I am asked to prove that the Friedman statistic has two equivalent forms, namely:

$$ \Large \begin{aligned} S &= \frac{12n}{k(k+1)}\sum_{j=1}^k\left(R_{.j}-\frac{k+1}{2}\right)^2\\ &= \left[\frac{12}{nk(k+1)}\sum_{j=1}^kR_j^2\right]-3n(k+1) \end{aligned} $$

This is my attempt at a solution starting from LHS and trying to arrive at RHS. I feel like there are some cancellations of terms I should be aware of. Any help is appreciated. Thanks.

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  • $\begingroup$ Please explain what the "$R_j$" and "$R_{\cdot j}$" mean and what distinction is represented by the dot. $\endgroup$
    – whuber
    Oct 21, 2018 at 1:19
  • $\begingroup$ @whuber Rj is the sum of the ranks for each column while R.j = Rj / n (i.e. the sum of the ranks divided by the number of observations in that column). The dot distinguishes the sum (Rj) from the average (R.j) $\endgroup$
    – Jin Yu Li
    Oct 21, 2018 at 1:46

1 Answer 1


The last item should be

$$\frac {12n}{k(k+1)}\sum_{j=1}^k\frac {(k+1)^2}4=\frac {12n}{k(k+1)}k\frac {(k+1)^2}4 = 3n(k+1)$$ instead of $\frac {3n(k+1)}k$ as written on that yellow paper.

The middle item:

$$\frac {12}k\sum_{j=1}^kR_j = \frac {12}k\sum_{j=1}^k\sum_{i=1}^nr_{ij} = \frac {12}k\sum_{i=1}^n\sum_{j=1}^kr_{ij}=\frac {12}k\sum_{i=1}^n\frac {1+k}2k=\frac {12}kn\frac {1+k}2k = 6n(k+1)$$

Combine last two items, $$-6n(k+1)+3n(k+1)=-3n(k+1)$$


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