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I actually know that the answer is $N(k-1)$ (where $k$ is the minimum between number of rows and number of columns).

However, I can not seem to find a simple proof for why the statistic is bounded by this. Any suggestions (or references?)

Tal Galili
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1 Answers1

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For some intuition about this, consider the square case ($k$ rows and columns), with $N=nk$. Then the maximal Chi Square occurs when all the marginal total are equal (in this case $n$), and the values in the table are $n$ along the diagonal and $0$ for the off diagonal, so that you have perfect association between the row and column variables. Then the Chi Square statistic is $$ \sum (O-E)^2/E = k\cdot(n-n/k)^2/(n/k)+k\cdot(k-1)\cdot(0-n/k)^2/(n/k)$$ where the first part represents the sum of the k diagonal elements and the second part is the sum of the off diagonal elements. You can show that this sum is $nk(k-1)=N(k-1)$. Similar reasoning extends to the case where the number of rows and columns are not the same.

vbm
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CDX
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    One might add that for different number of rows and columns, $k$ then must be the minimum of these two. – caracal Jul 19 '11 at 08:17
  • I tried solving this myself, gave up, and wrote a follow up question here: http://stats.stackexchange.com/questions/13721/when-is-the-maximum-achieved-for-the-chi-square-statistic-for-a-nonsymetric-table – Tal Galili Aug 01 '11 at 17:50