Let $(X,\geq)$ be a finite, linearly ordered set, and $\Delta(X)$ the set of distributions over $X$, that is, $p \in \Delta(X) \Longrightarrow p:X \to [0,1]$ such that $\sum_{x \in X}p(x)=1$.
For $p,q\in \Delta(X)$, $p$ is said to monotone-likelihood-ratio-dominate $q$, denoted $p\geq_{MLRP} q$ iff $\forall x'>x$, $p(x')q(x)\geq q(x')p(x)$.
- Is $(\Delta(X),\geq_{MLRP})$ a lattice? Is it a complete lattice?
A reference would also be very welcome.
For clarification: $(\Delta(X),\geq_{MLRP})$ is a partially ordered set, since:
(1) [Reflexive] $p \geq_{MLRP} p$ as $p(x')p(x)=p(x')p(x)$;
(2) [Antisymmetric] $p \geq_{MLRP} q \geq_{MLRP} p \Longrightarrow p=q$.
$p \geq_{MLRP} q \geq_{MLRP} p$ implies $p(x')q(x)= p(x)q(x') \forall x,x'$. Let $\tilde X= > \text{support}(p)\cup \text{support}(q)$ and let $x_n$ denote the $n$th smallest element of $\tilde X$ according to $\geq$. Then $\sum_{x\geq x_1}p(x)q(x_1)=q(x_1)=p(x_1)=\sum_{x\geq x_1}q(x)p(x_1)$. For arbitrary $n>1$, assume that $p(x_m)=q(x_m)$ for all $m<n$. Then $p(x_n)\sum_{x< x_n}q(x)=q(x_n)\sum_{x< x_n}p(x)$. Since $p(x_1)=q(x_1)>0$ and $x_n>x_1$, $\sum_{x< x_n}q(x)=\sum_{x< > x_n}p(x)>0$, and then $p(x_n)=q(x_n)$, and, by induction, $p=q$.
(3) [Transitive] $p \geq_{MLRP} q \geq_{MLRP} r \Longrightarrow p \geq_{MLRP} r$.
Supposing otherwise implies $\exists x'>x$ s.t. and $r(x')p(x)> > p(x')r(x)$. $p(x')r(x)<r(x')p(x) \Longrightarrow r(x'),p(x)>0$.
Suppose $q(x)+q(x')>0$. Then $q(x')\frac{r(x)}{r(x')}\geq q(x)$ and $q(x)\frac{p(x')}{p(x)}\geq q(x')$, which in turn implies $q(x)\frac{p(x')r(x)}{r(x')p(x)}\geq q(x)$, and so $q(x)=q(x')=0$, a contradiction.
This implies that for any $x'>x$ such that $q(x)+q(x')>0$, $p(x')r(x)\geq r(x')p(x)$.
Now suppose that for $x'>x$, $p(x')r(x)< r(x')p(x)$, and so $q(x)=q(x')=0$. Since $0=q(x')r(y)\geq r(x')q(y)$ $\forall y<x'$ and $r(x')>0$, then $q(y)=0, \forall y\leq x'$.
Let $x'':=\min\{y\in X| q(y)>0\} \Longrightarrow x''>x'$. $\forall > z\geq x''$, $p(z)q(x'')\geq q(z)p(x'') \Longrightarrow \sum_{z\geq > x''}p(z)q(x'')\geq p(x'') \Longrightarrow q(x'')>p(x'')\geq 0$. Moreover, $\forall z\leq x''$, $p(x'')q(z)\geq q(x'')p(z) > \Longrightarrow p(x'')q(x'')\geq q(x'')\sum_{z\leq x''}p(z) > \Longrightarrow p(x'')\geq \sum_{z\leq x''}p(z) \Longrightarrow p(z)=0 > \forall z<x'' \Longrightarrow p(x)=0$, a contradiction.
What the question is asking is:
(a) if $p\wedge q:=\sup\{r\in \Delta(X)| p\geq_{MLRP} r, q\geq_{MLRP} r\}$ and $p\vee q:=\inf\{r\in \Delta(X)| r\geq_{MLRP} p, r\geq_{MLRP} q\}$ exist; i.e. if meets, $p\wedge q$, and joins, $p \vee q$, are well-defined distributions; and
(b) if, for any $S \subseteq \Delta(X)$,
$\wedge S:=\sup\{r\in \Delta(X)| p\geq_{MLRP} r, \forall p \in S\}$ and
$\vee S:=\inf\{r\in \Delta(X)| r\geq_{MLRP} p, \forall p \in S\}$ exist.