$\newcommand{\ep}{\varepsilon}$Let us write $n$ instead of $N$ and $k$ instead of $n$. For brevity, let us also write $X_i$ and $Y_i$ instead of $X_{n,i}$ and $Y_{n,i}$.

For any function $f$ of several variables, let \begin{equation*} E_n f(X,Y,\ldots):=\frac1n\sum_{i\in[n]}f(X_i,Y_i,\ldots), \end{equation*} where $[n]:=\{1,\dots,n\}$.

It is given that the $X_i$'s are permutation invariant, the $Y_i$'s are permutation invariant, the $Y_i$'s are independent of the $X_i$'s, \begin{equation*} E_n X^k\to a_k,\quad E_n Y^k\to b_k \tag{10}\label{10} \end{equation*} for $k=1,2,\dots$. Here and in what follows, the convergence is in probability as $n\to\infty$.

We have to show that then \begin{equation*} E_n XY\overset{\text{(?)}}\to a_1 b_1. \tag{20}\label{20} \end{equation*} Note that $E_n XY=E_n(X-a_1)Y+a_1E_n Y$ and $E_n(X-a_1)^k=\sum_{j=0}^k\binom kj (-a_1)^j E_nX^{k-j}$. So, replacing $X_i$ by $X_i-a_1$, we see that without loss of generality (wlog) $a_1=0$. Similarly, wlog $b_1=0$. So, \begin{equation*} a_1=b_1=0, \tag{25}\label{25} \end{equation*} and \eqref{20} becomes \begin{equation*} E_n XY\overset{\text{(?)}}\to 0. \tag{20a}\label{20a} \end{equation*}

For each small enough real $\ep>0$, let \begin{equation*} M:=M_\ep:=\ep^{-2/3}. \tag{27}\label{27} \end{equation*} Let \begin{equation*} \hat X_i:=X_i\,1(|X_i|\le M),\quad \check X_i:=X_i-\hat X_i=X_i\,1(|X_i|>M), \end{equation*} \begin{equation*} \hat Y_i:=Y_i\,1(|Y_i|\le M),\quad \check Y_i:=Y_i-\hat Y_i=Y_i\,1(|Y_i|>M). \end{equation*}

Note that \begin{equation*} P(|E_N XY|>4\ep)\le p_1+\cdots+p_4, \tag{30}\label{30} \end{equation*} where \begin{equation*} p_1:=P(|E_N \hat X\hat Y|>\ep),\quad p_2:=P(|E_N \hat X\check Y|>\ep), \end{equation*} \begin{equation*} p_3:=P(|E_N \check X\hat Y|>\ep),\quad p_4:=P(|E_N \check X\check Y|>\ep). \end{equation*} Next, \begin{equation*} |\check Y_i|\le Y_i^4/M^3, \quad |\check X_i|\le X_i^4/M^3, \tag{40}\label{40} \end{equation*} and hence \begin{equation*} p_2\le P(E_N|\check Y|>\ep/M)\le P(E_N Y^4>M^2\ep)\to0, \tag{50}\label{50} \end{equation*} by \eqref{10} with $k=4$. Similarly, \begin{equation*} p_3\to0. \tag{60}\label{60} \end{equation*} Next, in view of the Cauchy--Schwarz inequality and the inequalities \begin{equation*} \check X_i^2\le X_i^4/M^2, \quad \check Y_i^2\le Y_i^4/M^2, \tag{40a}\label{40a} \end{equation*} we get \begin{equation*} p_4\le P(E_N \check X^2\,E_N \check Y^2>\ep^2) \le P(E_N \check X^2>\ep)+P(E_N \check Y^2>\ep) \\ \le P(E_N X^4>M^2\ep)+P(E_N Y^4>M^2\ep) \to0. \tag{70}\label{70} \end{equation*} So, \eqref{20a} reduces to \begin{equation*} p_1\overset{\text{(?)}}\to0. \tag{20b}\label{20b} \end{equation*}

