Actually, my question can be found at 
http://groups.google.com/group/sci.math/browse_thread/thread/e93e02b2004c2b04/c2f13ba350c80999?lnk=st&q=author%3Aroyliuk%40hotmail.com&rnum=1#c2f13ba350c80999

There are several replies there discussing this problem. However, this
problem was not completely solved. What I am interested in is to find
out whether we can say anything about the transition probability.

Please read the problem and all the previous discussion from the above link.
Thank you very much.

The question was answered in the thread. Apparently you don't like the answer.

But the thread only discusses the distribution of x or y. It does not
give the "transition probability" or the ratio of the transition
probability. Am I missing something?

//quote from the thread
If Y has a distribution with density f_Y, then X = k/Y has density 
f_X(x) = k/x^2 f_Y(k/x).  So if f_Y(a)/f_Y(b) -> 1 for every a and b, 
f_X(a)/f_X(b) -> b^2/a^2.
//

This actually computes the posterior probability f(x| f(x,y) = k), right?
Therefore f_X(a)/f_X(b) = f_X(x=a|f(x,y)= k) / f_X(x=b|f(x,y)= k) ->
b^2/a^2 tells us x may not be equally likely mapped to k.

But what does f(f(x,y)=k|x) imply?  What if f(f(x,y)=k|x=a) / f(f(x,y)=k|x=b) -->1?

Please don't use f(...) as a symbol to indicate probability - you're
only confusing the issue further. Indeed, I don't think that Robert
Israel should have used f_X and f_Y as symbols for the respective
distributions in the sci.math thread, since there's no connection
between these and the function f.

Be that as it may, the ratio given by Dr. Israel does tell us that x
is not (not "may not be") equally mapped to k, given that y is as
close to uniform as we can manage, with the associated caveats on what
we actually mean by saying something like that with continuous
variables.

I think what you're asking is what happens when we instead hold x
fixed and vary y. The answer is that it's exactly the same, since x
and y are symmetrical in the definition of f.

It may also be that you're asking about the form of the distribution
on y, if we insist that the function must be "equally likely" to map
two different values of x to k.

In this case we can reuse Dr. Israel's formula. In the region around
x=a, we get a density of k/a^2 f_Y(k/a), and around x=b we get k/b^2
f_Y(k/b). If we insist that these must match, we require f_Y(k/a) =
(a/b)^2 f_Y(k/b). For this to apply globally we would need f_Y(k/a) =
a^2 f_Y(k) for all a (setting b = 1) and thus f_Y(y) = (k/y)^2 f_Y(k).
Unfortunately, if you integrate this you get a divergent integral as
y->0. So no probability distribution on y can have this property
globally.