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?


1 Answer 1


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.

  • $\begingroup$ Thank you! It all makes much more sense now! $\endgroup$ Sep 11 at 16:38

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