Note now that \begin{equation*} P(|E_n\hat X|>2\ep)\le P(|E_n X|>\ep)+P(|E_n\check X|>\ep) \\ \le P(|E_n X|>\ep)+P(E_n X^4>M^3\ep)\to0, \end{equation*} by \eqref{10}, \eqref{25}, \eqref{40}, and \eqref{27}.
So, $E_n\hat X\to0$. Also, $|E_n\hat X|\le M$. So, by dominated convergence, \begin{equation*} E(E_n\hat X)^2\to0. \tag{80}\label{80} \end{equation*} On the other hand, by the permutation-invariance, \begin{equation*} E(E_n\hat X)^2=\frac1{n^2}\sum_{i,j\in[n]}E\hat X_i\hat X_j \\ =\frac n{n^2}E\hat X_1^2+\frac{n^2-n}{n^2}E\hat X_1\hat X_2. \end{equation*} Therefore and in view of \eqref{80} and because $|\hat X_1|\le M$, we see that \begin{equation} E\hat X_1\hat X_2\to0. \tag{90}\label{90} \end{equation} So, recalling that the $Y_i$'s are independent of the $X_i$'s, we get \begin{equation*} E(E_n\hat X\hat Y)^2 =\frac1{n^2}\sum_{i,j\in[n]}E\hat X_i\hat X_j\,E\hat Y_i\hat Y_j \\ =\frac n{n^2}E\hat X_1^2\,E\hat Y_1^2 +\frac{n^2-n}{n^2}E\hat X_1\hat X_2\,E\hat Y_1\hat Y_2. \end{equation*} Therefore and in view of \eqref{90} and because $|\hat X_1|\le M$ and $|\hat Y_1|\le M$, we see that \begin{equation} E(E_n\hat X \hat Y)^2\to0. \end{equation} Now \eqref{20b} follows by Markov's inequality. $\quad\Box$

$\newcommand{\ep}{\varepsilon}$Let us write $n$ instead of $N$ and $k$ instead of $n$. For brevity, let us also write $X_i$ and $Y_i$ instead of $X_{n,i}$ and $Y_{n,i}$.

For any function $f$ of several variables, let \begin{equation*} E_n f(X,Y,\ldots):=\frac1n\sum_{i\in[n]}f(X_i,Y_i,\ldots), \end{equation*} where $[n]:=\{1,\dots,n\}$.

It is given that the $X_i$'s are permutation invariant, the $Y_i$'s are permutation invariant, the $Y_i$'s are independent of the $X_i$'s, \begin{equation*} E_n X^k\to a_k,\quad E_n Y^k\to b_k \tag{10}\label{10} \end{equation*} for $k=1,2,\dots$. Here and in what follows, the convergence is in probability as $n\to\infty$.

We have to show that then \begin{equation*} E_n XY\overset{\text{(?)}}\to a_1 b_1. \tag{20}\label{20} \end{equation*} Note that $E_n XY=E_n(X-a_1)Y+a_1E_n Y$ and $E_n(X-a_1)^k=\sum_{j=0}^k\binom kj (-a_1)^j E_nX^{k-j}$. So, replacing $X_i$ by $X_i-a_1$, we see that without loss of generality (wlog) $a_1=0$. Similarly, wlog $b_1=0$. So, \begin{equation*} a_1=b_1=0, \tag{25}\label{25} \end{equation*} and \eqref{20} becomes \begin{equation*} E_n XY\overset{\text{(?)}}\to 0. \tag{20a}\label{20a} \end{equation*}

For each small enough real $\ep>0$, let \begin{equation*} M:=M_\ep:=\ep^{-2/3}. \tag{27}\label{27} \end{equation*} Let \begin{equation*} \hat X_i:=X_i\,1(|X_i|\le M),\quad \check X_i:=X_i-\hat X_i=X_i\,1(|X_i|>M), \end{equation*} \begin{equation*} \hat Y_i:=Y_i\,1(|Y_i|\le M),\quad \check Y_i:=Y_i-\hat Y_i=Y_i\,1(|Y_i|>M). \end{equation*}

