(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!

 A: 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.
