P(X = x | Y = y) = 0 for x < y.
... this example is for N = 3, using the same approach as here, http://www.mathhelpforum.com/math-he...on-169035.html.
The first stage gives the branches a weighting of 6, i.e. 3!, a common multiple of all the possible x's.
The second stage shares that weight between x possible y's.
So for example, P(X = 2 | Y = 2) = 3/5, while P(X = 3 | Y = 2) = 2/5. (Try to use the tree to see why that is.)
For P(X = x | Y = y) I guess you can then make a general numerator of
... and a denominator of
Hope this isn't too far off...
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