finding a conditional expectation It is an old exam problem about conditional expectation:
Let $(\xi_1,\xi_2)$ be a Gaussian vector with zero mean and covariance matrix A with $A_{11}=A_{22}=1, A_{12}=A_{21}=1/2.$ What is $E(\xi_1^2\xi_2|2\xi_1-\xi_2)$?
With the condition I know $E(\xi_1\xi_2)=1/2$ and $E(\xi_1^2)=E(\xi_2^2)=1$. And I tried to represent $\xi_1^2\xi_2$ with $2\xi_1-\xi_2$, like
$$
E(\xi_1^2\xi_2|2\xi_1-\xi_2)=\frac{1}{2}E(\xi_1\xi_2^2+\xi_1\xi_2(2\xi_1-\xi_2)|2\xi_1-\xi_2)
\\=\frac{1}{2}[E(\xi_1\xi_2^2|2\xi_1-\xi_2)+(2\xi_1-\xi_2)E(\xi_1\xi_2|2\xi_1-\xi_2)].
$$
However, I don't know how to continue. Is it a right path? Thanks for any help.
 A: Make the transformation $\eta_1 = \xi_1, \eta_2 = 2\xi_1 - \xi_2$, since 
$$\begin{pmatrix}
\xi_1 \\
\xi_2 \end{pmatrix} \sim \mathcal{N}\left(\begin{pmatrix} 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & 1\end{pmatrix}\right),$$
it follows that 
$$\begin{pmatrix}
\eta_1 \\
\eta_2 \end{pmatrix} \sim \mathcal{N}\left(\begin{pmatrix} 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 & \frac{3}{2} \\ \frac{3}{2} & 3\end{pmatrix}\right).$$
By the theorem of conditional distribution of Gaussian random vector, we have
$$\eta_1 \mid \eta_2 \sim \mathcal{N}\left(0 + \frac{3}{2}\times\frac{1}{3}(\eta_2 - 0), 1 - \frac{3}{2}\times\frac{1}{3}\times\frac{3}{2}\right) = \mathcal{N}\left(\frac{1}{2}\eta_2, \frac{1}{4}\right). \tag{1}$$
Now 
$$E(\xi_1^2 \xi_2 \mid 2\xi_1 - \xi_2) = E(\eta_1^2(2\eta_1 - \eta_2) \mid \eta_2) = 2E(\eta_1^3 \mid \eta_2) - \eta_2E(\eta_1^2 \mid \eta_2).\tag{2}$$
Can you proceed based on $(1)$ and $(2)$?
(After some algebra, you should get
\begin{align*}
& E(\eta_1^2 \mid \eta_2) = \frac{1}{4}(1 + \eta_2^2), \\
& E(\eta_1^3 \mid \eta_2) = \frac{1}{8}(3\eta_2 + \eta_2^3).
\end{align*}
You can get the final result by substituting $\eta_2$ by $\xi_1, \xi_2$.
)
