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Comment on your attempt: the idea looks great but unfortunately $\xi - \eta \perp \xi + \eta$ of course does not imply $\xi - 2\eta \perp \xi + 2\eta$. However, the "product-to-sum" identity of $\xi\eta$ is still very useful in approaching this problem. From there, you only need to apply some very basic properties of conditional expectation and the multivariate normal distribution to get the job done.

By the linearity and the "pulling out known factors" property of conditional expectation, \begin{align} E[\xi\eta|\xi - 2\eta] = \frac{1}{8}E[(\xi + 2\eta)^2|\xi - 2\eta] + \frac{1}{8}(\xi - 2\eta)^2. \tag{1} \end{align}

So it remains to evaluate the $E[(\xi + 2\eta)^2|\xi - 2\eta]$, which is tractable thanks to $(\xi, \eta) \sim N_2(0, I_{(2)})$. Because of it, it follows by the affine transformation property of the multivariate normal distribution that \begin{align} \begin{bmatrix} \xi + 2\eta \\ \xi - 2\eta \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} \xi \\ \eta \end{bmatrix} \sim N_2\left(\begin{bmatrix} 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 5 & -3 \\ -3 & 5 \end{bmatrix} \right), \end{align} which implies, by the conditional distribution of the multivariate normal distribution, that \begin{align} & E[\xi + 2\eta | \xi - 2\eta] = -\frac{3}{5}(\xi - 2\eta), \\ & \operatorname{Var}(\xi + 2\eta | \xi - 2\eta) = 5 - 9 \times \frac{1}{5} = \frac{16}{5}, \end{align} whence \begin{align} E[(\xi + 2\eta)^2|\xi - 2\eta] &= \operatorname{Var}(\xi + 2\eta | \xi - 2\eta) + (E[\xi + 2\eta | \xi - 2\eta])^2 \\ &= \frac{16}{5} + \frac{9}{25}(\xi - 2\eta)^2. \tag{2} \end{align} Substituting $(2)$ into $(1)$ gives \begin{align} E[\xi\eta|\xi - 2\eta] = \frac{2}{5} + \frac{9}{200}(\xi - 2\eta)^2 + \frac{1}{8}(\xi - 2\eta)^2 = \frac{2}{5} + \frac{17}{100}(\xi - 2\eta)^2. \end{align}

Now, to get the hang of the key operations in solving this problem, try resolving it using the decomposition \begin{align} \xi\eta = (\xi - 2\eta + 2\eta)\eta = \eta(\xi - 2\eta) + 2\eta^2. \end{align}

Comment on your attempt: the idea looks great but unfortunately $\xi - \eta \perp \xi + \eta$ of course does not imply $\xi - 2\eta \perp \xi + 2\eta$. However, the "product-to-sum" identity of $\xi\eta$ is still very useful in approaching this problem. From there, you only need to apply some very basic properties of conditional expectation and the multivariate normal distribution to get the job done.

By the linearity and the "pulling out known factors" property of conditional expectation, \begin{align} E[\xi\eta|\xi - 2\eta] = \frac{1}{8}E[(\xi + 2\eta)^2|\xi - 2\eta] -\frac{1}{8}(\xi - 2\eta)^2. \tag{1} \end{align}

So it remains to evaluate the $E[(\xi + 2\eta)^2|\xi - 2\eta]$, which is tractable thanks to $(\xi, \eta) \sim N_2(0, I_{(2)})$. Because of it, it follows by the affine transformation property of the multivariate normal distribution that \begin{align} \begin{bmatrix} \xi + 2\eta \\ \xi - 2\eta \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} \xi \\ \eta \end{bmatrix} \sim N_2\left(\begin{bmatrix} 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 5 & -3 \\ -3 & 5 \end{bmatrix} \right), \end{align} which implies, by the conditional distribution of the multivariate normal distribution, that \begin{align} & E[\xi + 2\eta | \xi - 2\eta] = -\frac{3}{5}(\xi - 2\eta), \\ & \operatorname{Var}(\xi + 2\eta | \xi - 2\eta) = 5 - 9 \times \frac{1}{5} = \frac{16}{5}, \end{align} whence \begin{align} E[(\xi + 2\eta)^2|\xi - 2\eta] &= \operatorname{Var}(\xi + 2\eta | \xi - 2\eta) + (E[\xi + 2\eta | \xi - 2\eta])^2 \\ &= \frac{16}{5} + \frac{9}{25}(\xi - 2\eta)^2. \tag{2} \end{align} Substituting $(2)$ into $(1)$ gives \begin{align} E[\xi\eta|\xi - 2\eta] = \frac{2}{5} + \frac{9}{200}(\xi - 2\eta)^2 - \frac{1}{8}(\xi - 2\eta)^2 = \frac{2}{5} - \frac{2}{25}(\xi - 2\eta)^2. \end{align}

Now, to get the hang of the key operations in solving this problem, try resolving it using the decomposition \begin{align} \xi\eta = (\xi - 2\eta + 2\eta)\eta = \eta(\xi - 2\eta) + 2\eta^2. \end{align}

Comment on your attempt: the idea looks great but unfortunately $\xi - \eta \perp \xi + \eta$ of course does not imply $\xi - 2\eta \perp \xi + 2\eta$. However, the "product-to-sum" identity of $\xi\eta$ is still very useful in approaching this problem. From there, you only need to apply some very basic properties of conditional expectation and the multivariate normal distribution to get the job done.

