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Consider a random variable $x$ and $\theta$. Suppose $f(\theta)$ and $f(\theta|x)$ are given. Suppose, in addition, that $f(\theta|x)$ satisfies the strict monotone likelihood ratio property. That is, for every $\theta >\theta'$ and $x > x'$, we have $f(\theta'|x')f(\theta|x)>f(\theta|x')f(\theta'|x).$ Will they uniquely determine the joint distribution $f(x, \theta)$?

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No. For example, if $f(\theta|x)=f(\theta)$ for every $x$, the distribution of $X$ can be any distribution.

Edit to answer the revised version of the question There is no reason to believe the SMLR property + the distribution of $\Theta$ + the conditional distribution of $\Theta$ conditional on $X$ determine the joint distribution of $(\Theta,X)$ or, what is equivalent in your context, the distribution of $X$. If you have reasons to believe they do, you could explain why.

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Yes but I was in the middle of editing the question while you posted an answer. So please take a look at the question again and I will very much appreciate your comments. – Thales May 4 '11 at 13:30
To continue on my edit: More generally, browsing through the questions you asked on math.SE, I have never seen you show anything about what you had tried, what was your intuition, or where you got stuck. This runs contrary to the specific instructions given on this site and cannot improve the quality of the answers you receive here. – Did May 7 '11 at 13:55
So you did not even try with 2*2 case? – Thales May 8 '11 at 17:49

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