I am trying to understand the the following paper <https://arxiv.org/pdf/1810.10971.pdf>, in particular Example 2:


If $ Y \sim N(0,1)$, the standard normal on $\mathbb{R}$, then   
  

$ \begin{align\*} \Big( \mathbb{E} \Big[ \frac{1}{m!} Y^{\otimes m }\Big]\Big)\_{m \geq 0 } = \exp\Big(\frac{1}{2} e\_1 \otimes e\_1 \Big) \in \prod\_{m \geq 0 } \mathbb{R}^{\otimes m} \end{align\*},$   
  
 where $e\_1$ is the unit basis vector of $\mathbb{R}$ and $\exp$ is defined in the following way:   
 \begin{align\*} \exp &: \prod\_{m \geq 0 } V^{\otimes m} \to \prod\_{m \geq 0 } V^{\otimes m},\\ exp(s) &:= \sum\_{m \geq 0} \frac{s^{\otimes m}}{m!} \end{align\*}


What does it mean to take a tensor product of random variables? In particular, what is $\mathbb{E} \Big[ Y^{ \otimes 3 }\Big]$, for example?



What is a tensor product of random variables?
In this context, I believe the tensor product on random variables is nothing other than the tensor product over the values of the RVs. (In other words, if $\Omega$ is a sample space and $X : \Omega \rightarrow V$ and $Y : \Omega \rightarrow W$ are RVs, then $X \otimes Y : \Omega \rightarrow V \otimes W$ is defined by $(X \otimes Y)(\omega) = X(\omega) \otimes Y(\omega)$.)


In Example 2, the RV $Y$ takes values in $\mathbb R$, which makes the tensor powers rather boring since $\mathbb R^{\otimes n} \simeq \mathbb R$ for any $n \geq 0$; we just have to take apart the notation to read out the usual (normalized) moments for the standard normal.


Thus we can expand (see Appendix A, for example)
$$
\exp \left ( \frac 12 e\_1 \otimes e\_1 \right ) = \left (1, 0, \frac 12, 0, \frac 1{2!} \left ( \frac12 \right )^2, 0, \ldots \right),
$$


noting that this notation gives the coefficients of $e\_1^{\otimes n}$ for $n \geq 0$.


Thus $\mathbb E [Y^{\otimes 3}] = 0 \in \mathbb R^{\otimes 3}$, since the standard normal has zero odd moments.


