Take two triangular arrays $X_{N,i}$ and $Y_{N,i}$ of random variables where $1 \le i \le N$. Suppose that the families $\{X_{N,i}\}$ and $\{Y_{N,i}\}$ are independent, and that the following conditions hold:

• Each row is permutation invariant, that is for $N$ fixed and any $\sigma \in S_N$ $$ (X_{N,1},\ldots,X_{N,N}) \stackrel{d}{=} (X_{N,\sigma (1)},\ldots,X_{N,\sigma(N)}) $$ $$ (Y_{N,1},\ldots,Y_{N,N}) \stackrel{d}{=} (Y_{N,\sigma (1)},\ldots,Y_{N,\sigma(N)}) $$
• There are finite constants $a_n,b_n$ such that the sample moments converge in probability $$ \frac1N \sum_{i=1}^N X_{N,i}^n \xrightarrow{\mathcal{P}} a_n $$ $$ \frac1N \sum_{i=1}^N Y_{N,i}^n \xrightarrow{\mathcal{P}} b_n $$

Then is it true that the sample covariance converges in probability to $a_1b_1$? $$ \frac1N \sum_{i=1}^N X_{N,i}Y_{N,i} \xrightarrow{\mathcal{P}} a_1b_1 \ \ ? $$ I make no assumptions about the dependence within the two families, only that they are independent from one another.

The motivation for the problem comes from the following observation. If the random variables have finite moments and if there were finite constants $a',b'$ such that $$ \mathbb{E}\left[ \frac1N \sum_{i=1}^N X_{N,i} \right] \to a' $$ $$ \mathbb{E}\left[ \frac1N \sum_{i=1}^N Y_{N,i} \right] \to b' $$ then by the permutation invariance of the families it follows that $$ \mathbb{E}\left[ \frac1N \sum_{i=1}^N X_{N,i}Y_{N,i} \right] = \frac1N \sum_{i=1}^N \mathbb{E}[X_{N,i}]\mathbb{E}[Y_{N,i}] $$ $$ = \frac1N \sum_{i=1}^N \mathbb{E}\left[\frac1N \sum_{j=1}^N X_{N,j}\right]\mathbb{E}\left[ \frac1N \sum_{k=1}^N Y_{N,k}\right] \to a'b' $$ I am asking if a similar phenomenon holds in the case where all quantities are described by convergence in probability rather than convergence of expectations.

Take two triangular arrays $X_{N,i}$ and $Y_{N,i}$ of random variables where $1 \le i \le N$. Suppose that the families $\{X_{N,i}\}$ and $\{Y_{N,i}\}$ are independent, and that the following conditions hold:

• Each row is permutation invariant, that is for $N$ fixed and any $\sigma \in S_N$ $$ (X_{N,1},\ldots,X_{N,N}) \stackrel{d}{=} (X_{N,\sigma (1)},\ldots,X_{N,\sigma(N)}) $$ $$ (Y_{N,1},\ldots,Y_{N,N}) \stackrel{d}{=} (Y_{N,\sigma (1)},\ldots,Y_{N,\sigma(N)}) $$
• There are finite constants $a_n,b_n$ such that the sample moments converge in probability $$ \frac1N \sum_{i=1}^N X_{N,i}^n \xrightarrow{\mathcal{P}} a_n $$ $$ \frac1N \sum_{i=1}^N Y_{N,i}^n \xrightarrow{\mathcal{P}} b_n $$

Then is it true that the sample covariance converges in probability to $a_1b_1$? $$ \frac1N \sum_{i=1}^N X_{N,i}Y_{N,i} \xrightarrow{\mathcal{P}} a_1b_1 \ \ ? $$ I make no assumptions about the dependence within the two families, only that they are independent from one another.

