How do you take the expectation of a quadratic form with an extra random vector? Let $F$ be a $3 \times 3$ matrix, $B$ a $1 \times 3$ vector, and $x$ a $3 \times 1$ random vector. How would I go about finding the following whilst retaining vector notation.
$E[x'FxBx]$
 A: Let $\mu_k^{(n)} = \mathbb{E}(x_k^n)$, where $x_k$ is the $k^{th}$ entry of $x$, also $F_{ij}$ is the $(i,j)^{th}$ entry of $F$. Let $N$ be the sizes of all matrices and vectors ($N =3$ in your case)
\begin{equation}
\begin{split}
x^TFxBx 
&=
\Big(\sum_{i,j}F_{ij}x_ix_j\Big).\Big( Bx \Big)
\\&=
\Big(\sum_{i,j}F_{ij}x_ix_j\Big).\Big( \sum_{k} B_{k}x_k\Big)
\\&=
\Big(\sum_{i,j,k} F_{ij}B_{k} x_ix_jx_k\Big)
\end{split}
\end{equation}
Taking expectations we get 
\begin{equation}
\begin{split}
\mathbb{E} x^TFxBx
&=
\mathbb{E}\Big(\sum_{i,j,k} F_{ij}B_{k} x_ix_jx_k\Big)
\\&=
\Big(\sum_{i,j,k} F_{ij}B_{k} x_ix_jx_k\Big)
\end{split}
\end{equation}
Assuming $x_i$ and $x_j$ are independent for $i \neq j$. 
\begin{equation}
\begin{split}
\mathbb{E} x^TFxBx
&=
\sum_i \mu_i^{(3)} F_{ii} B_{i}
+
\Big(
\sum_{i \neq j} \mu_i^{(2)} \mu_j^{(1)} \big[  F_{ii}B_j +F_{ij}B_{i} + F_{ji}B_{i}\big]
\Big)
+
\sum_{i \neq j \neq k} \mu_i^{(1)} \mu_j^{(1)}\mu_k^{(1)}F_{ij}B_{k} 
\\&=
\sum_i \Big(\mu_i^{(3)}  - 3 \mu_i^{(2)} \mu_i^{(1)} + 2\big(\mu_i^{(1)}\big)^3 \Big) F_{ii} B_{i} 
\\&+
\sum_{i  j} \Big(\mu_i^{(2)} \mu_j^{(1)} - \big(\mu_i^{(1)}\big)^2\mu_j^{(1)}\Big)\big[  F_{ii}B_j +F_{ij}B_{i} + F_{ji}B_{i}\big]
\\&+
\sum_{i j k} \mu_i^{(1)} \mu_j^{(1)}\mu_k^{(1)}F_{ij}B_{k} 
\end{split}
\end{equation}
In vector notation

From here on $B$ is a column vector. Also, define
  \begin{align}
f &=[F_{11} \ldots F_{NN}]\\
\mathcal{F} &= \text{diag }(f)\\
m_n &= [\mu_1^{(n)} \ldots \mu_N^{(n)}]^T\\
\sigma &= m_2 - (m_1 \odot m_1)
\end{align}
  where $\odot$ is the Hadamard product (i.e. pointwise product).
Sorting and arranging we get:
  \begin{equation}
\begin{split}
\mathbb{E} x^TFxBx
&=\Big(m_3 -3(m_1\odot m_2)+2(m_1\odot m_1\odot m_1) \Big)^T\mathcal{F}B
\\&+ \Big( \sigma^Tf m_1^TB+  2\big[B \odot \sigma \big]^T F m_1  \Big)
\\&
+
m_1^TFm_1m_1^TB
\end{split}
\end{equation}

A: The best I can come up with is
$V.E(xx'X_d).\bar1$
Where,
$V=BF$
if $x=[x_1,x_2,x_3]^T$, then $X_d$ is a $3\times3$ matrix that has the entries of $x$ on the main diagonal, with the same order.
$\bar1=[1,1,1]^T$
You would probably need more conditions to further simplify it.