Note that \begin{equation*} P(|E_n XY|>4\ep)\le p_1+\cdots+p_4, \tag{30}\label{30} \end{equation*} where \begin{equation*} p_1:=P(|E_n \hat X\hat Y|>\ep),\quad p_2:=P(|E_n \hat X\check Y|>\ep), \end{equation*} \begin{equation*} p_3:=P(|E_n \check X\hat Y|>\ep),\quad p_4:=P(|E_n \check X\check Y|>\ep). \end{equation*} Next, \begin{equation*} |\check Y_i|\le Y_i^4/M^3, \quad |\check X_i|\le X_i^4/M^3, \tag{40}\label{40} \end{equation*} and hence \begin{equation*} p_2\le P(E_n|\check Y|>\ep/M)\le P(E_n Y^4>M^2\ep)\to0, \tag{50}\label{50} \end{equation*} by \eqref{10} with $k=4$. Similarly, \begin{equation*} p_3\to0. \tag{60}\label{60} \end{equation*} Next, in view of the Cauchy--Schwarz inequality and the inequalities \begin{equation*} \check X_i^2\le X_i^4/M^2, \quad \check Y_i^2\le Y_i^4/M^2, \tag{40a}\label{40a} \end{equation*} we get \begin{equation*} p_4\le P(E_n \check X^2\,E_n \check Y^2>\ep^2) \le P(E_n \check X^2>\ep)+P(E_n \check Y^2>\ep) \\ \le P(E_n X^4>M^2\ep)+P(E_n Y^4>M^2\ep) \to0. \tag{70}\label{70} \end{equation*} So, \eqref{20a} reduces to \begin{equation*} p_1\overset{\text{(?)}}\to0. \tag{20b}\label{20b} \end{equation*}

Note now that \begin{equation*} P(|E_n\hat X|>2\ep)\le P(|E_n X|>\ep)+P(|E_n\check X|>\ep) \\ \le P(|E_n X|>\ep)+P(E_n X^4>M^3\ep)\to0, \end{equation*} by \eqref{10}, \eqref{25}, \eqref{40}, and \eqref{27}.
So, $E_n\hat X\to0$. Also, $|E_n\hat X|\le M$. So, by dominated convergence, \begin{equation*} E(E_n\hat X)^2\to0. \tag{80}\label{80} \end{equation*} On the other hand, by the permutation-invariance, \begin{equation*} E(E_n\hat X)^2=\frac1{n^2}\sum_{i,j\in[n]}E\hat X_i\hat X_j \\ =\frac n{n^2}E\hat X_1^2+\frac{n^2-n}{n^2}E\hat X_1\hat X_2. \end{equation*} Therefore and in view of \eqref{80} and because $|\hat X_1|\le M$, we see that \begin{equation} E\hat X_1\hat X_2\to0. \tag{90}\label{90} \end{equation} So, recalling that the $Y_i$'s are independent of the $X_i$'s, we get \begin{equation*} E(E_n\hat X\hat Y)^2 =\frac1{n^2}\sum_{i,j\in[n]}E\hat X_i\hat X_j\,E\hat Y_i\hat Y_j \\ =\frac n{n^2}E\hat X_1^2\,E\hat Y_1^2 +\frac{n^2-n}{n^2}E\hat X_1\hat X_2\,E\hat Y_1\hat Y_2. \end{equation*} Therefore and in view of \eqref{90} and because $|\hat X_1|\le M$ and $|\hat Y_1|\le M$, we see that \begin{equation} E(E_n\hat X \hat Y)^2\to0. \end{equation} Now \eqref{20b} follows by Markov's inequality. $\quad\Box$

$\newcommand{\ep}{\varepsilon}$Let us write $n$ instead of $N$ and $k$ instead of $n$. For brevity, let us also write $X_i$ and $Y_i$ instead of $X_{n,i}$ and $Y_{n,i}$.