The motivation for the problem comes from the following observation. If the random variables have finite moments and if there were finite constants $a',b'$ such that $$ \mathbb{E}\left[ \frac1N \sum_{i=1}^N X_{N,i} \right] \to a' $$ $$ \mathbb{E}\left[ \frac1N \sum_{i=1}^N Y_{N,i} \right] \to b' $$ then by the permutation invariance of the families it follows that $$ \mathbb{E}\left[ \frac1N \sum_{i=1}^N X_{N,i}Y_{N,i} \right] = \frac1N \sum_{i=1}^N \mathbb{E}[X_{N,i}]\mathbb{E}[Y_{N,i}] $$ $$ = \frac1N \sum_{i=1}^N \mathbb{E}\left[\frac1N \sum_{j=1}^N X_{N,j}\right]\mathbb{E}\left[ \frac1N \sum_{k=1}^N Y_{N,k}\right] \to a'b' $$ I am asking if a similar phenomenon holds in the case where all quantities are described by convergence in probability rather than convergence of expectations.

Take two triangular arrays $X_{N,i}$ and $Y_{N,i}$ of random variables where $1 \le i \le N$. Suppose that the families $\{X_{N,i}\}$ and $\{Y_{N,i}\}$ are independent, and that the following conditions hold:

• Each row is permutation invariant, that is for $N$ fixed and any $\sigma \in S_N$ $$ (X_{N,1},\ldots,X_{N,N}) \stackrel{d}{=} (X_{N,\sigma (1)},\ldots,X_{N,\sigma(N)}) $$ $$ (Y_{N,1},\ldots,Y_{N,N}) \stackrel{d}{=} (Y_{N,\sigma (1)},\ldots,Y_{N,\sigma(N)}) $$
• There are finite constants $a_n,b_n$ such that the sample moments converge in probability $$ \frac1N \sum_{i=1}^N X_{N,i}^n \xrightarrow{\mathcal{P}} a_n $$ $$ \frac1N \sum_{i=1}^N Y_{N,i}^n \xrightarrow{\mathcal{P}} b_n $$

Then is it true that the sample covariance converges in probability to $a_1b_1$? $$ \frac1N \sum_{i=1}^N X_{N,i}Y_{N,i} \xrightarrow{\mathcal{P}} a_1b_1 $$ I make no assumptions about the dependence within the two families.

Take two triangular arrays $X_{N,i}$ and $Y_{N,i}$ of random variables where $1 \le i \le N$. Suppose that the families $\{X_{N,i}\}$ and $\{Y_{N,i}\}$ are independent, and that the following conditions hold:

• Each row is permutation invariant, that is for $N$ fixed and any $\sigma \in S_N$ $$ (X_{N,1},\ldots,X_{N,N}) \stackrel{d}{=} (X_{N,\sigma (1)},\ldots,X_{N,\sigma(N)}) $$ $$ (Y_{N,1},\ldots,Y_{N,N}) \stackrel{d}{=} (Y_{N,\sigma (1)},\ldots,Y_{N,\sigma(N)}) $$
• There are finite constants $a_n,b_n$ such that the sample moments converge in probability $$ \frac1N \sum_{i=1}^N X_{N,i}^n \xrightarrow{\mathcal{P}} a_n $$ $$ \frac1N \sum_{i=1}^N Y_{N,i}^n \xrightarrow{\mathcal{P}} b_n $$

Then is it true that the sample covariance converges in probability to $a_1b_1$? $$ \frac1N \sum_{i=1}^N X_{N,i}Y_{N,i} \xrightarrow{\mathcal{P}} a_1b_1 \ \ ? $$ I make no assumptions about the dependence within the two families, only that they are independent from one another.

