0
$\begingroup$

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.

enter image description here

Improve this question
$\endgroup$
2
  • $\begingroup$ Please explain what the "$R_j$" and "$R_{\cdot j}$" mean and what distinction is represented by the dot. $\endgroup$
    – whuber ♦
    Commented 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
    Commented Oct 21, 2018 at 1:46

1 Answer 1

Reset to default
1
$\begingroup$

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, $$-6n(k+1)+3n(k+1)=-3n(k+1)$$

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.