What does Cayley's hyperdeterminant of a 2x2x2 mixed-product moment tensor tell us about how two variables are related? Suppose we have a collection of random variables $S = \{ X_0, X_1 \}$ encoded into the $2 \times 2 \times 2$ tensor
$$\mathcal{C}[i, j, k] = \mathbb{E}[X_i X_j X_k]$$
where $X_i, X_j, X_k \in S$ and $i,j,k \in \{ 0,1 \}$.
Cayley's hyperdeterminant for $\mathcal{C}$ can be (verbosely) expanded to:
\begin{align}
\det (\mathcal{C}) &= (\mathbb{E}[X_0^3]^2\mathbb{E}[X_1^3]^2+ 3\mathbb{E}[X_0^2X_1]^2 \mathbb{E}[X_0X_1^2]^2) \\
&-2(3\mathbb{E}[X_0^3] \mathbb{E}[X_0^2X_1]\mathbb{E}[X_0X_1^2]\mathbb{E}[X_1^3] + 3\mathbb{E}[X_0^2X_1]^2\mathbb{E}[X_0X_1^2]^2) \\
&+4(\mathbb{E}[X_0^3]\mathbb{E}[X_0X_1^2]^3 + \mathbb{E}[X_0^2X_1]^3 \mathbb{E}[X_1^3])
\end{align}
What does Cayley's hyperdeterminant of a 2x2x2 mixed-product moment tensor tell us about how two $X_0$ and $X_1$ are related?
 A: A partial answer is to assume that $(X_0,X_1)$ has a joint distribution function is symmetric, and mean vector $\vec\mu = (0,0)$, then $\det \mathcal{C}=0$.
So for a given distribution we can construct a contrapositive argument that if $\lnot (\det \mathcal{C}=0)$ then the join distribution is not symmetric.
A: A partial answer relies on assuming statistical independence between $X_0$ and $X_1$. This entails that many of the expectations distribute across the products:

*

*$\mathbb{E}[X_0^2X_1] = \mathbb{E}[X_0^2]\mathbb{E}[X_1]$

*$\mathbb{E}[X_0X_1^2] = \mathbb{E}[X_0]\mathbb{E}[X_1^2]$
This suggests that we can infer certain forms of statistical independence using $\det \mathcal{C}$. If the assumption of statistical independence entails that $\det \mathcal{C} = \alpha$ where $\alpha$ is a constant, then we can use the contrapositive argument that $\lnot (\det \mathcal{C} = \alpha)$ implies that $X_0$ and $X_1$ are not independent.
This assumes that we have the correct probability model to quantify such an $\alpha$.
