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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 $Z$?

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\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}

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  • $\begingroup$ Thats great thanks! $\endgroup$ – Bertus101 Dec 1 '20 at 16:02
  • $\begingroup$ I just asked a related question here. Basically is it possible to separate the $X$ from the $Y$ and $Z$ for higher powers. $\endgroup$ – Bertus101 Dec 4 '20 at 15:18

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