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I would like to find a generalization of the monotone likelihood ratio ordering that does not require that the probability distributions admit a density and have the same support, which would allow me to deal with degenerate probability distributions, and with the case where the supports are disjoint.

For instance, if $T_0$ is a degenerate probability distribution localized at $0$ and $T_1$ is a probability distribution with support included in $[1,+\infty)$, I would like the definition to guarantee that $T_1$ dominates $T_0$ according to this ordering. This seems to be a natural extension of the standard definition.

I would also like to the definition to boil down to the standard criterion if the distributions admit a density and have a common support.

Are there any existing generalizations of the monotone likelihood ratio order that would be appropriate for this purpose? Thanks!

Oliv
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2 Answers2

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A general definition is the following (see Def. 1.C.1 in Skahed and Shuntikumar, Stochastic Orders and Applications, 2007). Let $X$ and $Y$ be continuous [discrete] r.v.'s with densities [probability distributions] $f$ and $g$; then $Y \succeq X$ (in the likelihood ratio order) is $$f(u)g(v) \ge f(v)g(u)$$ for all $u \le v$. An equivalent formulation is to require that the ratio $$\frac{f(t)}{g(t)}$$ is decreasing over the union of supports of $X$ and $Y$ (interpreting $a/0$ as $\infty$ when $a > 0$).

mlc
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  • Thanks! But isn't this condition vacuous if $f$ and $g$ have disjoint supports? For instance, if $f$ has support $[1,+\infty)$ and $g$ has support $(-\infty,0]$, it seems to me that this definition implies both $Y \succeq X$ and $X \succeq Y$, whereas a more sensible ordering would have $X \succ Y$. – Oliv Apr 01 '17 at 09:47
  • @Oliv: take $u < 0 < 1 < v$ and you get $0<f(u)g(v) \not\le f(v)g(u) = 0$. So $X \not,\succeq Y$. – mlc Apr 01 '17 at 10:06
  • Why is it that $0<f(u)g(v)$ for $u<0$? Don't we have $f(u)=0$ since the support of $f$ is $[1,+\infty)$? – Oliv Apr 01 '17 at 10:10
  • Sorry: I switched $X$ and $Y$ (and their densities) in my comment. I should have written: take $u < 0 < 1 < v$ and you get $0<g(u)f(v) \not\le g(v)f(u) = 0$. So $Y \not,\succeq X$. – mlc Apr 01 '17 at 10:14
  • I see, thanks a lot. Also, I guess that the condition should be $f(u)g(v) \geq f(v)g(u)$ in your answer? – Oliv Apr 01 '17 at 10:18
  • Yes. Clearly, today I am not in my best shape (sorry). Will correct my answer. – mlc Apr 01 '17 at 10:19
  • Not in your best shape you mean :) Great answer, thanks a lot for your help! – Oliv Apr 01 '17 at 10:20
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If we take in full generality two probability measures $\mu , \nu$ on $\Bbb R$ we can define the likelihood ratio order as in Def. 1.C.1 in Stochastic Orders by Shaked and Shanthikumar (2007):

For two measurable sets $A,B \subset \Bbb R$ we write $A\leq B$, if $(x,y) \in A\times B$ implies $x\leq y$ for any such pair. We say $\mu$ is smaller than $\nu$ in the likelihood ratio order, if $$A \leq B \Rightarrow \mu (A) \nu (B) \geq \mu (B) \nu (A).$$

If $\mu , \nu$ have Lebesgue densities $f, g$, this definition is equivalent to $$f(u)g(v) \geq f(v) g(u) \qquad \text{for almost all } u\leq v.$$

Falrach
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