For any function $f$ of several variables, let \begin{equation*} E_n f(X,Y,\ldots):=\frac1n\sum_{i\in[n]}f(X_i,Y_i,\ldots), \end{equation*} where $[n]:=\{1,\dots,n\}$.

It is given that the $X_i$'s are permutation invariant, the $Y_i$'s are permutation invariant, the $Y_i$'s are independent of the $X_i$'s, \begin{equation*} E_n X^k\to a_k,\quad E_n Y^k\to b_k \tag{10}\label{10} \end{equation*} for $k=1,2,\dots$. Here and in what follows, the convergence is in probability as $n\to\infty$.

We have to show that then \begin{equation*} E_n XY\overset{\text{(?)}}\to a_1 b_1. \tag{20}\label{20} \end{equation*} Note that $E_n XY=E_n(X-a_1)Y+a_1E_n Y$ and $E_n(X-a_1)^k=\sum_{j=0}^k\binom kj (-a_1)^j E_nX^{k-j}$. So, replacing $X_i$ by $X_i-a_1$, we see that without loss of generality (wlog) $a_1=0$. Similarly, wlog $b_1=0$. So, \begin{equation*} a_1=b_1=0, \tag{25}\label{25} \end{equation*} and \eqref{20} becomes \begin{equation*} E_n XY\overset{\text{(?)}}\to 0. \tag{20a}\label{20a} \end{equation*}

For each small enough real $\ep>0$, let \begin{equation*} M:=M_\ep:=\ep^{-2/3}. \tag{27}\label{27} \end{equation*} Let \begin{equation*} \hat X_i:=X_i\,1(|X_i|\le M),\quad \check X_i:=X_i-\hat X_i=X_i\,1(|X_i|>M), \end{equation*} \begin{equation*} \hat Y_i:=Y_i\,1(|Y_i|\le M),\quad \check Y_i:=Y_i-\hat Y_i=Y_i\,1(|Y_i|>M). \end{equation*}

Note that \begin{equation*} P(|E_N XY|>4\ep)\le p_1+\cdots+p_4, \tag{30}\label{30} \end{equation*} where \begin{equation*} p_1:=P(|E_N \hat X\hat Y|>\ep),\quad p_2:=P(|E_N \hat X\check Y|>\ep), \end{equation*} \begin{equation*} p_3:=P(|E_N \check X\hat Y|>\ep),\quad p_4:=P(|E_N \check X\check Y|>\ep). \end{equation*} Next, \begin{equation*} |\check Y_i|\le Y_i^4/M^3, \quad |\check X_i|\le X_i^4/M^3, \tag{40}\label{40} \end{equation*} and hence \begin{equation*} p_2\le P(E_N|\check Y|>\ep/M)\le P(E_N Y^4>M^2\ep)\to0, \tag{50}\label{50} \end{equation*} by \eqref{10} with $k=4$. Similarly, \begin{equation*} p_3\to0. \tag{60}\label{60} \end{equation*} Next, in view of the Cauchy--Schwarz inequality and the inequalities \begin{equation*} \check X_i^2\le X_i^4/M^2, \quad \check Y_i^2\le Y_i^4/M^2, \tag{40a}\label{40a} \end{equation*} we get \begin{equation*} p_4\le P(E_N \check X^2\,E_N \check Y^2>\ep^2) \le P(E_N \check X^2>\ep)+P(E_N \check Y^2>\ep) \\ \le P(E_N X^4>M^2\ep)+P(E_N Y^4>M^2\ep) \to0. \tag{70}\label{70} \end{equation*} So, \eqref{20a} reduces to \begin{equation*} p_1\overset{\text{(?)}}\to0. \tag{20b}\label{20b} \end{equation*}

