V-statistics

A few days ago I wrote about U-statistics, statistics which can be expressed as the average of a symmetric function over all combinations of elements of a set. V-statistics can be written as an average of over all products of elements of a set.

Let S be a statistical sample of size n and let h be a symmetric function of r elements. The average of h over all subsets of S with r elements is a U-statistic. The average of h over the Cartesian product of S with itself r times

\underbrace{S \times S \times \cdots \times S}_{n \text{ times}}

is a V-statistic.

As in the previous post, let h(x, y) = (xy)²/2. We can illustrate the V-statistic associated with h with Python code as before.

    import numpy as np
    from itertools import product

    def var(xs):
        n = len(xs)
        h = lambda x, y: (x - y)**2/2
        return sum(h(*c) for c in product(xs, repeat=2)) / n**2

    xs = np.array([2, 3, 5, 7, 11])
    print(np.var(xs))
    print(var(xs))

This time, however, we iterate over product rather than over combinations. Note also that at the bottom of the code we print

   np.var(xs)

rather than

   np.var(xs, ddof=1)

This means our code here is computing the population variance, not the sample variance. We could make this more explicit by supplying the default value of ddof.

   np.var(xs, ddof=0)

The point of V-statistics is not to calculate them as above, but that they could be calculated as above. Knowing that a statistic is an average of a symmetric function is theoretically advantageous, but computing a statistic this way would be inefficient.

U-statistics are averages of a function h over all subsamples of S of size r without replacement. V-statistics are averages of h over all subsamples of size r with replacement. The difference between sampling with or without replacement goes away as n increases, and so V-statistics have the same asymptotic properties as U-statistics.

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