# $E[X^T (Y-Z)] = E[X^T] E[Y-Z]$ but what about $E[(X^T (Y-Z))^2]$?

Let $$X, Y$$, and $$Z$$ be random vectors with $$X$$ independent of $$Y$$ and $$Z$$. Due to the independence we have $$E[X^T (Y-Z)] = E[X^T] E[Y-Z].$$ But what what $$E[(X^T (Y-Z))^2]$$? Is it possible to decompose this in a similar fashion so that we separate the $$X$$ and from the $$Y$$ and $$Z$$?

\begin{align} \mathbf{E} \left[ (X^T (Y-Z))^2 \right] &= \mathbf{E} \left[ X^T (Y-Z) (Y - Z)^T X \right] \\ &= \mathbf{E} \left[ \textbf{Tr} \left\{ X^T (Y-Z) (Y - Z)^T X \right\} \right] \\ &= \mathbf{E} \left[ \textbf{Tr} \left\{ X X^T (Y-Z) (Y - Z)^T \right\} \right] \\ &= \textbf{Tr} \left\{\mathbf{E} \left[ X X^T (Y-Z) (Y - Z)^T \right] \right\} \\ &= \textbf{Tr} \left\{\mathbf{E} \left[ X X^T \right] \mathbf{E}\left[(Y-Z) (Y - Z)^T \right] \right\} \end{align}
• I just asked a related question here. Basically is it possible to separate the $X$ from the $Y$ and $Z$ for higher powers. – Bertus101 Dec 4 '20 at 15:18