Note now that \begin{equation*} P(|E_n\hat X|>2\ep)\le P(|E_n X|>\ep)+P(|E_n\check X|>\ep) \\ \le P(|E_n X|>\ep)+P(E_n X^4>M^3\ep)\to0, \end{equation*} by \eqref{10}, \eqref{25}, \eqref{40}, and \eqref{27}.
So, $E_n\hat X\to0$. Also, $|E_n\hat X|\le M$. So, by dominated convergence, \begin{equation*} E(E_n\hat X)^2\to0. \tag{80}\label{80} \end{equation*} On the other hand, by the permutation-invariance, \begin{equation*} E(E_n\hat X)^2=\frac1{n^2}\sum_{i,j\in[n]}E\hat X_i\hat X_j \\ =\frac n{n^2}E\hat X_1^2+\frac{n^2-n}{n^2}E\hat X_1\hat X_2. \end{equation*} Therefore and in view of \eqref{80} and because $|\hat X_1|\le M$, we see that \begin{equation} E\hat X_1\hat X_2\to0. \tag{90}\label{90} \end{equation} So, recalling that the $Y_i$'s are independent of the $X_i$'s, we get \begin{equation*} E(E_n\hat X\hat Y)^2 =\frac1{n^2}\sum_{i,j\in[n]}E\hat X_i\hat X_j\,E\hat Y_i\hat Y_j \\ =\frac n{n^2}E\hat X_1^2\,E\hat Y_1^2 +\frac{n^2-n}{n^2}E\hat X_1\hat X_2\,E\hat Y_1\hat Y_2. \end{equation*} Therefore and in view of \eqref{90} and because $|\hat X_1|\le M$ and $|\hat Y_1|\le M$, we see that \begin{equation} E(E_n\hat X E_n\hat Y)^2\to0. \end{equation} Now \eqref{20b} follows by Markov's inequality. $\quad\Box$

$\newcommand{\ep}{\varepsilon}$Let us write $n$ instead of $N$ and $k$ instead of $n$. For brevity, let us also write $X_i$ and $Y_i$ instead of $X_{n,i}$ and $Y_{n,i}$.

For any function $f$ of several variables, let \begin{equation*} E_n f(X,Y,\ldots):=\frac1n\sum_{i\in[n]}f(X_i,Y_i,\ldots), \end{equation*} where $[n]:=\{1,\dots,n\}$.

It is given that the $X_i$'s are permutation invariant, the $Y_i$'s are permutation invariant, the $Y_i$'s are independent of the $X_i$'s, \begin{equation*} E_n X^k\to a_k,\quad E_n Y^k\to b_k \tag{10}\label{10} \end{equation*} for $k=1,2,\dots$. Here and in what follows, the convergence is in probability as $n\to\infty$.

We have to show that then \begin{equation*} E_n XY\overset{\text{(?)}}\to a_1 b_1. \tag{20}\label{20} \end{equation*} Note that $E_n XY=E_n(X-a_1)Y+a_1E_n Y$ and $E_n(X-a_1)^k=\sum_{j=0}^k\binom kj (-a_1)^j E_nX^{k-j}$. So, replacing $X_i$ by $X_i-a_1$, we see that without loss of generality (wlog) $a_1=0$. Similarly, wlog $b_1=0$. So, \begin{equation*} a_1=b_1=0, \tag{25}\label{25} \end{equation*} and \eqref{20} becomes \begin{equation*} E_n XY\overset{\text{(?)}}\to 0. \tag{20a}\label{20a} \end{equation*}

For each small enough real $\ep>0$, let \begin{equation*} M:=M_\ep:=\ep^{-2/3}. \tag{27}\label{27} \end{equation*} Let \begin{equation*} \hat X_i:=X_i\,1(|X_i|\le M),\quad \check X_i:=X_i-\hat X_i=X_i\,1(|X_i|>M), \end{equation*} \begin{equation*} \hat Y_i:=Y_i\,1(|Y_i|\le M),\quad \check Y_i:=Y_i-\hat Y_i=Y_i\,1(|Y_i|>M). \end{equation*}

