Separating $X$ from $Y$ in $E[(X^T Y))^p]$ for $p = 3$ and $4$?

Question

Let $X$ and $Y$ be random vectors with $X$ independent of $Y$. When dealing with terms of the form $E[(X^T Y)^p]$ for $p \ge 1$, it is very useful to be able to separate the $X$ from the $Y$.

For $p=1,2$, we have $$ \begin{align} E[X^T Y] &= E[X^T] E[Y], \\ E[(X^T Y)^2] &= Tr \left\{E \left[ X X^T \right] E\left[YY^T \right] \right\}. \end{align} $$ The second expression was proven here. I am wondering it is possible to obtain similar expressions for the cases $p=3$ and $p=4$ as some applications require us to go to fourth order: $$ \begin{align} E[(X^T Y)^3] &= \quad ? \\ E[(X^T Y)^4] &= \quad ? \\ \end{align} $$ Does anyone know if it is possible to derive expressions that separate the $X$ from $Y$ in these cases? (Maybe a pattern will then become clear for all $p \ge 1$).

Answer 1

The answer is that for nonnegative integer values $P$, it is in principle possible, but gets progressively more complicated as $P$ increases. To this end, one can write

\begin{align} \mathbf{E} \left[ \left( X^T Y \right)^P\right] &= \mathbf{E} \left[ \left( \sum_{i = 1}^D X_i Y_i \right)^P \right] \\ &= \mathbf{E} \left[ \sum_{i_1 = 1}^D \sum_{i_2 = 1}^D \cdots \sum_{i_P = 1}^D \prod_{p = 1}^P X_{i_p} Y_{i_p} \right] \\ &= \sum_{i_1 = 1}^D \sum_{i_2 = 1}^D \cdots \sum_{i_P = 1}^D \mathbf{E} \left[ \prod_{p = 1}^P X_{i_p} Y_{i_p} \right] \\ &= \sum_{i_1 = 1}^D \sum_{i_2 = 1}^D \cdots \sum_{i_P = 1}^D \mathbf{E} \left[ \prod_{p = 1}^P X_{i_p} \right] \cdot \mathbf{E} \left[ \prod_{p = 1}^P Y_{i_p} \right]. \end{align}

However, the expressions like $\prod_{p = 1}^P X_{i_p}$ become more challenging to characterize concisely in closed form as $P$ grows.

For a concrete example, consider when $P = 3$, and assume that the coordinates of $X$ and $Y$ are each iid. The above expression can then be simplified to

\begin{align} \sum_{i_1 = 1}^D \sum_{i_2 = 1}^D \sum_{i_3 = 1}^D \mathbf{E} \left[ \prod_{p = 1}^3 X_{i_p} \right] \cdot \mathbf{E} \left[ \prod_{p = 1}^3 Y_{i_p} \right] &= D \cdot \mathbf{E} \left[ X^3 \right] \cdot \mathbf{E} \left[ Y^3 \right] \\ &+ 3D(D - 1) \cdot \mathbf{E} \left[ X \right] \cdot \mathbf{E} \left[ X^2 \right] \cdot \mathbf{E} \left[ Y \right] \cdot \mathbf{E} \left[ Y^2 \right] \\ &+ D(D-1)(D-2) \cdot \mathbf{E} \left[ X \right] ^3 \cdot \mathbf{E} \left[ Y \right]^3. \end{align}