| dbp:proof
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- (en)
- WLOG, we reorder the data pairs, so that . By assumption of independence, the order of is a permutation sampled uniformly at random from , the permutation group on .
For each permutation, its unique inversion code is such that each is in the range . Sampling a permutation uniformly is equivalent to sampling a -inversion code uniformly, which is equivalent to sampling each uniformly and independently.
Then we have
The first term is just . The second term can be calculated by noting that is a uniform random variable on , so and , then using the sum of squares formula again. (en)
- Use a result from A class of statistics with asymptotically normal distribution Hoeffding . (en)
- Define the following quantities.
*
* is a point in .
In the notation, we see that the number of concordant pairs, , is equal to the number of that fall in the subset . That is, .
Thus,
Since each is an independent and identically distributed sample of the jointly normal distribution, the pairing does not matter, so each term in the summation is exactly the same, and so and it remains to calculate the probability. We perform this by repeated affine transforms.
First normalize by subtracting the mean and dividing the standard deviation. This does not change . This gives us where is sampled from the standard normal distribution on .
Thus, where the vector is still distributed as the standard normal distribution on . It remains to perform some unenlightening tedious matrix exponentiations and trigonometry, which can be skipped over.
Thus, iff where the subset on the right is a “squashed” version of two quadrants. Since the standard normal distribution is rotationally symmetric, we need only calculate the angle spanned by each squashed quadrant.
The first quadrant is the sector bounded by the two rays . It is transformed to the sector bounded by the two rays and . They respectively make angle with the horizontal and vertical axis, where
Together, the two transformed quadrants span an angle of , so and therefore (en)
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