By the linearity and the "pulling out known factors" property of conditional expectation, \begin{align} E[\xi\eta|\xi - 2\eta] = \frac{1}{8}E[(\xi + 2\eta)^2|\xi - 2\eta] + \frac{1}{8}(\xi - 2\eta)^2. \tag{1} \end{align}

So it remains to evaluate the $E[(\xi + 2\eta)^2|\xi - 2\eta]$, which is tractable thanks to $(\xi, \eta) \sim N_2(0, I_{(2)})$. Because of it, it follows by the affine transformation property of the multivariate normal distribution that \begin{align} \begin{bmatrix} \xi + 2\eta \\ \xi - 2\eta \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} \xi \\ \eta \end{bmatrix} \sim N_2\left(\begin{bmatrix} 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 5 & -3 \\ -3 & 5 \end{bmatrix} \right), \end{align} which implies, by the conditional distribution of the multivariable normal distribution, that \begin{align} & E[\xi + 2\eta | \xi - 2\eta] = -\frac{3}{5}(\xi - 2\eta), \\ & \operatorname{Var}(\xi + 2\eta | \xi - 2\eta) = 5 - 9 \times \frac{1}{5} = \frac{16}{5}, \end{align} whence \begin{align} E[(\xi + 2\eta)^2|\xi - 2\eta] &= \operatorname{Var}(\xi + 2\eta | \xi - 2\eta) + (E[\xi + 2\eta | \xi - 2\eta])^2 \\ &= \frac{16}{5} + \frac{9}{25}(\xi - 2\eta)^2. \tag{2} \end{align} Substituting $(2)$ into $(1)$ gives \begin{align} E[\xi\eta|\xi - 2\eta] = \frac{2}{5} + \frac{9}{200}(\xi - 2\eta)^2 + \frac{1}{8}(\xi - 2\eta)^2 = \frac{2}{5} + \frac{17}{100}(\xi - 2\eta)^2. \end{align}

Now, to get the hang of the key operations in solving this problem, try resolving it using the decomposition \begin{align} \xi\eta = (\xi - 2\eta + 2\eta)\eta = \eta(\xi - 2\eta) + 2\eta^2. \end{align}

Comment on your attempt: the idea looks great but unfortunately $\xi - \eta \perp \xi + \eta$ of course does not imply $\xi - 2\eta \perp \xi + 2\eta$. However, the "product-to-sum" identity of $\xi\eta$ is still very useful in approaching this problem. From there, you only need to apply some very basic properties of conditional expectation and the multivariate normal distribution to get the job done.

By the linearity and the "pulling out known factors" property of conditional expectation, \begin{align} E[\xi\eta|\xi - 2\eta] = \frac{1}{8}E[(\xi + 2\eta)^2|\xi - 2\eta] + \frac{1}{8}(\xi - 2\eta)^2. \tag{1} \end{align}

So it remains to evaluate the $E[(\xi + 2\eta)^2|\xi - 2\eta]$, which is tractable thanks to $(\xi, \eta) \sim N_2(0, I_{(2)})$. Because of it, it follows by the affine transformation property of the multivariate normal distribution that \begin{align} \begin{bmatrix} \xi + 2\eta \\ \xi - 2\eta \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} \xi \\ \eta \end{bmatrix} \sim N_2\left(\begin{bmatrix} 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 5 & -3 \\ -3 & 5 \end{bmatrix} \right), \end{align} which implies, by the conditional distribution of the multivariate normal distribution, that \begin{align} & E[\xi + 2\eta | \xi - 2\eta] = -\frac{3}{5}(\xi - 2\eta), \\ & \operatorname{Var}(\xi + 2\eta | \xi - 2\eta) = 5 - 9 \times \frac{1}{5} = \frac{16}{5}, \end{align} whence \begin{align} E[(\xi + 2\eta)^2|\xi - 2\eta] &= \operatorname{Var}(\xi + 2\eta | \xi - 2\eta) + (E[\xi + 2\eta | \xi - 2\eta])^2 \\ &= \frac{16}{5} + \frac{9}{25}(\xi - 2\eta)^2. \tag{2} \end{align} Substituting $(2)$ into $(1)$ gives \begin{align} E[\xi\eta|\xi - 2\eta] = \frac{2}{5} + \frac{9}{200}(\xi - 2\eta)^2 + \frac{1}{8}(\xi - 2\eta)^2 = \frac{2}{5} + \frac{17}{100}(\xi - 2\eta)^2. \end{align}

Now, to get the hang of the key operations in solving this problem, try resolving it using the decomposition \begin{align} \xi\eta = (\xi - 2\eta + 2\eta)\eta = \eta(\xi - 2\eta) + 2\eta^2. \end{align}

Comment on your attempt: the idea looks great but unfortunately $\xi - \eta \perp \xi + \eta$ of course does not imply $\xi - 2\eta \perp \xi + 2\eta$. However, the "product-to-sum" identity of $\xi\eta$ is still very useful in approaching this problem. From there, you only need to apply some very basic properties of conditional expectation and the multivariate normal distribution to get the job done.