Note that \begin{equation*} P(|E_N XY|>4\ep)\le p_1+\cdots+p_4, \tag{30}\label{30} \end{equation*} where \begin{equation*} p_1:=P(|E_N \hat X\hat Y|>\ep),\quad p_2:=P(|E_N \hat X\check Y|>\ep), \end{equation*} \begin{equation*} p_3:=P(|E_N \check X\hat Y|>\ep),\quad p_4:=P(|E_N \check X\check Y|>\ep). \end{equation*} Next, \begin{equation*} |\check Y_i|\le Y_i^4/M^3, \quad |\check X_i|\le X_i^4/M^3, \tag{40}\label{40} \end{equation*} and hence \begin{equation*} p_2\le P(E_N|\check Y|>\ep/M)\le P(E_N Y^4>M^2\ep)\to0, \tag{50}\label{50} \end{equation*} by \eqref{10} with $k=4$. Similarly, \begin{equation*} p_3\to0. \tag{60}\label{60} \end{equation*} Next, in view of the Cauchy--Schwarz inequality and the inequalities \begin{equation*} \check X_i^2\le X_i^4/M^2, \quad \check Y_i^2\le Y_i^4/M^2, \tag{40a}\label{40a} \end{equation*} we get \begin{equation*} p_4\le P(E_N \check X^2\,E_N \check Y^2>\ep^2) \le P(E_N \check X^2>\ep)+P(E_N \check Y^2>\ep) \\ \le P(E_N X^4>M^2\ep)+P(E_N Y^4>M^2\ep) \to0. \tag{70}\label{70} \end{equation*} So, \eqref{20a} reduces to \begin{equation*} p_1\overset{\text{(?)}}\to0. \tag{20b}\label{20b} \end{equation*}

Note now that \begin{equation*} P(|E_n\hat X|>2\ep)\le P(|E_n X|>\ep)+P(|E_n\check X|>\ep) \\ \le P(|E_n X|>\ep)+P(E_n X^4>M^3\ep)\to0, \end{equation*} by \eqref{10}, \eqref{25}, \eqref{40}, and \eqref{27}.
So, $E_n\hat X\to0$. Also, $|E_n\hat X|\le M$. So, by dominated convergence, \begin{equation*} E(E_n\hat X)^2\to0. \tag{80}\label{80} \end{equation*} On the other hand, by the permutation-invariance, \begin{equation*} E(E_n\hat X)^2=\frac1{n^2}\sum_{i,j\in[n]}E\hat X_i\hat X_j \\ =\frac n{n^2}E\hat X_1^2+\frac{n^2-n}{n^2}E\hat X_1\hat X_2. \end{equation*} Therefore and in view of \eqref{80} and because $|\hat X_1|\le M$, we see that \begin{equation} E\hat X_1\hat X_2\to0. \tag{90}\label{90} \end{equation} So, recalling that the $Y_i$'s are independent of the $X_i$'s, we get \begin{equation*} E(E_n\hat X\hat Y)^2 =\frac1{n^2}\sum_{i,j\in[n]}E\hat X_i\hat X_j\,E\hat Y_i\hat Y_j \\ =\frac n{n^2}E\hat X_1^2\,E\hat Y_1^2 +\frac{n^2-n}{n^2}E\hat X_1\hat X_2\,E\hat Y_1\hat Y_2. \end{equation*} Therefore and in view of \eqref{90} and because $|\hat X_1|\le M$ and $|\hat Y_1|\le M$, we see that \begin{equation} E(E_n\hat X \hat Y)^2\to0. \end{equation} Now \eqref{20b} follows by Markov's inequality. $\quad\Box$