Suppose we have $X_1, ..., X_n$ are iid N(0,1), and the rank is defined as below $$ r(X_i) = \sum_{i \neq j} I(X_j \geq X_i) $$ What is the expected rank?

I get that it is $(n-1)/2$ but Mood in "On the Asymptotic Efficiency of Certain Nonparametric Two-Sample Tests" states it to be (n+1)/2.

Am I missing something here?

  • 2
    $\begingroup$ As a clue the average of $0,1,2,\ldots,n-1$ has an average of $\frac{n-1}{2}$ while the average of $1,2,3,\ldots,n$ has an average of $\frac{n+1}{2}$ $\endgroup$ – Henry Jan 13 at 22:38

There is a very strong symmetry along the variables that we can exploit; because all variables are iid the probability that the rank is uniformy distributed among the possibilities.

There are $n$ possible ranks, between $0$ and $n-1$. Hence the average rank is $E[r(X_i)] = \sum_{i=0}^{n-1} \frac{i}{n} = \frac{n(n-1)}{2n} = \frac{n-1}{2}$

  • $\begingroup$ Could you explain why the following paper states it as $(n+1)/2$ researchgate.net/publication/… $\endgroup$ – math111 Jan 13 at 20:12
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    $\begingroup$ @math111 in Mood's original paper a different definition of rank is used, which results in a rank between $1$ and $n$ (section 4 here projecteuclid.org/download/pdf_1/euclid.aoms/1177728719). I assume in your paper they are using this other definition $\endgroup$ – Jsevillamol Jan 13 at 20:45
  • $\begingroup$ Where does he give the definition of rank? $\endgroup$ – math111 Jan 13 at 21:13
  • $\begingroup$ @math111 that section 4 has $m$ and $n$ observations "ranked from $1$ to $m+n$" $\endgroup$ – Henry Jan 13 at 22:36

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