The motivation for the problem comes from the following observation. If the random variables have finite moments and if there were finite constants $a',b'$ such that $$ \mathbb{E}\left[ \frac1N \sum_{i=1}^N X_{N,i} \right] \to a' $$ $$ \mathbb{E}\left[ \frac1N \sum_{i=1}^N Y_{N,i} \right] \to b' $$ then by the permutation invariance of the families it follows that $$ \mathbb{E}\left[ \frac1N \sum_{i=1}^N X_{N,i}Y_{N,i} \right] = \frac1N \sum_{i=1}^N \mathbb{E}[X_{N,i}]\mathbb{E}[Y_{N,i}] $$ $$ = \frac1N \sum_{i=1}^N \mathbb{E}\left[\frac1N \sum_{j=1}^N X_{N,j}\right]\mathbb{E}\left[ \frac1N \sum_{k=1}^N Y_{N,k}\right] \to a'b' $$ I am asking if a similar phenomenon holds in the case where all quantities are described by convergence in probability rather than convergence of expectations.

Take two triangular arrays $X_{N,i}$ and $Y_{N,i}$ of random variables where $1 \le i \le N$. Suppose that the families $\{X_{N,i}\}$ and $\{Y_{N,i}\}$ are independent, and that the following conditions hold:

• Each row is permutation invariant, that is for $N$ fixed and any $\sigma \in S_N$ $$ (X_{N,1},\ldots,X_{N,N}) \stackrel{d}{=} (X_{N,\sigma (1)},\ldots,X_{N,\sigma(N)}) $$ $$ (Y_{N,1},\ldots,Y_{N,N}) \stackrel{d}{=} (Y_{N,\sigma (1)},\ldots,Y_{N,\sigma(N)}) $$
• There are finite constants $a_n,b_n$ such that the sample moments converge in probability $$ \frac1N \sum_{i=1}^N X_{N,i}^n \xrightarrow{\mathcal{P}} a_n $$ $$ \frac1N \sum_{i=1}^N Y_{N,i}^n \xrightarrow{\mathcal{P}} b_n $$

Then is it true that the sample covariance converges in probability to $a_1b_1$? $$ \frac1N \sum_{i=1}^N X_{N,i}Y_{N,i} \xrightarrow{\mathcal{P}} a_1b_1 $$ I make no assumptions about the dependence within the two families.

I am trying to formulate the problem in $L^0(\Omega)$, over the space of measurable function with topology given by convergence in probability. The equivalent form in $L^\omega(\Omega) = \bigcap_{p\ge 1} L^p(\Omega)$, where convergence is with respect to all $\mathbb{E}[|\cdot|^n]$, is easy since $$ \mathbb{E}\left[ \frac1N \sum_{i=1}^N X_{N,i}Y_{N,i} \right] = \frac1N \sum_{i=1}^N \mathbb{E}\left[ \frac1N\sum_{j=1}^N X_{N,j} \right] \mathbb{E}\left[ \frac1N\sum_{k=1}^N Y_{N,k} \right] \to a_1b_1 $$

Take two triangular arrays $X_{N,i}$ and $Y_{N,i}$ of random variables where $1 \le i \le N$. Suppose that the families $\{X_{N,i}\}$ and $\{Y_{N,i}\}$ are independent, and that the following conditions hold:

• Each row is permutation invariant, that is for $N$ fixed and any $\sigma \in S_N$ $$ (X_{N,1},\ldots,X_{N,N}) \stackrel{d}{=} (X_{N,\sigma (1)},\ldots,X_{N,\sigma(N)}) $$ $$ (Y_{N,1},\ldots,Y_{N,N}) \stackrel{d}{=} (Y_{N,\sigma (1)},\ldots,Y_{N,\sigma(N)}) $$
• There are finite constants $a_n,b_n$ such that the sample moments converge in probability $$ \frac1N \sum_{i=1}^N X_{N,i}^n \xrightarrow{\mathcal{P}} a_n $$ $$ \frac1N \sum_{i=1}^N Y_{N,i}^n \xrightarrow{\mathcal{P}} b_n $$

Then is it true that the sample covariance converges in probability to $a_1b_1$? $$ \frac1N \sum_{i=1}^N X_{N,i}Y_{N,i} \xrightarrow{\mathcal{P}} a_1b_1 $$ I make no assumptions about the dependence within the two families.