By the linearity and the "pulling out known factors" property of conditional expectation, \begin{align} E[\xi\eta|\xi - 2\eta] = \frac{1}{8}E[(\xi + 2\eta)^2|\xi - 2\eta] + \frac{1}{8}(\xi - 2\eta)^2. \tag{1} \end{align}

So it remains to evaluate the $E[(\xi + 2\eta)^2|\xi - 2\eta]$, which is tractable thanks to $(\xi, \eta) \sim N_2(0, I_{(2)})$. Because of it, it follows by the affine transformation property of the multivariate normal distribution that \begin{align} \begin{bmatrix} \xi + 2\eta \\ \xi - 2\eta \end{bmatrix} \sim N_2\left(\begin{bmatrix} 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 5 & -3 \\ -3 & 5 \end{bmatrix} \right), \end{align} which implies, by the conditional distribution of the multivariable normal distribution, that \begin{align} & E[\xi + 2\eta | \xi - 2\eta] = -\frac{3}{5}(\xi - 2\eta), \\ & \operatorname{Var}(\xi + 2\eta | \xi - 2\eta) = 5 - 9 \times \frac{1}{5} = \frac{16}{5}, \end{align} whence \begin{align} E[(\xi + 2\eta)^2|\xi - 2\eta] &= \operatorname{Var}(\xi + 2\eta | \xi - 2\eta) + (E[\xi + 2\eta | \xi - 2\eta])^2 \\ &= \frac{16}{5} + \frac{9}{25}(\xi - 2\eta)^2. \tag{2} \end{align} Substituting $(2)$ into $(1)$ gives \begin{align} E[\xi\eta|\xi - 2\eta] = \frac{2}{5} + \frac{9}{200}(\xi - 2\eta)^2 + \frac{1}{8}(\xi - 2\eta)^2 = \frac{2}{5} + \frac{17}{100}(\xi - 2\eta)^2. \end{align}

Now, to get the hang of the key operations in solving this problem, try resolving it using the decomposition \begin{align} \xi\eta = (\xi - 2\eta + 2\eta)\eta = \eta(\xi - 2\eta) + 2\eta^2. \end{align}

Comment on your attempt: the idea looks great but unfortunately $\xi - \eta \perp \xi + \eta$ of course does not imply $\xi - 2\eta \perp \xi + 2\eta$. However, the "product-to-sum" identity of $\xi\eta$ is still very useful in approaching this problem. From there, you only need to apply some very basic properties of conditional expectation and the multivariate normal distribution to get the job done.

By the linearity and the "pulling out known factors" property of conditional expectation, \begin{align} E[\xi\eta|\xi - 2\eta] = \frac{1}{8}E[(\xi + 2\eta)^2|\xi - 2\eta] + \frac{1}{8}(\xi - 2\eta)^2. \tag{1} \end{align}

So it remains to evaluate the $E[(\xi + 2\eta)^2|\xi - 2\eta]$, which is tractable thanks to $(\xi, \eta) \sim N_2(0, I_{(2)})$. Because of it, it follows by the affine transformation property of the multivariate normal distribution that \begin{align} \begin{bmatrix} \xi + 2\eta \\ \xi - 2\eta \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 1 & -2 \end{bmatrix} \begin{bmatrix} \xi \\ \eta \end{bmatrix} \sim N_2\left(\begin{bmatrix} 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 5 & -3 \\ -3 & 5 \end{bmatrix} \right), \end{align} which implies, by the conditional distribution of the multivariable normal distribution, that \begin{align} & E[\xi + 2\eta | \xi - 2\eta] = -\frac{3}{5}(\xi - 2\eta), \\ & \operatorname{Var}(\xi + 2\eta | \xi - 2\eta) = 5 - 9 \times \frac{1}{5} = \frac{16}{5}, \end{align} whence \begin{align} E[(\xi + 2\eta)^2|\xi - 2\eta] &= \operatorname{Var}(\xi + 2\eta | \xi - 2\eta) + (E[\xi + 2\eta | \xi - 2\eta])^2 \\ &= \frac{16}{5} + \frac{9}{25}(\xi - 2\eta)^2. \tag{2} \end{align} Substituting $(2)$ into $(1)$ gives \begin{align} E[\xi\eta|\xi - 2\eta] = \frac{2}{5} + \frac{9}{200}(\xi - 2\eta)^2 + \frac{1}{8}(\xi - 2\eta)^2 = \frac{2}{5} + \frac{17}{100}(\xi - 2\eta)^2. \end{align}

Now, to get the hang of the key operations in solving this problem, try resolving it using the decomposition \begin{align} \xi\eta = (\xi - 2\eta + 2\eta)\eta = \eta(\xi - 2\eta) + 2\eta^2. \end{align}