In an article, I recently came across the mention of first and second order U-type statistics without further detail.

Does anyone know what U-type statistics are?
References will be highly appreciated.

  • 5
    $\begingroup$ Do you mean in this sense? $\endgroup$ – whuber Jul 25 '12 at 19:42
  • $\begingroup$ Huber has found the Wikipedia description of U-statistics that come up in nonparametric statistics and were original found by Hoeffding. That is what I assumed you meant also when I saw the question. I don't think the term U-type is common though. $\endgroup$ – Michael R. Chernick Jul 25 '12 at 20:26
  • $\begingroup$ It looks like it. Any hint at what first and second order could be? $\endgroup$ – gui11aume Jul 25 '12 at 20:30
  • 2
    $\begingroup$ Probably averages over functions taking one argument vs. averages over all pairs for functions taking two arguments. A. van der Vaart's Asymptotic Statistics has a chapter that provides a lovely introduction to this topic. $\endgroup$ – cardinal Jul 25 '12 at 20:43
  • $\begingroup$ @cardinal yes that would make sense in the context. I will look it up. Thanks! $\endgroup$ – gui11aume Jul 25 '12 at 20:46

From the comments and the answer I got that "U-type statistics" is jargon for "U-statistics".

Here are a couple of elements taken from the reference provided by @cardinal, and in the previous answer. A U-statistics of degree or order $r$ is based on a permutation symmetric kernel function $h$ of arity $r$

$$ h(x_1, ..., x_r): \mathbb{X}^r \rightarrow \mathbb{R}, $$

and is the average of that function taken over all possible subsets of observations from the sample. More formally

$$ U = \frac{1}{\left( \array{n\\r} \right)} \sum_{\Pi_r(n)}h(x_{\pi_1}, ..., x_{\pi_r}), $$

where the sum is taken over $\Pi_r$, the set of all unordered subsets chosen from $\{1, ..., n\}$. The interest of U-statistics is that they are asymptotically Gaussian provided $E \{ h^2(X_1, ..., X_r) \} < \infty$.

Example 1: The sample mean is a first order U-statistics with $h(x) = x$.

Example 2: The signed rank statistic is a second order U-statistics with $h(x_1, x_2) = 1_{\mathbb{R}^+}(x_1+x_2)$ (the function that is equal to $1$ if $x_1 + x_2 > 0$, and $0$ otherwise).

$$ U = \frac{1}{\left( \array{n\\2} \right)} \sum_{i=1}^{n-1} \sum_{j=i+1}^n 1_{\mathbb{R}^+}(x_i+x_i) $$

is the sum of pairs $(x_i, x_j)$ from the sample with positive sum $x_i+x_j > 0$ and can be used as test statistic for investigating whether the distribution of the observations is located at 0.

Example 3: The unit definition space $\mathbb{X}$ of $h$ need not be real. Kendall's $\tau$ statistics is a second order U-statistics with $\frac{1}{2} h((x_1, y_1), (x_2, y_2)) = 1_{\mathbb{R}^+}((y_2-y_1)(x_2-x_1)) - 1$.

$$ \tau = \frac{2}{\left( \array{n\\2} \right)} \sum_{i=1}^{n-1} \sum_{j=i+1}^n 1_{\mathbb{R}^+}((y_2-y_1)(x_2-x_1)) - 1 $$

is a measure of dependence between $X$ and $Y$ and counts the number of concordant pairs $(x_i, y_i)$ and $(x_j, y_j)$ in the observations.

| cite | improve this answer | |

We have established that U-statistics are what the OP is looking for. I will address his second question about orer of U-statistics. The theory of U-statistics can be found in many books on nonparametrics and I am sure also in the various statistical encyclopedias. Here is a nice article by Tom Ferguson that summarizes the theory. I think it is actually a class tutorial on it. Here is what he says about order. The rest you can find in the paper

5. Degeneracy. When using U-statistics for testing hypotheses, it occasionally happens that at the null hypothesis, the asymptotic distribution has variance zero. This is a degenerate case, and we cannot use Theorem 2 to find approximate cutoff points. The general definition of degeneracy for a U-statistic of order $m$ and variances, $\sigma_1^2 \leq \sigma_2^2 \leq ... \leq \sigma_m^2$ given by (19) is as follows. Definition 3. We say that a U-statistic has a degeneracy of order $k$ if $\sigma_1^2 = · · · = \sigma_k^2 = 0$ and $\sigma^2_{k+1} > 0$.


| cite | improve this answer | |
  • $\begingroup$ @gui11aume Thanks for the nice editing job. I just fixed one thing (an extra 1 at the end after k+1). $\endgroup$ – Michael R. Chernick Jul 25 '12 at 21:19
  • 1
    $\begingroup$ @gu11aume. Do you know Tom Ferguson? He was a UCLA professor way back in the late 1970s when I was a graduate student. At Stanford would used his book in our graduate math stat course. It was a really good text and I think he writes very well. $\endgroup$ – Michael R. Chernick Jul 26 '12 at 18:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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