# Extension of a monotone likelihood ratio property (revised)

Suppose the conditional distribution of $\theta$ and $x$ satisfies the monotone likelihood ratio property: for every real numbers $\bar{\theta} \ge \theta$ and $\bar{x} \ge x$, we have $f(\bar{\theta}|\bar{x})/f(\theta|\bar{x}) \ge f(\bar{\theta}|x)/f(\theta|x).$

Now suppose I extend this property as follows: for each real numbers $\bar{x} \ge x$ and each given $\bar{A}$ and $A$ that are subsets of [0,1] with the property that $E[\theta|\bar{x}, \theta \in \bar{A}] \ge E[\theta|\bar{x}, \theta \in A]$, we have $Pr(\bar{A}|\bar{x})/Pr(A|\bar{x}) \ge Pr(\bar{A}|x)/Pr(A|x).$

If $A$ and $\bar{A}$ are singleton, it is exactly the monotone likelihood ratio property. But I now want to require this property for sets $A$.

What additional condition on the joint distribution of $x$ and $\theta$ will provide this property?

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Presumably $A$ is a (measurable) subset of $[0,1]$. More importantly, your question is unclear to me: $E[\theta|x,\theta\in A]$ is always a random variable $Z$ measurable with respect to $x$, hence $Z=\phi(x,A)$ and $Z\in[0,1]$ almost surely. Are you asking when $Z$ depends on $A$ only? – Did Mar 26 '11 at 16:42
My previous comment applied to the first version of your question. I do not understand the second version. – Did Mar 26 '11 at 16:56
And I forgot to mention: my last statement extends to the title of your post. – Did Mar 26 '11 at 17:06
Thank you for your comments. I was editing my questions. Does it make sense to you now? Thank you very much. – Thales Mar 26 '11 at 17:17
See the answer below. – Did Mar 26 '11 at 18:49

## 1 Answer

(This is really a comment but seemed too long to be posted as one.) I understand the setting of (the third version of) your question as follows. One considers $[0,1]$-valued random variables $X$ and $T$. For every measurable subset $A$ of $[0,1]$, one considers measurable functions $f_A$ and $g_A$ defined on $[0,1]$ such that $$f_A(X)=E(T|X,T\in A)\quad\mbox{and}\quad g_A(X)=P(T\in A|X)\quad\mbox{almost surely}.$$ One says that $(X,T)$ satisfies property (P) if the following holds: for every $x$ in $[0,1]$ and every measurable subsets $A$ and $B$ of $[0,1]$ such that $f_A(x)\ge f_B(x)$, and for every $y\le x$ in $[0,1]$, $$\frac{g_A(x)}{g_B(x)}\ge\frac{g_A(y)}{g_B(y)}.$$ You ask how strong is property (P) and I do not know what you mean by that exactly. I think that, to get meaningful answers to your post, first you could confirm that indeed the previous paragraph describes the setting you are interested in and second, you could precise your question.

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OK, you modified the question again while I was writing this. However it seems that what I wrote still applies. (Except that I do not understand the last sentence of your post: what is $f$?) – Did Mar 26 '11 at 18:56
T is not a random variable. – Thales Mar 26 '11 at 19:14
Great. Since my $T$ stands for your $\theta$, you might wish to explain what is the joint distribution of $x$ and $\theta$ (last sentence of your post). – Did Mar 26 '11 at 19:38
I meant in your $f_A(x)$ is a function of x that is a realization of X. I meant what additional assumption on f will give this property. – Thales Mar 26 '11 at 22:25
@Thales Sorry but I do not understand your last comment. However, let me mention that my post ends with a question and with a request and that you did not answer the question nor fulfill the request. – Did Mar 26 '11 at 22:56