I'm following an equation from a published paper in order to calculate probabilities using a Markov Chain. The equation says:

Construct a set $U$ that consists of all items that appear in the top-$k$ in at least one list.

For each pair of items $i$ and $j$ in $U$, let the preference for $i$ over $j$, $m_i{_j}$, equal $1$ if the majority of the lists ($>=50$%) that rank both $i$ and $j$ rank $j$ above $i$ and $0$ otherwise. Let $m_i{_j} = m_j{_i} = 0.5$ if items $i$ and $j$ are never directly compared in any list.

My problem with the above is: $U$ is composed of the top-k items in at least one list. Wouldn't this mean that $m_i{_j} = 0.5$ would never be possible because if $U$ is composed of items that must all appear in one list, each item is directly compared in at least one list.

I just want someone to confirm my reading of this.

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It is hard to read without having the article, but I am reading it in the same way as you are. – utdiscant Jul 13 '12 at 10:35
@utdiscant the article gives no other relevant information it's exactly as I've typed it out. The article isn't about this specific equation, it's just one of the equations they are using for their proof – adohertyd Jul 13 '12 at 10:37
Maybe you could post a link to the article, or at least a title. – utdiscant Jul 13 '12 at 12:35
If there are at least two lists, the convention is necessary if i appears in some lists, j appears in some other lists and no list contains them both. – Did Jul 13 '12 at 13:39
@did after reading your point and going over my data again, with a pen and paper for more than an hour!, I can see that you are correct and that there are in fact a few instances of this state. Thanks for that! – adohertyd Jul 13 '12 at 15:07

If there are at least two lists, the convention is necessary if item $i$ appears in some lists, item $j$ appears in some other lists, and no list contains both items $i$ and $j$.