The following problem resists my understanding, depending on the approach I get a different result:
Let $\mathbf{T}$ be a fully symmetric random tensor of order $3$ and size $N$. Its components can be represented as $T_{ijk}$ for all $1\leq i,j,k\leq N$. By symmetric I mean that if I permute any indices the value stays the same: \begin{equation} T_{ijk}=T_{ikj}=\dots=T_{jki} \end{equation}
Its components $T_{ijk}$ are following a normal distribution with mean $0$ and the following variance $\sigma_{ijk}^2$:
- $\sigma_{iij}^2=\text{Var}(T_{iij})=\langle T_{iij}^2\rangle=\langle T_{ijj}^2\rangle=\frac{1}{N^2}$ with ($i\neq j$)
- $\sigma_{ijk}^2=\text{Var}(T_{ijk})=\langle T_{ijk}^2\rangle=\dots=\langle T_{jki}^2\rangle=\frac{1}{2N^2}$ ($i\neq j \neq k$)
- $\sigma_{iii}^2=\text{Var}(T_{iii})=\langle T_{iii}^2\rangle=\frac{3}{N^2}$
This can be encapsulated by the following probability distribution:
$$f(\mathbf{T})=\int\left(\prod_{i \leq j \leq k} \mathrm{~d} T_{i j k}\right) \exp \left\{-\frac{1}{6} \sum_{i j k}^N T_{i j k}^{2}\right\}$$
- Now I define the matrix $\mathbf{M}$ where its components are given by
$$M_{ij}=\sum_{k}^N T_{ijk}c_k$$ Where $c_k\in\mathbb{R}$ and we impose $\sum_k c_k^2=1$.
I know that the variance of $M_{ij}$ will be equal to $\sum_k \sigma_{ijk}^2c_k^2$ and that as $N\to \infty$ the following relation holds: $$\sum_{j}^N M_{ij}^2\sim\sum_j\sum_{k}\sigma_{ijk}^2c_k^2$$
- Where I am stuck: What will the following sum converge to:
\begin{equation} \boxed{\sum_{\substack{j=1 \\ (j\neq k,i)}}^N\sum_{\substack{k=1 \\ (k\neq j,i)}}^NM_{ij}M_{ki}=?} \end{equation}
In the case of really symmetric random matrices this term will be 0, however here $M_{ij}$ is not independent from $M_{ki}$ although $i\neq j\neq k$. Indeed they both depend on $T_{ijk}$. Note that $T_{ijk}$ equals $T_{kij}$, due to the symmetry of the tensor: they are not different random variables with equal mean and variance. They are the same variable. However will this dependence be enough to contribute in this large sum?
Please let me know if you need any clarification or more details. Thank you