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

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 $$p \ge 1$$).

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}

• Do you know if there are nice closed form expressions for the $P=3$ and $4$ cases? – Bertus101 Dec 4 '20 at 16:01
• This general result simplifies by means of multinomial coefficients. That reduces the expression to sums over integer partitions of $P.$ This rapidly gets more and more complicated as $P$ increases. – whuber Dec 4 '20 at 16:53
• The pattern for general $P$ is very interesting but my main concern is the $P=3$ and $4$ cases. I'm wondering are there 'nice' expressions for these cases like the $P=1$ and $2$ cases. – Bertus101 Dec 4 '20 at 19:40
• @Bertus101: Finite summations of finite products is already "closed form" (though not particulary simple looking in this case). – Ben Dec 4 '20 at 23:54
• I have made the assumption that the coordinates of $X$ and $Y$ are each iid, so $\mathbf{E} \left[ X^3 \right] = \mathbf{E} \left[ X_1^3 \right]$, etc. – πr8 Dec 5 '20 at 11:55