concrete example of degenracy I was reading  on degeneracy. Can someone give concrete example of following definition.
A kernel $h$ is symmetric if it is invariant under permutations of its inputs.  $S_m$ be symmetric group containing all permutations of $m$ elements. For any  $\sigma_1,\sigma_2 \in S_m$ then we have that $$h(x_{\sigma_1(1)},x_{\sigma_1(2)} , \dots ,x_{\sigma_1(m)}) = h(x_{\sigma_2(1)},x_{\sigma_2(2)} , \dots, x_{\sigma_2(m)})$$

Let $P$ be probability measure on $\mathcal{X}$. A symmetric kernel $h$ is P-degenerate of order $q-1$ , $1<q \leq m$ if for all $x_1,x_2,\dots,x_{q-1} \in \mathcal{X}$ following two conditions hold


*

*$\displaystyle \int h(x_1,\dots,x_m) dP^{m-q+1}(x_{q},\dots,x_{m}) = \int h(x_1,\dots,x_m) dP^{m}(x_{1},\dots,x_{m}) $

*$\displaystyle (x_1.\dots,x_{q}) \mapsto \int h(x_1,x_2,\dots,x_m) dP^{m-q}(x_{q+1},\dots,x_m) $ is not constant function

 A: The concept of "degeneracy" is context-dependent, and so it can be formulated in different ways depending on what object and purposes you are talking about.  Since you have not specified the context, and you have not specified what your "symmetric kernel" is, I'll try to give a general answer showing how you can examine the conditions and formulate and example.  Before constructing a concrete example, let's see if we can simplify this presentation to get an intuitive feel for what the conditions are saying.  For simplicity, we will use the index $k \equiv q-1$ to denote the order of the degeneracy ($0 \leqslant k < m$).
To facilitate this analysis, we can use the vector notation $\mathbf{x} = (x_1,...,x_m)$, $\mathbf{x}_{i} = (x_1,...,x_i)$ and $\mathbf{x}_{-i} = (x_{i+1},...,x_m)$ to denote the full vector and subvectors of initial and final values.  We also let $\mathcal{X}_i$ denote the support of $\mathbf{x}_{i}$.  Taking $\mathbf{x} \sim P$ we can translate the integrals into expectation notation as follows:$^\dagger$
$$\begin{align}
&(1) \quad \quad \quad 
\mathbb{E}(h(\mathbf{x})) = \mathbb{E}(h(\mathbf{y}_k, \mathbf{x}_{-k})) 
\quad \quad \quad \quad \quad \ 
\text{for all } \mathbf{y}_{k} \in \mathcal{X}_k, \\[12pt]
&(2) \quad \quad \quad 
\mathbb{E}(h(\mathbf{x})) \neq \mathbb{E}(h(\mathbf{y}_{k+1}, \mathbf{x}_{-(k+1)})) 
\quad \quad \quad 
\text{for some } \mathbf{y}_{k+1} \in \mathcal{X}^{k+1}. \\[12pt]
\end{align}$$
Intuitively, the first condition says that repacing the first $k$ values in the input does not change the expected value of the kernel output (which establishes degeneracy of order at least $k$), and the second condition says that this does not hold at any higher order.  Since we are applying this to symmetric kernels, we can replace reference to the first $k$ values with reference to any $k$ values.  Consequently, in the context of a symmetric kernel, degeneracy at order $k$ means that the expected value of the output is not affected if we change the values of up to $k$ of the inputs.
I cannot presently give you a concrete example because it is not clear from your question what your "symmetric kernel" $h$ is in this context.  (Specifically, what does the symmetry property require here.)  If you can clarify that aspect of your question I will try to add a concrete example of degeneracy of the kernel.

$^\dagger$ For the second condition I have used the fact that constancy of the conditional expectation implies that the conditional expectation is equal to the marginal expectation.
