# (From van der Vaart's Asymptotic Statistics, page 161, U-statistic) Why we can always replace the function $h$ with a symmetric one?

I'm reading the following Chapter from van der Vaart's Asymptotic Statistics, Section 12.1 page 161 (see the screenshot below). For the $$h$$ function that it mentioned, I have two questions regarding its permutation symmetry:

Question 1. I don't quite understand why we could always replace an asymmetric one with a symmetric one, and how do we do it? For example, suppose my $$h(X_1,X_2)=X_1^2+X_2$$, how to replace it with a new symmetric function $$h^*(\cdot,\cdot)$$ satisfying $$h^*(X_1,X_2)=h^*(X_2,X_1)$$).

Question 2. Related to the first question, I feel the highlighted sentence is a bit incomplete, and I feel the complete sentence should be something like: a given $$h$$ could always be replaced by a symmetric one without affecting a certain property. So I 'm wondering what property it is. I have guessed the three ones for its full meaning, and I'm wondering which one is correct:

Let $$U_h=\frac{1}{{n}\choose{r}}\underset{\beta}{\sum}h(X_{\beta_1},\dots,X_{\beta_r})$$, and $$h$$ is not necessarily symmetric.

Interpretation (1). If $$h$$ is not symmetric, we can always replace it with a symmetric $$h^*$$ such that $$U_{h^*}=U_{h}$$, i.e., they are always numerically equal in any sample;

Interpretation (2). If $$h$$ is not symmetric, we can always replace it with a symmetric $$h^*$$ such that $$U_{h^*}$$ is still unbiased for $$\theta$$, just like $$U_{h}$$;

Interpretation (3). If $$h$$ is not symmetric, we can always replace it with a symmetric $$h^*$$ such that $$U_{h^*}$$ and $$U_{h}$$ are asymptotically equivalent, in the sense that they are both consistent for $$\theta$$ and have exactly the same limiting distribution;

Thanks! Consider the $$r=2$$ case to simplify the notation. Because $$X_1$$ and $$X_2$$ are iid and thus exchangeable $$Eh(X_1,X_2)=Eh(X_2,X_1)$$ so the expectation is symmetric because of the exchangeability of $$X_1$$ and $$X_2$$ rather than because of any property of $$h$$. But since the expectation is symmetric, we can choose $$h$$ symmetric. Start with an arbitrary and and define $$h_s(x,y) =(h(x,y)+h(y,x))/2$$. By definition $$Eh_s(X_1,X_2)=(Eh(X_2,X_1)+Eh(X_1,X_2)/2$$ and by exchangeability this is $$2Eh(X_1,X_2)/2=Eh(X_1,X_2)=\theta$$

For your second part: none of the above.

$$U_h$$ is not the same, it's $$\theta$$ that's the same. What he's saying is that we are interested in $$\theta$$; that for defining $$\theta$$ there is no loss of generality in taking $$h$$ symmetric; and that if we take $$h$$ symmetric, we can define $$U_h$$ as the average over unordered partitions.

If we wanted to allow non-symmetric $$h$$ in the definition of $$U_h$$, we coud. We'd need to define $$U_h$$ as an average over ordered partitions. In the $$r=2$$ case above with a non-symmetric $$h$$ $$\frac{1}{n(n-1)} \sum_{i\neq j} h(X_i,X_j)=\frac{1}{n(n-1)} \sum_{i\neq j} h_s(X_i,X_j)= \frac{1}{n\choose 2}\sum_{i< j} h_s(X_i,X_j)$$

So, since $$U$$-statistics only give us symmetric $$\theta$$s, there's no loss of generality in symmetrising the $$h$$. Basically, the question is whether you demand symmetric $$h$$ in advance or construct the symmetrised $$h_s$$ inside your proofs.

If you wanted to study U-statistics on non-exchangeable data (eg, in my case, multistage samples) then there would be a loss of generality in taking $$h$$ to be symmetric.

• Thanks, this is very helpful! One question, the last long equation you wrote should be $$\frac{1}{n(n-1)} \sum_{i\neq j} h(X_i,X_j)=\frac{2}{n(n-1)} \sum_{i< j} h_s(X_i,X_j)= \frac{1}{n\choose 2}\sum_{i< j} h_s(X_i,X_j)$$, as we defined $h_s(X_i,X_j)$ as half the sum of a pair, right? Sep 6, 2020 at 6:49
• That would also be true, but I meant what I wrote. The first equality says that if you use all the pairs the ordering doesn't matter, and the second says that if you then have symmetry you can go to just the ordered pairs. Sep 6, 2020 at 7:04
• Thanks! I see, and agree the first equality is true. But I have difficulty in seeing the second equality, as the summation is identical, but the factor in front differ by 2, that is $\frac{1}{{n}\choose{2}}=\frac{2}{n(n-1)}$, you must half the summation for the third expression (by using $i<j$ instead of $i\neq j$), otherwise it won't equal to the second expression, right? Sep 6, 2020 at 7:18
• Sorry, yes. The last term should have $i<j$. Fixed. Sep 6, 2020 at 7:23
• Thanks! You really deepened my understanding of U-statistics! What I got in my problem is also an asymmetric h (summed over all possible permutations), but thanks to your explanation, I guess I know how to handle it now. Sep 6, 2020 at 7:28