problem of distances between points in $[0,1]$ 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 $\rho<\frac{16}{17}$?)

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.
