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I'm reading about Friedman's statistic from the 3rd edition of Nonparametric Statistical Methods (Hollander, Wolfe and Chicken). On page 302 they have an exercise problem:

  1. Show directly, or illustrate by means of an example, that the maximum value of $S$ is $S_{\text{max}} = n ( k − 1 )$ . For what configuration is this maximum achieved?

They define $S$ as follows:

$\frac{12}{nk(k+1)}\sum_{j=1}^k R_j^2 - 3n(k+1)$,

where $R_j$ is the sum of group $j$ ranks, there are $k$ groups and $n$ blocks. Looking at the definition of $S$, I need to maximize $\sum R_j^2$. My uneducated guess is, that the maximum is achieved when the ranks of each column sum to most extreme values. But doing the algebra with groups $Rj$ having only rank $j$ $k$ times or distributing the ranks so that they start from group one having ranks from 1 to $n$ and continue in order to group two up to group $k$, which has ranks from $(k-1)n+1$ to $kn$ does not give me $n(k-1)$. So there must be some statistical expertise at play in the result? Could someone clarify the idea of how to maximize $S$?

They tell that with a large number of blocks, $S$ follows "asymptotic chi-squared distribution with $(k-1)$ degrees of freedom." What does that mean and is there a connection to the proposed maximum value?

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    $\begingroup$ It's & Chicken, not $. $\endgroup$ Commented Dec 2, 2024 at 19:48

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The general form of the Friedman statistic using the ranks $R_1,...,R_k$ is:

$$Q(\mathbf{R}) = \frac{12n}{k(k+1)} \sum_{i=1}^k \bigg( R_i - \frac{k+1}{2} \bigg)^2.$$

The statistic is maximised when the ranks are $1,...,n$ (in any order), which gives the sums:

$$\begin{align} \sum_{i=1}^k R_i &= \frac{k(k+1)}{2}, \\[6pt] \sum_{i=1}^k R_i^2 &= \frac{k(k+1)(2k+1)}{6}. \\[6pt] \end{align}$$

This gives the maximised test statistic:

$$\begin{align} Q(\mathbf{R}) &= \frac{12n}{k(k+1)} \sum_{i=1}^k \bigg( R_i - \frac{k+1}{2} \bigg)^2 \\[6pt] &= \frac{12n}{k(k+1)} \bigg[ \sum_{i=1}^k R_i^2 - (k+1) \sum_{i=1}^k R_i + \frac{k(k+1)^2}{4} \bigg] \\[6pt] &= \frac{12n}{k(k+1)} \bigg[ \frac{k(k+1)(2k+1)}{6} - (k+1) \frac{k(k+1)}{2} + \frac{k(k+1)^2}{4} \bigg] \\[6pt] &= \frac{12n}{k(k+1)} \bigg[ \frac{k(k+1)(2k+1)}{6} - \frac{k(k+1)^2}{2} + \frac{k(k+1)^2}{4} \bigg] \\[6pt] &= \frac{n}{k} \bigg[ 2k(2k+1) - 6k(k+1) + 3k(k+1) \bigg] \\[6pt] &= \frac{n}{k} \bigg[ 4k^2 + 2k - 6k^2 - 6k + 3k^2 + 3k \bigg] \\[8pt] &= \frac{n}{k} (k^2 - k) \\[14pt] &= n(k-1). \\[6pt] \end{align}$$

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