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I posted a related question also to math.SE.

Basically, I would like to know how to show the following, which is part of this paper:

\begin{align} &\frac{1}{m} \mathbb E_{X,X'}\left[\sum_{i=1}^m\sum_{j=1}^m\left(k(x_i, x_j) + k(x_i', x_j') -k(x_i,x_j') - k(x_i',x_j)\right)\right]^\frac{1}{2} \\ \leq & \frac{1}{m}\left[2m\mathbb E_xk(x,x) + 2m(m-1)\mathbb E_{x,x'}k(x,x')-2m^2\mathbb E_{x,x'}k(x,x')\right]^\frac{1}{2} \end{align}

As part of this, I am asking:

How to proof that for samples $X,X'$ of size $m$ and with distribution $p$ we have that

\begin{align} &\mathbb E_{X,X'} \left[\sum_{i=1}^m\sum_{j=1}^mk(x_i,x_j) \right]\\ = & 2m\mathbb E_{x\sim p} [k(x,x)] \end{align}

where $x_i \in X, x_i'\in X'$ for all $i=1,\dots,m$, and $\mathbb E_{X,X'}$ denotes the expectation over all possible i.i.d. samples $X,X'$.

$k(\cdot,\cdot)$ is a (real-valued) kernel function, i.e., it is symmetric in its arguments and zero if both arguments are the same.

Thanks in advance!

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1 Answer 1

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Ok, the question is ill-posed. In fact, to follow the proof, it suffices to reorder the terms in the sum as to arrive at the proper solution (handling the diagonal-elements separately).

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    Dec 15, 2021 at 10:27

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